Physics 137B: Quantum Mechanics II, Lecture notes of Quantum Mechanics

Download Physics 137B: Quantum Mechanics II and more Lecture notes Quantum Mechanics in PDF only on Docsity! Physics 137B: Quantum Mechanics II U.C. Berkeley Spring 2014 Semester Professor: Petr Hořava GSI: Kevin Grosvenor Last Edited: August 20, 2014 b 0.1. LECTURE 1 1 0.1 Lecture 1 Introduction: • The universe is (entirely) quantum mechanical: you can’t just choose to apply quantum mechanics (QM) to some phenomena and not to others! • QM contains some of the most exciting puzzles (e.g. quantum gravity). • There is something strangely “fundamental” about QM – Other subjects like general relativity (GR) predict their own demise (e.g. for GR, near singularities, or when curvature radius approaches the Planck length). This is not so for QM. – No clue how to view QM as an approximation to something more fundamental. Is QM absolute? This should make you nervous. – Glimpses of ideas to move beyond standard QM from ’t Hooft, Gell-Mann, Hartle, Adler. Approximation Techniques: • In applying QM rules, such as the Schrödinger equation (SEq), one often gets stuck since most realistic systems are not exactly solvable. • Need approximation techniques (art, ideas). • Two groups: t-dep (hard), and t-indep (easier). • Main examples: – Perturbation theory in a small parameter, λ << 1, in the Hamiltonian, such that the limit λ → 0 is exactly solvable. – Variational Method: make an educated guess for the wave function, with a sufficiently rich, but not too rich (hence the art), set of tunable parameters. Usually used for ground state, for which one minimizes the energy with respect to (wrt) the parameters. – WKB (Wentzel, Kramers, Brillouin) is a semiclassical approximation. – Large-N expansion due to ’t Hooft. Really, this is just like a small λ = 1/N perturbation expansion. Background on SEq: • Also sometimes called Rayleigh-Schrödinger procedure. • SEq: i~ψ̇ = Hψ, where ψ is the wavefunction, H is the Hamiltonian operator (sometimes denoted with a hat, Ĥ), and ψ̇ ≡ dψ dt . • ~ is Planck’s constant and is a fundamental dimensionful constant. • Fundamental dimensional constants are associated with a major paradigm shift in physics and are really just conversion factors between dimensionful quantities: – Special Relativity (SR): speed of light, c. This is a conversion factor between time and space and setting c = 1 means measuring time and space using the same units. – Statistical Mechanics (SM): Boltzmann’s constant, kB . This is a conversion factor between tempera- ture and energy and setting kB = 1 means measuring temperature and energy using the same units. – QM: Planck’s constant, ~. This is a conversion factor between inverse time (frequency) and energy and setting ~ = 1 means measuring frequency and energy using the same units. – Gravity: Newton’s constant, GN . This case is yet unknown. • More precisely, |ψ〉 ∈ H , an infinite-dimensional Hilbert space over the complex numbers, C, with a countable basis. 2 CONTENTS • Often, this is usefully represented as square-integrable functions, ψ(x) = 〈x|ψ〉 on configuration space. Here, x can include time and space: x = (t,x). • The Hamiltonian is an operator (we will usually omit the hat) and is a function of time and phase space, H(t,q,p). Here, q includes spatial coordinates, but could also include all sorts of other quantum degrees of freedom such as spin, and p are the conjugate momenta. • Time is not an operator here. It is merely a real parameter. However, people have tried to promote time to an operator (e.g. Galapon). Non-degenerate Time-independent Perturbation Theory, Bransden and Joachim (BJ) sec. 8.1: • Assume H is not explicitly dependent on t. • SEq reduces to t-indep SEq (tiSEq), Hψn = Enψn, where En are the eigenvalues or eigenenergies (real since H is Hermitian) and ψn are the eigenfunctions, labeled by n. • If ψn is unique up to a phase eiα for some real constant α, then En is a non-degenerate energy level. • General time-dependent solution: Ψ = ∑ n cne −iEnt/~ψn, where cn are constants and ∑ n |cn|2 = 1. • Suppose H is “close” to being solvable: H = H0 + λH ′, where H0 is exactly solvable, λ << 1 is a small real positive parameter, and H ′ is independent of λ. • H0ψ (0) n = E (0) n ψ (0) n , with ψ (0) n and E (0) n known, since H0 is exactly solvable. 0.2. LECTURE 2 3 0.2 Lecture 2 Non-degenerate Time-independent Perturbation theory: • Perturbation theory is not just a technical procedure. The entire concept of a particle is a perturbation in a field. • tiSEq: Hψn = Enψn and suppose ψn is unique eigenfunction up to phase. • Focus on some En in a discrete ladder of energy levels. There could be chunks of the energy spectrum where levels become continuous. Neglect that for now. E En • H = H0 + λH ′ where H0 is exactly solvable and λ << 1 (and H ′ is λ-indep). • Hopefully, spectrum from E (0) n of H0 is changed only by a small amount. E λ • ∣∣∣ψ(0) i 〉 where i runs over all Hilbert space. We can use these to get an orthonormal basis for the Hilbert space, H . Assume, we have such a basis, 〈ψ(0) i |ψ (0) j 〉 = δij , where δ is actually a Dirac delta when ψ is in the continuous part of the spectrum. • En λ→0−−−→ E (0) n and ψn λ→0−−−→ ψ (0) n by assumption. If fails then procedure is internally inconsistent. • Expand En and ψn around E (0) n and ψ (0) n , assuming we can Taylor expand (consistent with the assumption of smooth variation with λ): – En = ∑∞ j=0 λ jE (j) n . – ψn = ∑∞ j=0 λ jψ (j) n . • (H0 + λH ′)(ψ (0) n + λψ (1) n + λ2ψ (2) n + · · · ) = (E (0) n + λE (1) n + λ2E (2) n + · · · )(ψ(0) n + λψ (1) n + λ2ψ (2) n + · · · ). – Order λ0: H0ψ (0) n = E (0) n ψ (0) n is true by assumption. – Order λ1: H0ψ (1) n +H ′ψ (0) n = E (0) n ψ (1) n + E (1) n ψ (0) n . – Order λ2: H0ψ (2) n +H ′ψ (1) n = E (0) n ψ (2) n + E (1) n ψ (1) n + E (2) n ψ (0) n . • To get E (1) n : (pre-? according to BJ) multiply by 〈ψ(0) n | and integrate over space: 〈ψ(0) n |H0 − E(0) n |ψ(1) n 〉+ 〈ψ(0) n |H ′ − E(1) n |ψ(0) n 〉 = 0. Since H0 is Hermitian, 〈ψ(0) n |H = 〈ψ(0) n |H† = ( H|ψ(0) n 〉 )† = ( E (0) n |ψ(0) n 〉 )† = 〈ψ(0) n |E(0) n . Therefore, first term vanishes: E(1) n = 〈ψ(0) n |H ′|ψ(0) n 〉 with |ψ(0) n 〉 properly normalized. 6 CONTENTS 0.3 Lecture 3 Non-degenerate Time-independent Perturbation theory: • H = H0 + λH ′, where λ is “small” and H ′ is λ-independent. • Anharmonic oscillator: H0 = p2 2m + 1 2kx 2 and H ′ = ax3 + bx4. Consider a = 0 and b ≥ 0. If b < 0 then the energy is unbounded from below and there isn’t even a good ground state. • E(1) n = 3 4α4 (2n2 + 2n+ 1) where α4 = mk/~. Perturbation theory breaks down for large enough n. General picture and asymptotic series: • More general picture: 0th order is a collection of harmonic oscillators and 1st order, you introduce some anharmonicity, λbx4. • E(k) n for all k = 1, . . .∞. Suppose you have En = ∑∞ k=0(bλ)kE (k) n . Are you done? NO!!! This series is an asymptotic series (zero radius of convergence)! • Extend λ ∈ C. The radius of convergence is the distance from λ = 0 to the nearest singularity on C. Analytically continue to λ < 0. But, if λ < 0, then the energy is unbounded from below and there is no stable ground state. Therefore, the whole negative axis is singular and the radius of convergence is 0. λ • Asymptotic vs. summable/convergent. Let ∆ be the absolute value of the difference between the kth order result for some observable (e.g. energy) and the real value (known perhaps by experiment). For a convergent series, ∆ k→∞−−−−→ 0 (black dots), whereas for an asymptotic series, a minimum is reached and then ∆ may diverge after that point (unfilled circles). ∆ k • Consequences: – Don’t have to go too far in k. – Something (about the real system) is missing in perturbation theory: Non-perturbative effects. Often these take the form of new degrees of freedom (e.g. solitons, instantons, magnetic monopoles, D- branes). Degenerate Time-independent Perturbation theory: • diag 3. ψ (0) nr with r = 1, . . . , α, where α is some integer. α = 1 means non-degenerate. All of these α linearly independent state have the same energy. • Even though they share the same energy at zeroth order, there is no reason why they should share the same perturbations away from that. Thus, E (0) nr , such that all of these (for all values of r) go back to E (0) n as λ→ 0. • Pick the ψ (0) nr ’s to be orthonormal (Gram-Schmidt if need be). • ψnr = χ (0) nr + λψ (1) nr + λ2ψ (2) nr + · · · , where χ (0) nr is some linear combination of the ψ (0) nr . diag 4 0.3. LECTURE 3 7 • χnr = ∑α s=1 cnrsψ (0) ns . We will have to figure out which linear combinations to choose as starting points. • Enr = E (0) n + λE (1) nr + λ2E (2) nr + · · · . – Order λ1: H0ψ (1) nr +H ′χnr = E (0) n ψ (1) nr + E (1) nr χnr. – Expand ψ (1) nr = ∑ ks a (1) nr,ksψ (0) ks . – Multiply by 〈ψ(0) nu | to get ∑ ks a (1) nr,ks(E (0) k − E (0) n )〈ψ(0) nu |ψ(0) ks 〉 + ∑ s cnrs [ H ′nu,ns − E (1) nr δus ] = 0, where H ′nu,ns = 〈ψ(0) nu |H ′|ψ(0) ns 〉. – The first term vanishes since the inner product is zero unless k = n, in which case the coefficient vanishes. – The rest is a linear homogeneous system of equations for cnrs. The coefficients H ′nu,ns are known. cnrs 6= 0 for some rs iff det(H ′nu,ns −E (1) nr δus). This is an α× α matrix in the space of the degenerate states corresponding to a particular n. There are α roots, one for each r. – Think of the equation as a matrix H ′n,n − E (1) nr I, where I is the identity in the degeneracy space, multiplying a vector of c’s equals zero. – Note: degeneracies do not have be completely broken. We can have partial degeneracy breaking. • Comments: – In retrospect, χnr’s were those specific linear combinations of the original ψ (0) nr for which H ′ is diagonal. In this case the H ′ part of the equation is ∼ δus and the equation decomposes into α different equations. Then, 〈ψ(0) nr |H ′|χ(0) ns 〉 = E (1) nr δrs. – Often, there are good symmetry reasons why particular linear combinations should diagonalize H ′. Example: Doubly-degenerate energy level: • Drop the index n. Let ψ (0) 1 and ψ (0) 2 be degenerate. • If you can travel back in time, you would see what I just erased on the whiteboard. That’s one of the benefits of time travel. • Then, we have the matrix ( H ′11 − E (1) r H ′12 H ′21 H ′22 − E (1) r )( cr1 cr2 ) = 0. • Since H is Hermitian, H ′11 and H ′22 are real and H ′12 = (H ′21)∗. • The secular equation is det ( H ′11 − E (1) r H ′12 H ′21 H ′22 − E (1) r ) = 0. • This is a quadratic equation with solutions E (1) 1,2 = 1 2 (H ′11 +H ′22)± √ (H ′11 −H ′22)2 + 4|H ′12|2. • We also find cr1 cr2 = − H ′12 H ′11 − E (1) r = −H ′ 22 − E (1) r H ′21 . • Special case 1: H ′11 = H ′22 = 0 and H ′12 6= 0. Then, E (1) 1,2 = ±|H ′12| and c11 c12 = − c21c22 = H′12 |H′12| . Therefore, χ1 = ψ (0) 1 + ψ (0) 2 , χ2 = ψ (0) 1 − ψ(0) 2 . • Special case 2: H ′12 = H ′21 = 0, then E (1) 1 = H ′11 and E (1) 2 = H ′22. 8 CONTENTS Example: Fine Structure of Hydrogen: • H0 = p2 2m + V (r), where m is the reduced mass, which is approximately the mass of the electron. • Here, V (r) = V (r) = − 1 r in units where e = 1 and 4πε0 = 1. • This is non-relativistic: kinetic energy is non-relativistic. Also, this does not take into account the electron spin. (Note: electrons are described by the Dirac equation). • Suppose H = H0 and add first order relativistic corrections. H ′1,2,3 turns out three terms all come in at the same order in some expansion parameter. • What is a sensible expansion parameter? Well, v2 c2 , where v is the speed of the electron and c is the speed of light. • H ′1 = − p4 8m3c2 and H ′2 = 1 2m2c2 1 r dV dr L · S and H ′3 = −π~ 2Ze2 2m2c2 δ (3)(r). 0.4. LECTURE 4 11 Nearly degenerate states: • Punchline: “eigenvalues repel each other”. • 2-fold near degeneracy: ψ (0) i with E (0) i = E(0) ± ε. • Non-degenerate perturbation theory gives crossing eigenvalues. diag 2 • Near-degenerate perturbation theory gives diag 3. • Crossing eigenvalues is difficult to set up and is associated with quantum phase transitions. 12 CONTENTS 0.5 Lecture 5 Special case: nearly degenerate Example: two-fold near degeneracy using non-degenerate perturbation theory: • Can’t we use non-degenerate perturbation theory? The problem is that the result involves the difference in the energy of the two states (in the denominator). If that difference becomes too small, then the perturbation becomes too large and is invalid. • WIth “fuzzy glasses” (due to A. Zee, c.f. QFT in a Nutshell), they look degenerate. • ψ(0) 1 and ψ (0) 2 with E (0) 1 = E(0) + ε and E (0) 2 = E(0) − ε, with ε > 0. • (H0 + λH ′)ψr = Erψr with r = 1, 2. • Assume non-degenerate perturbation theory: Er ≈ E(0) r + λH ′rr. • Define Ē = 1 2 (E1 + E2) = E(0) + λ 2 (H ′11 +H ′22) and ∆E = E1 − E2 = 2ε+ λ(H ′11 −H ′22). • Prediction: diag1. Suggests that eigenvalues cross! If one of these is the ground state, then the ground state jumps from one state to the other in a non-analytic manner (non-smooth, i.e. there is a kink). This suggests a phase transition. Two-fold near degeneracy using degenerate perturbation theory: • ψr = ∑ k arkψ (0) k = ar1ψ (0) 1 + ar2ψ (0) 2 + ∑ k 6=1,2 arkψ (0) k , assuming ψ (0) k complete and orthonormal. • ar`(E(0) ` − Er) = −λ ∑ k arkH ′ `k. • ar1(E (0) 1 + λH ′11 − Er) + λar2H ′ 12 = −λ ∑ k 6=1,2 arkH ′ 1k. • λar1H ′21 + ar1(E (0) 2 + λH ′22 − Er) = −λ ∑ k 6=1,2 arkH ′ 2k. • The RHS of these equations are at least O(λ2) and may be neglected. • For which values of Er is this system solvable? det ( E (0) 1 + λH ′11 − Er λH ′12 λH ′12 E (0) 2 + λH ′22 − Er ) = 0. • Er = Ē ± 1 2 √( ∆E 2 )2 + λ2|H ′12|2 . diag 2 • Pithy comment: “eigenvalues are like fermions”. Time-independent perturbations: two more techniques: Ch. 8.3: variational principle: • “trial and error”. Ch. 8.4: WKB (Wentzel-Kramers-Brillouin) approximation (semi-classical): • 1926 problems only in 1D. But this has been generalized since. • Small ~. We have to find something else, S, whose dimension is the same as ~. Then, ~ is small compared to S, which will turn out to be the classical action. • Bohr: complementarity principle: “quantum turns classical at large quantum numbers”. • Best formulated in the path integral approach to quantum mechanics (see Ch. 5.9). Or see “QFT in a Nutshell” by A. Zee. 0.5. LECTURE 5 13 • diag 3. We want the transition amplitude 〈f |i〉, where i and f are initial and final states. Dirac: this is approximated well by eiS/~, where S is the action functional. • Feynman put intermediate stages by small increments. diag 4. Classically, the particle follows a path, which extremizes (minimizes) the classical action S[q(t), q̇(t)] = ∫ L[q(t), q̇(t)] dt. Feynman says that all paths contribute: 〈f |i〉 = ∫ Dq(t) exp [ iS[q, q̇]/~ ] . • How does the classical trajectory arise as ~→ 0. This is the one that extremizes the action, namely, if we choose a trajectory near but not quite the same as the classical trajectory, δq, then δS would be 0. • If S/~ >> 1, the saddle-points dominate (i.e. δS = 0, which is the same as the classical equation of motion). • Simplify to 1D: − ~2 2m d2ψ dx2 + V (x)ψ(x) = Eψ(x). Assume that V is “slowly varying” (pending clarification). • Ex. 1: V (x) = V0 is the most slowly-varying potential. Solutions are plane waves: Ae±ip0x/~, where p0 is the momentum, which is well-defined in this state, where p0 = √ 2m(E − V0) (assuming E > V0). • V slowly varying in x, approximate by exponential: p(x) = √ 2m(E − V (x)). This particle would have a de Broglie wavelength is λ(x) = h/p(x). Notice: λ→ 0 as h→ 0. • Ansatz: ψ(x) = AeiS(x)/~. tiSEq becomes − i~ 2m d2S dx2 + 1 2m ( ∂S ∂x )2 + V (x)− E = 0. • Expand in ~ as S(x) = S0(x) + ~S1(x) + 1 2~ 2S2(x) + · · · . • Zeroth order: 1 2 ( dS0 dx )2 + V (x) − E = 0 is the Hamilton-Jacobi equation. This accomplishes Bohr’s corre- spondence principle. The reverse of this is actually how Schrd̈ingier derived his equation! • diag 5 • First order: dS0 dx dS1 dx − i 2 d2S0 dx2 = 0. • Second order: dS0 dx dS2 dx + ( dS1 dx )2 − id2S1 dx2 = 0. • diag 6. If E > V (x), then S0 = ± ∫ x p(x′) dx′ (well-defined up to a constant; hence, lower bound of integral is unspecified). • S1 = i 2 log p(x). • ψ(x) = A√ p(x) exp [ ± i ~ ∫ x p(x′) dx′ ] . • S2 = 1 2m[p(x)]−3 dV dx − 1 4m 2 ∫ x [p(x′)]−5 [dV (x′ dx′ ]2 dx′. Indeed, if V is slowly varying, these derivatives are small and this expansion is sensible. 16 CONTENTS 0.7 Lecture 7 Time-dependent perturbation theory: • H = H0 + λH ′(t) where H0 is static and solvable with exact energy levels E (0) n and eigenfunctions, ψ (0) n , assumed to be orthonormal and complete. • No good notion of energy eigenvalues because energy is not conserved. We are interested in the full wavefunction as it depends on time, Ψ(t). • Ψ(t) = ∑ k ck(t)ψ (0) k exp [ − i ~E (0) k t ] where |ck(t)|2 is the probability of finding the system at some time t in the state ψ (0) k . • Obviously, ∑ k |c (0) k (t)|2 = 1. • Plug this into the Schrödinger equation and take the inner product with some ψ (0) b : ċb(t) = λ i~ ∑ k H ′bk(t) eiωbktck(t). Here, ωbk = 1 ~ (E (0) b − E (0) k ) is Bohr’s angular frequency. • ck = c (0) k + λc (1) k + λ2c (2) k + · · · . • O(λ0) equation: ċ (0) b = 0. This is obvious since there is no perturbation in this case. • O(λ1) equation: ċ (1) b = 1 i~ ∑ kH ′ bk(t) eiωbktc (0) k . • In general, ċ (s+1) b = 1 i~ ∑ kH ′ bk(t) eiωbktc (s) k . • Let’s play... 1. Let’s take c (0) b = δba, where a is a specific state. If continuous spectrum, δ(k − a). Then, let this evolve. 2. Evolution: ċb = 1 i~H ′ ba(t) eiωbatδba. – If b = a, then ca(t) = 1 i~ ∫ t t0 H ′aa(t′) dt′ . This is reminiscent of time-independent perturbation theory since the diagonal element, H ′aa, is what shows up in that case. – If b 6= a, then cb(t) = 1 i~ ∫ t t0 H ′ba(t) eiωbat ′ dt′ . 3. Pba(t) = |cb(t)|2 is probability to transition from a to b. If b 6= a, then P (0) ba = 0 and P (1) ba (t) = 1 ~2 ∣∣∣∣∫ t t0 H ′ba(t) eiωbat ′ dt′ ∣∣∣∣2 4. If b = a, then ca(t) ≈ 1 + c(1) a (t) + · · · ≈ 1 + 1 i~ ∫ t t0 H ′aa(t′) dt′ ≈ exp [ − i ~ ∫ t t0 H ′aa(t′) dt′ ] . Therefore, Paa(t) ≈ 1. • Special case: Time-independent perturbation. This means, H = H0 for all t < t0 and then H0 + λH ′ for t ≥ t0, where H ′ is time-independent. H isn’t really time-independent; it has a θ function time dependence. Then, suppose that at t = t0, the state is ψ (0) a . This is equivalent to just saying H = H0 + λH ′ but that the system is prepared in that state. This is a special property called the Markovian property: you don’t need to know the full history before some initial condition. 0.7. LECTURE 7 17 – Now, c (1) a (t) = 1 i~H ′ aat and c (1) b = H′ba ~ωba [ 1− eiωbat ] . This is sensible, the ωba factor in the denominator suppresses transition to states that are very far apart in energy from the initial state. – Note: ca(t) ≈ exp [ − i ~H ′ aat ] . Therefore, Ψ ≈ ca(t)ψ (0) a e−iE (0) a t/~ = ψ (0) a exp [ − i ~ ( E (0) a + H ′aa ) t ] . This agrees with time-independent perturbation theory. • First-order transition probabilities: P (1) ba (t) = 2 ~2 |H ′ba|2F (t, ωba), where this function is so important that it deserves its own blackboard: F (t, ω) ≡ 1− cos(ωt) ω2 = 2 sin2(ωt/2) ω2 . F stands for “fancy” or Founding father or something really important. • plot: diag 1 for fixed (generic) t. As t→ this resembles a Dirac delta function. • Properties of F : – We have F (t, ω = 0) = t2/2. – Sharp peak at ω = 0 with width ≈ 2π/t. – The integral can be done exactly: ∫∞ −∞ F (t, ω) dω = πt. – The limit as t→∞ is F (t, ω)→ πtδ(ω). • Consequences: – Sharp peak implies transitions to states whose ωba does not deviate from zero too much are enhanced: ωba ≤ 2π/t. Therefore, energy is conserved up to δE ∼ 2π~t. This is reminiscent of the uncertainty principle ∆E∆t ≥ ~. – For b 6= a, we have P (1) ba (t) = 4|H′ba| 2 ~2ω2 ba sin2 ( ωbat 2 ) . oscillates about the mean value 2|H′ba| 2 ~2ω2 ba with period T = 2π/|ωba|. – Total probability to transition from a into anything other than a: P (1) not a(t) = ∑ b 6=a P (1) ba (t) = ∑ b6=a 4|H ′ba|2 (E (0) b − E (0) a )2 sin2 [ (E (0) b − E (0) a )t 2 ] . Validity: P (1) not a(t) << 1. A sufficient bound:∑ b6=a 4|H ′ba|2 (E (0) b − E (0) a )2 << 1. All of this assumes non-degenerate energy levels. • Suppose a is in a degenerate level and b 6= a but E (0) b = E (0) a , then P (1) ba (t) = |H ′ba|2 ~2 t2. This grows indefinitely and at some time the validity breaks down! Example: 2-level QM system, degenerate: • For example, NMR, MRI is based on response of two spins of water molecules. • E(0) a = E (0) b = E(0). Assume H ′aa = H ′bb = 0, but H ′ab is real and > 0. • Before, we derived χa = 1√ 2 (ψ (0) a + ψ (0) b ) with energy E(0) − H ′ab and χb = 1√ 2 (ψ (0) a − ψ(0) b ) with energy E(0) +H ′ab. • Assume Ψ(t) is given at t = t0 by Ψ(t0) = ψ (0) a (t0). 18 CONTENTS • In general, Ψ(t) = c1χa exp [ − i ~ (E(0) +H ′ba)t ] + c2χ (0) b exp [ − i ~ (E(0) −H ′ba)t ] with c=c2 = 1√ 2 fixed by the initial condition at t = t0. • This can be written Ψ(t) = e− i ~E (0)t [ ψ (0) a cos (H′bat ~ ) − iψ (0) b sin (H′bat ~ )] . Therefore, the system oscillates between the two degenerate levels with frequency ν = H ′ba/π~. This is the “resonance” between the two degenerate levels. • So, we get oscillatory behavior instead of the power growth predicted by non-degenerate time-dependent perturbation theory. Example: Perturbations periodic in time • H ′(t) = Ĥ ′ sin(ωt) = Aeiω +A†e−iωt where A = 1 2iĤ ′. Note: Ĥ ′ is some constant in time. • At t = 0, pick the intial state ca = 1, cb = 0 for all b 6= a. “We are the gods of the lab, so we can pick the initial conditions. Unless you get into cosmology, in which case no one has a clue what the initial conditions are.” • c(1) b (t) = 1 i~ { Aba ∫ t 0 dt′ ei(ωba+ω)t′ +A†ba ∫ t 0 dt′ ei(ωba−ω)t′ } . Doing the integrals yields c (1) b (t) = Aba 1− e i~ (E (0) b −E (0) a +~ω)t E (0) b − E (0) a + ~ω +A†ba 1− e− i ~ (E (0) b −E (0) a −~ω)t E (0) b − E (0) a − ~ω . • So, we see large P (1) ba when the denominator of one or the other term blows up. We only really need to consider one term at a time in this order: no need to consider the cross-terms. Thus, P (1) ba (t) = { 2 ~ |A † ba|2F (t, ωba − ω), E (0) a + ~ω ≈ E(0) b , 2 ~ |Aba| 2F (t, ωba − ω), E (0) a − ~ω ≈ E(0) b . This reproduces Bohr’s early results. 0.9. LECTURE 9 21 0.9 Lecture 9 Atoms: • Techniques we will use: – Perturbation theory to first order. Mostly concerned with ground state. – Variational principle. Again, focus on ground state. • Hydrogen is done. Consider Helium: two electrons at positions r1 and r2: H = − ~2 2m (∇2 1 +∇2 2)− Ze2 4πε0r1 − Ze2 4πε0r2 + e2 4πε0r12 , where r12 = |r1 − r2|. • Generalities about systems with two electrons: – The operator that switches the positions of the two fermions commutes with the Hamiltonian. There- fore, the state of the system can be either symmetric or antisymmetric under this switch. – For fermions, the FULL wavefunction (including spin) must be antisymmetric. Let qi denote the combination (position, spin) of the ith particle. Ψ(q1, q2) = −Ψ(q2, q1). – Notation of the book: α is spin up and β is spin down. α(i) is spin up for ith particle, and so on. • Write Ψ(q1, q2) = φ+(r1, r2) 1√ 2 [ α(1)β(2) − α(2)β(1) ] where φ+(r1, r2) = φ+(r2, r1). This is called the singlet state. The spin part is often written 1√ 2 ( |↑↓〉 − |↓↑〉 ) . This has total spin, S = 0. We will need to solve Hφ+ = Eφ+. • Another option: Ψ(q1, q2) = φ−(r1, r2) 1√ 2 [ α(1)β(2) + α(2)β(1) ] where φ−(r1, r2) = −φ−(r2, r1). Or, we could choose the spin state to be α(1)α(2) or β(1)β(1). This is the triplet state, with total spin S = 1. Again, we have to solve Hφ− = Eφ−. Perturbation theory: • Not too good here because there is no really clear small part of the Hamiltonian. We choose the perturbation to be the Coulomb repulsions between the two electrons and the zeroth order Hamiltonian to be the rest. • E(0) = En1 + En2 just the sum of the Bohr energy levels, one for each electron. This is given by E(0) = − Z2e2 (4πε0)2a0 ( 1 n2 1 + 1 n2 2 ) , where a0 is the Bohr radius: a0 = 4πε0~2 me2 . • Therefore, E (0) gs = −2Z2 × 13.6 eV. • The non-normalized wavefunctions of the system are φ (0) ± (r1, r2) = ψn1`1m1 (r1)ψn2`2m2 ± ψn1`1m1 (r2)ψn2`2m2 (r1). • Total angular momentum: L = |`1 − `2|, |`1 − `2|+ 1, . . . , `1 + `2. • Notation: S stands for L = 0, then P for L = 1, then D for L = 2, then F for L = 3, then G for L = 4, then H for L = 5, and so on. • How to measure the ground state energy: find minimum energy of two photons with which we hit the atom in order to ionize it twice. For Helium, we find Egs ≈ −79 eV. 22 CONTENTS • However, our first estimate (neglecting electron repulsion) gives −108.85 eV. • For the ground state wavefunction, the lowest we can get is ψ100 for both electrons. Therefore, the orbital wavefunction must be symmetric: φ (0) + = ψ100(r1)ψ100(r2). Thus, the spin is the antisymmetric one and we are in the singlet state. ψ100 is given by ψ100(r) = 1√ π ( Z a0 )3/2 e−Zr/a0 . • Note: since the electron-electron repulsion is about 1/Z times the electron-proton attraction, we expect this perturbation result to be better and better for higher and higher Z. • First order perturbation theory: the integral is in the book: E(1) = 1 π ( Z a0 )3 e2 4πε0 〈φ+|r−1 12 |φ+〉 = ... = 5 8 e2Z 4πε0a0 . Really, only 5/8 is unknown. The rest you can do by dimensional analysis and a bit of logic. • The numerical result is E(0) + E(1) = −74.83 eV. Not bad! • For Z = 6 (Carbon), exact is Egs = −881.93 eV and perturbation gives E(0) +E(1) = −877.57 eV whereas E(0) = −979.62 eV. Variational principle: • E[φ] = 〈φ|H|φ〉/〈φ|φ〉. Pick test functions that depend on some parameter λ, then minimize E[φ] as a function of λ since E(λ) ≥ Egs. • Need some physical intuition to guide our choice of test function. One electron sees an effective Coulomb potential that is screened by the other electron. This suggests changing Z to λ between 1 and 2: φλ = 1 π ( λ a0 ) e−λ(r1+r2)/a0 . You could say this is a lazy choice because we have done the calculation already. Altogether, we get E(λ) = ( λ2 − 2Zλ+ 5 8λ ) e2 4πε0a0 . The minimum value of this is reached at λbest = Z − 5 16 with result Ev.p. = − ( Z − 5 16 )2 e2 4πε0a0 . For Helium, Ev.p. = −77.50 eV. Many electron atoms: • H = ∑N i=1 ( − ~2 2m∇ 2 i − Ze2 4πε0r ) + ∑N i>j=1 e2 4πεrij . • Spin only enters by forcing symmetry or antisymmetry of the orbital wavefunctions. • Effectively, N−1 electrons screen: − Ze2 4πε0r1 turns into V (r1). Expect V (r) to look like the unscreened result for very small r. For very large r, we expect Coulomb with effective Z given by Z − (N − 1). • H = Hc +H1 where Hc = ∑N i=1 ( − ~2 2m∇ 2 i + V (ri) ) and H1 = N∑ i>j=1 e2 4πε0rij − N∑ i=1 ( Ze2 4πε0ri + V (ri) ) . Here, the subscript c on Hc stands for “central field approximation”. 0.9. LECTURE 9 23 • Denote hi = − ~2 2m∇ 2 i + V (ri). We know that hiun`m(r) = En`un`m(r), where un`m(r) = Rn`(r)Y`m(θ, φ). We have ` = 0, 1, . . . , n− 1 and m = −`, . . . , `. • Note that the energy can depend on `. The Bohr model for the hydrogen atom is very special in that the energy is independent of `. • In the central field approximation, Ec = ∑N i=1Eni`i . Shells: • K shell: n = 1 and ` = 0: this is the 1s states (two states for spin up and down). • L shell: 2s has two states (singlet times two spins) and 2p as six states (triplet times two spins). • M shell: 3s (two states), 3p (six states), 4s (two states), 3d (ten states). Note: 4s and 3d are very close in energy and then the thing that matters most is screening. Being further away from the nucleus increases screening and higher ` means higher average radial position. This is usually enough to bump 3d above 4s. • Next, is 4p (six states), then 5s (two states), then 4d (ten states), then 5p, then 6s, then 4f , then 5d. • This is summarized by the Aufbau rule. (See diag online). • We see this structure in the periodic table since how many electrons are in the outer most shell has a strong effect on the chemical properties of the element. 26 CONTENTS • ψ1s is ground state of hydrogen. • χg = ψ1s(rA) + ψ1s(rB) not normalized. (g stands for gerade (German for straight) • What about χu = ψ1s(rA)− ψ1s(rB). And u stands for ungerade (not straight). • χg is parity-even and χu is parity-odd. It turns out that χg binds, but χu does not. • Eg = 〈χg|H|χg〉/〈χg|χg〉. We find Eg(R) = E1s + |E1s|fg(x), where x = R/a0 and fg(x) = (1 + x)e2x + (1 +− 2 3x 2)e−x 1 + (1 + x+ 1 3x 2)e−x . Minimum occurs at xmin = 2.49 or Rmin = 2.49a0 = 1.32 angstroms. Emin = −13.6− 1.77 eV. Eexpt = −13.6 − 2.8 eV. Diag 2. As expected by variational principle, the actual result is lower than the variational one. Radiation from and to molecules: • Electronic: transitions from χ2 to χ1 with typical energy ∆Ee ∼ eV, which is in the visible to UV spectrum. The form of the energy is ∼ ~2/ma2. • Vibrational: ∆Ev ∼ 0.1 eV in the IR and is the electronic suppressed by (m/M)1/2 since ω ∼ √ k/m and k is basically the same for protons and electrons (they feel the same force by Newton’s 3rd law). • Rotational: ∆Er ∼ 0.001 eV in microwave and far-IR with L(L+1)~2 2I ∼ ~2 Ma2 ∼ m MEe. Nuclei: • Short range force. Modeled by potential V ∼ 1 r e −r/a called the Yukawa potential. Here, a determines the typical scale of the distance between nucleons: a ∼ 10−15 m = fm. • Replace this potential with a potential well. The binding energy for deuteron (one proton and one neutron) is EB = 2.22 MeV. 0.11. LECTURE 11 27 0.11 Lecture 11 10.1: Identical Particles • Two different statistics of particles empirically. You cannot track each particle in a large ensemble in principle in quantum mechanics. • Consider N particles: ri where i = 1, . . . , N positions of the particles. Can also have pi (momentum), and si (spin), ei (charge), etc. • i~Ψ̇(q1, . . . , qN , t) = HΨ(q1, . . . , qN , t), where q denotes the collection of the quantum numbers. • Assume H = T + V , where T = ∑N i=1 Ti and Ti = − 1 2mi p2 i . • Assume V is independent of time, so Ψ(q1, . . . , qN , t) = ψ(q1, . . . , qN ) e−iEt/~, where E is the total energy. • Consider symmetries and conserved quantities (Noether’s theorem relates the two): – Total orbital angular momentum: L = ∑ i Li. – Total spin: S = ∑ i Si. – Total angular momentum: J = ∑ i Ji, where J = L + S. • Special case: H = ∑ i hi when V = ∑ i Vi, for an external potential. – Each particle has its own Schrödinger equation: hiuλ(qi) = Eλuλ(qi). – The total wavefunction is a product: ψ(q1, . . . , qN ) = uλ1 (q1) · · ·uλN (qN ). This is an eigenfunction of H with eigenvalue E = Eλ1 + · · ·+ EλN . – This wavefunction is “uncorrelated”. – Two possible opposites to “uncorrelated”: (1) “correlated” (usually when strongly interacting); (2) “entangled” (linear combination of direct product terms). Indistinguishable particles: • [Pij , H] = 0, where Pij switches the particles i and j. Note that Pijψ(· · · qi · · · qj · · · ) = eiϕψ(· · · qj · · · qi · · · ), where ϕ is some phase since a phase does not affect the physics. • This is generalizable to larger permutations rather than just a switch of two particles: [Pπ(1,...,N), H] = 0. The permutation group, ΠN . • P 2 ij = 1. There’s an implicit assumption here! (Topology; anyone). This implies that the eigenvalues of Pij are ±1. There are only two options, which are consistent for all permutations: (1) completely symmetric (Bose-Einstein statistics; bosons); or (2) completely antisymmetric (+1 eigenvalue under even permutations and −1 for odd permutations) (Fermi-Dirac statistics; fermions). • Spin-statistics theorem: In Poincaré invariant theories of many particles (relativistic QFT), all particles of spin 2n+1 2 , for some integer n, are fermions and all with spin n are bosons. Construction of symmetric and antisymmetric states: • Symmetric: uα(q1) · · ·uν(qN )→ 1 |ΠN | ∑ π∈ΠN uα(qπ(1)) · · ·uν(qπ(N). • Antisymmetric: uα(q1) · · ·uν(qN )→ 1 |ΠN | ∑ π∈ΠN (−1)|π|uα(qπ(1)) · · ·uν(qπ(N), where |π| = 0 if it is an even permutation and 1 if it is odd. Ch. 11: Matter + Electromagnetism: 28 CONTENTS • Proper setting: QED, relativistic, treats matter and radiation quantum mechanically (particles = quantum of a field). • Why do we split matter as quantum but radiation as classical? (1) Historical context: spontaneous emission, stimulated emission, and absorption of radiation by a particle; (2) Physics: often the number of photons involved in the process is huge (compared to the matter), so you can replace photons with an electromagnetic field, Aµ (potential). • Electromagnetic field: E and B. Potentials: π, A in terms of which E = −∇φ− Ȧ, B = ∇×A. • Gauge “symmetry” (more appropriately redundancy): A→ A +∇χ and φ→ φ− χ̇. • Lagrangian for coupling of electromagnetism to matter: L ∝ ∫ (ρφ− j ·A). • This is crucial to derive the Hamiltonian, which we derive canonically. 0.12. LECTURE 12 31 • Alternative way to derive this approximation: B/E ∼ 1/c and so set B = 0 when c is large. So, this approximation should be good as long as E is not accidentally zero. Time-dependent perturbation theory: • H0ψk = Ekψk. • ψ(t, r) = ∑ ck(t)ψk(r) e−iEkt/~ with ck(t ≤ 0) = δk,a. Then, ċ (1) b = 1 i~ E0〈ψb|ε̂ ·D|ψa〉 ∫ t 0 sin(ωt′ − δω) eiωbat ′ dt′ = 1 i~ E0〈ψb|ε̂ ·D|ψa〉 { 1 2 e−iδω [ 1− ei(ωba+ω)t ωba + ω ] − 1 2 eiδω [ 1− ei(ωba−ω)t ωba − ω ]} . • Stimulated emission: ωba ≈ −ω. Absorption: ωba ≈ +ω. Absorption: • Eb > Ea and we keep only the large piece: |c1b |2 = 1 2 ( E0(ω) ~ )2 |ε̂ ·Dba|2 F (t, ωba − ω), where Dba is the dipole operator matrix element from a to b, and F (t, ω) = 1− cosωt ω2 . • We have |ε̂ ·Dba|2 = |Dba|2 cos2 θ, where θ is the angle between ε̂ and Dba. • This gives 1 2E 2 0 = 1 ε0c I(ω). Thus, |c(1) b (t)|2 = I(ω) ~2cε0 cos2 θ |Dba|2 F (t, ωba − ω). • Choosing some I(ω) for the radiation, we can integrate this over ω. Given the assumptions – I(ω) is approximately constant in the region where F is peaked • The result is P (1) ba (t) ≈ πI(ωba) ~cε0 cos2 θ |Dba|2 t. • The transition rate is the time derivative, which just gets rid of the t at the end of the above formula. • Incoherent radiation will have all sorts of possible values of θ. We must average cos2 θ over all angles (in 3D). This gives a factor of 1/3: Wba = πI(ωba) 3~cε0 . • Absorption cross-section: σba = ~ωbaWba I(ωba) . 32 CONTENTS 0.13 Lecture 13 Light shining on atom: • H0ψa = Eaψa and same for b. Let ψ(0) = caψa + cbψb. Then ψ(t) = ca(t)ψae iEat + cb(t)ψbe iEbt. • ċa = − i ~H ′ ab(t)e −iω0tcb where ω0 = (Eb − Ea)/~. • Assume: H ′ii = 0 (usually happens) and H ′ is small. • Zeroth order: ca(0) = 1 and cb(0) = 0. Then, c (0) a (t) = 1 and c (0) b (t) = 0. • First order: ċa = 0, but ċb = − i ~H ′ ab(t) e −iω0t. • Solve: c (1) a (t) = 1 and c (1) b (t) = − i ~ ∫ t 0 H ′ab(t ′) e−iω0t ′ dt′. Example: Sinusoidal perturbation: • H ′(r, t) = V (r) cosωt. Then, H ′ab = Vab cosωt. • cb = − 1 2~Vab [ ei(ω0+ω)t−1 ω0+ω + ei(ω0−ω)t−1 ω0−ω ] . • Real part of cb as a function of omega: diag 1 • Cross terms don’t matter much: so take ω ≈ ±ω0. • Take ω ∼ ω0 (shine light with ω ∼ ω0 on some atom that has a gap of energy states somewhere of the order of ω0). Then, Pa→b(t) = |cb(t)|2 = |Vab|2 ~2 sin2(ω−ω0)t/2 (ω−ω0)2 . • Plot as a function of t: diag 2. This is called Rabi flopping. • Plot as a function of ω: diag 3. Shine light at two-state system: • Assume: λ >> system size (e.g. atom). We can treat the EM field to be approximately constant over space. • Usually, E much more significant than B. So neglect magnetic effects. • Monochromatic polarized light: E = E0 cos(ωt) ẑ. • U = −q ∫ E · dr for electrostatics. We must assume that ν−1 >> time scale appropriate to the system. • H ′ = qE0z cosωt, then H ′ab = −PE0 cosωt where P = q〈ψa|z|ψb〉 is the dipole moment. • Usually, our wavefunctions will be even or odd with respect to parity in z (or any other axis). Therefore, the expectation value of z in one state will usually vanish. • This is like our previous result with |Vab| = −PE0. • Energy density: u = 1 2ε0E 2 0 . • The transition probability for absorption is Pa→b(t) = 2|P |2u ~2 sin2[(ω − ω0)t/2] (ω0 − ω)2 . • For stimulated emission: switch a and b and ω0 → −ω0. Stimulated emission is how LASER’s work. • We find detailed balance: Pb→a = Pa→b. • Spontaneous emission is like stimulated emission by virtual photons. 0.13. LECTURE 13 33 Non-monochromatic and incoherent: • ρ(ω)dω is the energy density for radiation with frequency between ω and ω + dω. We might call this uω. Pb→a(t) = 2|P |2 ~2 ∫ ∞ 0 ρ(ω) [ sin2[(ω0 − ω)t/2] (ω0 − ω)2 ] dω. This is incoherent since we squared and then integrated. We really should integrate then square. It would make a different for coherent light. Pb→(t) ≈ π|P |2 ε0~2 ρ(ω0)t. The rate is just the time derivative. • We want 〈 (n̂ · p)2 〉 averaged over incident directions n̂. • We sum over two polarizations (say y and z) and divide by 2 to average over polarizations: 1 2 [P 2 z + P 2 y ] = 1 2P 2 sin2 θ. We must average this over angles. Einstein’s argument: • Na in state ψa and Nb in state ψb. What is Ṅb? Ṅb = Naρ(ω0)Ba→b −Nbρ(ω0)Bb→a −NbA where B = Rb→a = π 3ε0~2 P 2ρ(ω0). • Static equilibrium: Ṅb = 0. Then, ρ(ω0) = A (Na/Nb)Bab −Bba . By stat mech, we have at thermal equilibrium Na Nb = e~ω0/kT . Also, by stat mech, ρ(ω) = ~ π2c3 ω3 e~ω/kT − 1 . Putting this together, we get Bab = Bba and A = ω3~ ~2c3 Bba = ω3|P |2 3πε0~c3 . • A is the lifetime of an atom if it is excited and then sits in zero temperature: Ṅb = −ANb =⇒ Nb = Nb(0) e−At. Example: Harmonic oscillator: • a |n〉 = √ n |n− 1〉 and a† |n〉 = √ n+ 1 |n〉. x = √ ~ 2mΩ (a+ a†). • Dipole: P = q〈nf |x|ni〉 = √ ~ 2mΩ (√ nf δnf ,ni−1 + √ ni δnf ,ni+1 ) . 36 CONTENTS Thus, under parity, we get I→ (−1)`+` ′+1 ∫ 2π 0 dφ ∫ π 0 dθ sin θ Y ∗`′m′ r̂Y`m. Therefore, transitions always change the parity. This just says that ` and `′ differ by an odd integer. This odd integer is not necessarily ±1, which what it should be. That requires more work (done last lecture). • Selection rules: ∆` = ±1 and ∆m = 0 or ∆m = ±1. • Therefore, only non-zero matrix element is between 200 and 210. This is actually easy to argue: electric field breaks full rotations, but maintains rotations in the x-y plane. That is, L2 does not commute with H ′, but Lz does and so m is still a good quantum number). So, states with different m do not mix! −E(1) r cr1 +H ′12cr2 = 0, H ′21cr1 − E(1) r cr2 = 0, for r = 1, 2. Then, E (1) r = ±|H ′12| and the two corresponding states are the symmetric (E (1) 1 = +|H ′12|) and antisym- metric (E (1) 2 = −|H ′12|) combinations. • Consider n = 3: diag 1. • Transitions between n = 3 and n = 2: ∆m = 0 called π lines and ∆m = ±1 called σ lines. Weak Stark Effect: i.e., need to consider fine structure as well: • Still treat these non-degenerate levels as quasi-degenerate: ∆EStark ∼ √ (∆Efs)2 + |Dba|2E2. Stark effect example 2: • Molecule with permanent dipole moment (e.g., salt). • H0 = L2 2I and H ′ = −D ·E. • E(0) ` = 1 2I ~ 2`(`+ 1) for ` = 0, 1, 2, . . .. ψ (0) `m = Y`m(θ, φ0). E (1) `m = 〈`,m|H ′|`′m′〉 = 0 by parity. E (2) `m = D2E2I ~2 [ `(`+1)−3m2 `(`+1)(2`−1)(2`+3) ] for ` 6= 0 and = − 1 3 D2EI ~2 for ` = 0. Note that this only depends on m2 and not just m itself. This is because of a discrete symmetry of the system. Particles in magnetic field: • Recall the experimental fact that particles have intrinsic magnetic moment: M ∝ S. • Electron: Ms = gsµB S ~ , where µB is the Bohr magneton µB = e~/2me. This gs is the gyromagnetic ratio of the electron and experimentally is gs ≈ 2. This is explained by the Dirac equation (assuming electron theory is Lorentz invariant). • In a weak magnetic field, the coupling is ∝M ·B. • Proton: Mp = gpµN σ 2 with µN = e~/2mp and gp ≈ 5.5883 . . .. • Neutron: Mn = gnµN σ 2 with gn = −3.8263 . . .. 0.15. LECTURE 15 37 0.15 Lecture 15 Landau Levels: • Exact solutions for a charged particle in a constant magnetic field (arbitrary strength; not just perturbative). • Intrinsic magnetic moment M interacts with magnetic field via −M · B at linear level. But we do not want to resort to this linear approximation. • Back to Schrödinger equation: L = 1 2mṙ2 + qA · ṙ gave us the canonical momentum pcan = mṙ + qA. • Hamiltonian: H = ṙ · pcan − L = 1 2m (p− qA)2. We have set φ = 0. • B constant, we want to pick a gauge for the vector potential. Suppose B = Bẑ. A(r) = 1 2 B× r = 1 2 B(−yx̂ + xŷ). This maintains rotation invariance in the x-y plane. • Spinless: i~Ψ̇ = [ − ~2 2m∇ 2 + i~ q mA · ∇+ q2 2mA 2 ] Ψ. Note: the order of A and ∇ is immaterial because A is divergenceless! i~ q m A · ∇ = i~q 2m B · (r×∇) = i~q 2m B · L. Here, L is the usual angular momentum operator. Let ML = q 2m L. The, this linear coupling is −ML ·B. In general, spin can add angular momentum: −Mtot ·B, where Mtot = ML + MS . • The quadratic piece is q2 2m A2 = q2 8m [B2r2 − (B · r)2]. • Altogether: [ − ~2 2m ∇2 −M ·B + e2 8m [B2r2 − (B · r)2] ] Ψ = EΨ. • For the electron, and B along z:[ − ~2 2m ∇2 − µBB ~ (Lz + 2Sz) + e2B2 8m (x2 + y2) ] Ψ = EΨ. • Symmetries: H commutes with Lz (not Lx or Ly). Also, with Sz. Also, with pz = −i~∂z. We can simultaneously make our solutions eigenstates of all of these operators. HΨq = EqΨq. Here, q stands for all quantum numbers. • LzΨ(r, σ) = ~m`Ψq(r, σ). Note: L2 does not commute with H and so ` is not a good quantum number. So, it does not really make sense to group the m`’s into ones associated with a fixed `. • SzΨ = ~msΨ where ms = ±1/2. • PzΨ = ~kΨ where k ∈ R. • q = {E,m`,ms, k} although some of these depend on others. 38 CONTENTS • Ψq(r, σ) = eikzφq(x, y)× ( 1 0 ) for ms = + 1 2 or ( 0 1 ) for ms = − 1 2 . • Plug into Schrödinger equation:[ − ~2 2m ( ∂2 ∂x2 + ∂2 ∂y2 ) + 1 2 mω2 L(x2 + y2) ] φq(x, y) = ( E − ~2k2 2m − ~ωL(m` + 2ms) ) φq(x, y). Here, ωL ≡ µBB/~ is the Larmor angular frequency. • This is just a 2D harmonic oscillator. Therefore, the shifted energy of the 2D oscillator, E′ are E′ = (nx + ny + 1)~ωL. • Thus, the total energy is E = ~2k2 2m + ~ωL(nx + ny +m` + 2ms + 1). However, there are constraints: nx + ny +m` has to be even (check textbook). Therefore, nx + ny +m` + 2ms + 1 = 2r + 1 where r ≥ 0. These levels are called Landau levels. Different gauge: • Break rotation invariance in x-y plane: A = −Byx̂. • tiSEq: [ − ~2 2m ∇2 + i~ q m ( −By ∂ ∂x ) + q2 2m B2y2 ] Ψ = EΨ. • H commutes with px (in addition to pz): pzψq = ~kzψq, pxψq = ~kxψq. Therefore, Ψq = ei(kxx+kzz)φ(y).[ − ~2 2m (k2 x + k2 z + ∂2 y) + ~qBkx m y + q2B2 2m y2 ] Ψ = EΨ. This is just a 1D shifted SHO. • E = ~2k2z 2m + ~ωL(2r + 1) where r now is just the ny. • Each level is technically infinitely degenerate (degeneracy labeled by kx). How does this compare with other gauge: compactly R2 to T 2, then kx must be discrete. Berry’s Phase (Sir Michael Berry, 1984, but done earlier by Pancharatnam in 1956): • Nontrivial topology of things matters! • Ψ(x, t)→ eiαΨ(x, t), the phase is unobservable. However, these phases can pick up dependence on external parameters such as an adiabatically changed magnetic field. Locally, as functions of these parameters, these phases are unobservable. However, when integrated over nontrivial loops, the integral itself becomes observable! • Aharonov-Bohm effect: Predicted in 1959 and measured in 1960. B and E are observable. Wwitch to A and φ. Usually, these are not observable because of gauge redundancy. But, not all gauge invariant features of A and φ reduce to the information contained in B and E. • θ ∈ S1 with θ ∈ [0, 2π). Set the radius of the circle to 1 for simplicity. Let A just be constant and in one dimension and set φ = 0. Then E and B vanish identically. Here, A just points along the circle. Let it be constant and nonzero along the circle. Can we gauge this away?: A → A′ = A + ∂θχ. But this is not possible for a single-valued well-defined function χ. 0.16. LECTURE 16 41 • Generalization 2: Abelian (E&M) to non-abelian Yang-Mills. Degenerate states: |ψα〉 for α = 1, . . . , n for some n. Suppose this degeneracy is not lifted by the adiabatic perturbation. • See: Zee Appendix to Chapter IV.5. (QFT). • Pick an orthonormal basis of states. Another basis given by a unitary transformation: Uβα (λ(t)) |ψβ(λ(t))〉. • Now, we need the matrix 〈 ψα(λ) ∣∣∣∣∂ψβ∂λi 〉 = Aαβi (λj). • Compare different bases: Ai → U†AiU + U† ∂ ∂λi U. This is precisely the gauge transformation for non-Abelian Yang-Mills. Aharonov-Bohm Effect: • Predicted by theory in 1959 and verified in 1960 by Chambers. • Confine a charge in a box. Put some magnetic flux through some confined region where charge cannot go. Turn on just a vector potential (no scalar since no electric field anyway). We have ∇×A = 0 outside the region, but this does not mean A = 0. We have A = ∇χ locally, but not globally! • Gauge transformation on wavefunction: ψ′ = eiqχ/~ψ. • Here, χ(r) = ∫ r r0 A dr. Total phase around the closed contour: δ = i ~ q ∮ C A · dr = i ~ q ∫ D B · dΣ = i ~ qΦ, where Φ is the magnetic flux. • The experiment of Chambers involved a double-slit experiment with a thin whisker containing magnetic flux between the two slits. Ψend = exp [ iq ~ ∫ CB A ] [ exp [ iq ~ ∮ A ] ΨA + ΨB ] . We can get rid of the overall exponential since it is just an integral over a line, not a loop. But we cannot get rid of the loop part! Ψend ∼ e iq ~ ΦΨA + ΨB . The interference depends on Φ! Different situations of particles in external electromagnetic fields: • Stark is E 6= 0. • Zeeman is B 6= 0. • H = − ~2 2m∇ 2 − Ze2 4πε0r + µB ~ (Lz + 2Sz)B + ξ(r)L · S. • Zeeman types: – Strong-field: B dominates over spin-orbit. – Paschen-Back: B strong but comparable to spin-orbit. – Anomalous: B is much weaker than spin-orbit. 42 CONTENTS 0.17 Lecture 17 Scattering (Quantum Collision Theory): • Central concept: the “S-matrix”. This is a unitary matrix if we believe a probability interpretation and probability conservation. • Probes in physics: – Transitions between energy levels of bound states (Spectroscopy) – Scattering known probes off of the system of interest. Scattering two systems of interest off of each other (Collider physics). Most of the Standard Model was developed using these techniques. – Observational cosmology. (Like “scattering inside out”). • Scattering experiments: the fundamental observables are “cross-sections”. Reference: S. Weinberg, “Lectures on Quantum Mechanics”, Ch. 7. Beam of known particles. Mono-energetic. Non-interacting (among themselves). Collimated. Collimated plus mono-energetic seem contradictory of uncertainty principle. More exactly, we should make wave-packets. See Weinberg. This beam is aimed at a target (e.g., particle, many particles, potential, etc.) For now, assume beam and target are distinguishable. • Elastic scattering: internal quantum numbers of beam (but often also target) don’t change. Non-elastic scattering usually requires QFT since number of particles and species may change. • Let beam move in the z direction. Deflection angle of beam is θ. If beam and target are rotationally invariant about the z axis, then the cross-section ought to be independent of φ. • Detector at θ, φ. Target at origin of coordinates. • We want to measure the number of particles entering the detector is NdΩ, where dΩ is the angular extent of the detector. In 3 dimensions, dΩ = sin θ dθ dφ. • Assume that the beam is in “steady-state”, so that the rate is constant and we needn’t worry about when we’re doing the measurement. • Beam has an incoming flux: F = number of incoming particles in the beam, per unit time, per unit area transverse to the beam. • Differential scattering cross-section: dσ dΩ = N F . • Total cross-section is simply the integral over all angles. • Scattering in the lab frame: target is at rest. But in LHC and others, the scattering is usually where the actual lab is near the center of momentum frame. • How to compare the two frames? In the center of momentum frame, set pA = −pB = p. In the lab frame, only A is moving with qA. The scattering angles in the com frame will be θcom, φcom and similarly for lab frame angles. ( dσ dΩ ) L = ∣∣∣∣d(cos θcom) d(cos θL) ∣∣∣∣( dσdΩ ) com = ( 1 + m2 A m2 B + 2mAmB cos θ )3/2∣∣1 + mA mB cos θ ∣∣ ( dσ dΩ ) com . • Derivation: Tcom = p2 2µ where µ = mAmB mA+mB is the reduced mass. Lab momentum: qA = (mA + mB)vcom where vcom is the velocity of the center of momentum frame relative to the lab frame. 0.17. LECTURE 17 43 Longitudinal momentum: p′A cos θcm = q′A cos θL −mAvcm. The transverse components are equal to each other. tan θL = sin θcm cos θcm + mA mB . We derive the relation between the differential cross-sections by forcing the total cross-sections to be the same. The total cross-section cannot care about which frame we’re in! Potential Scattering: • i~ψ̇ = [ − ~2 2m∇ 2 + V (r) ] ψ. We can often reduce to this case, where V goes to zero sufficiently quickly as r →∞. • Steady state: time-independent equations: Ψ(r, t) = ψ(r) e−iEt/~. • E = p2 2m = 1 2mv 2 = ~2k2 2m (this is the eigenvalue near spatial infinity, but the energy will be the same near the potential, just harder to measure; this is just energy conservation).[ ∇2 + k2 − U(r) ] ψ(r) = 0, where U(r) = 2m ~2 V (r). We only care about asymptotics at large r, where our detectors are located. Near infinity, this is just a free equation because V vanishes. ψ(r) = ψin(r) + ψscatt(r). • ψin = eikz oriented along the z direction. |ψin|2 = 1, one particle per unit volume. Velocity ~k/m. Flux is just equal to v because this is just one particle per unit volume: F = ~k/m. • ψscatt(r) = f(k, θ, φ) e irk r since eirk/r is a solution of the free equation in spherical coordinates. Here, f(k, θ, φ) is the “scattering amplitude”. • Probability current: j(r) = ~ 2mi [ ψ∗(∇ψ)− (∇ψ∗)ψ ] . • ∇ = (∂r, (1/r)stuff, (1/r)stuff) ≈ (∂r, 0, 0) near r =∞. Thus, jr = ~k mr2 |f | 2. • Rate of arrival of particles at detector with area r2dΩ: NdΩ = ~k m |f |2 dΩ. • Differential cross-section: dσ dΩ = |f |2. Method of partial waves: • Usually expand in plane waves. Good for incoming beam, but not for outgoing ones. Switch to spherical coordinates. • Detour: Free particle in d = 3 in spherical coordinates. H0 = − ~2 2m∇ 2 • H0 = − ~2 2m [ r−2∂r(r 2∂r)− L2 ~2r2 ] . • Switching from eigenfunctions of H0 and p (plane waves), to H0 and L2 and Lz (spherical waves with spherical harmonics). • E = ~2k2/2m. Note that H0 and p ¯ only contain three independent operators. If we know the energy and two of the momentum components, then we know the third. This agrees with the counting of independent operators for the spherical case, also 3. • ψk`m = Rk`(r)Y`m(Ω) with L2 = ~2`(`+ 1). • Recall: Y`m ∼ eimφ and so only Y`0 are φ-independent. For spherically symmetric systems, this is the only possible Y that can contribute since the diff. scatt. cs. cannot depend on φ. 46 CHAPTER 1. DISCUSSION 1.1 Discussion 1 Preliminary Quiz Exercise in harmonic oscillator raising and lowering operators: Let |n〉 be the 1D harmonic oscillator energy eigenstate with energy En = ~ω(n + 1 2 ) (note: BJ call this state |En〉). Calculate the expectation values of p, p2, H and H2 in this state, where p and H are the momentum and Hamiltonian operators, respectively. Use your results to calculate the standard deviations, σp and σH , in momentum and energy (note: the standard deviation of an operator, O, is defined via σ2 O = 〈O2〉 − 〈O〉2). SOLUTION: The raising and lowering operators are defined as a± = 1√ 2 [( mω ~ )1/2 x∓ i p (m~ω)1/2 ] . (1.1.1) Therefore, p = i ( m~ω 2 )1/2 (a+ − a−). (1.1.2) The action of a± on |n〉 is a+ |n〉 = √ n+ 1 |n+ 1〉 , a− |n〉 = √ n |n− 1〉 . (1.1.3) Therefore, p |n〉 = i ( m~ω 2 )1/2(√ n+ 1 |n+ 1〉 − √ n |n− 1〉 ) . (1.1.4) Both terms on the RHS have vanishing overlap with |n〉. Thus, 〈p〉 = 〈n|p|n〉 = 0 . (1.1.5) Next, we have p2 |n〉 = i ( m~ω 2 )1/2(√ n+ 1 p |n+ 1〉 − √ n p |n− 1〉 ) = −m~ω 2 (√ (n+ 1)(n+ 2) |n+ 2〉 − (n+ 1 + n) |n〉+ √ n(n− 1) |n− 2〉 ) . (1.1.6) Only the middle term contributes to the expectation value: 〈p2〉 = 〈n|p2|n〉 = m~ω ( n+ 1 2 ) . (1.1.7) We could calculate 〈 x2 〉 and use it and 〈 p2 〉 to calculate 〈H〉 since H is simply given by H = 1 2mω 2x2 + 1 2mp 2. However, we can also simply recall that H = ~ω ( a+a− + 1 2 ) . (1.1.8) This shows that |n〉 is an eigenstate of H, and so 〈H〉 = 〈n|H|n〉 = ~ω ( n+ 1 2 ) ,〈 H2 〉 = 〈n|H2|n〉 = ~2ω2 ( n+ 1 2 )2 . (1.1.9) The standard deviations are σp = √ m~ω(n+ 1 2 ), σH = 0. (1.1.10) 1.1. DISCUSSION 1 47 Perturbations to the potential energy do not have to “look” small: Example 1 (Anharmonic oscillator): We can see this already in the example in homework 1, where H0 is the harmonic oscillator and H ′ = ax3 + bx4. How on earth can perturbation theory work when H ′ becomes large for sufficiently large x? It does not matter how small a and b are, eventually, far enough away from the origin (x = 0), H ′ will dominate H0. Here’s a clue: You saw the second order correction in lecture and in BJ and you have to calculate the third order correction in homework 1. The corrections are all given by polynomials in n, the principal quantum number of the harmonic oscillator. Therefore, no matter how small a and b are, as long as they are not both strictly zero, these corrections will become huge for sufficiently large n! So we have two facts: (1) H ′ ≈ 0 within a region around x = 0 (how large the region depends on how small a and b are), but eventually, H ′ diverges; and (2) perturbation theory breaks down for sufficiently large n. Might these points actually be linked. Indeed, they are! The qualitative behavior of the wavefunctions of the harmonic oscillator as a function of n is that they wiggle more and more as n increases and they spread out reaching further and further away from x = 0 as n increases. This makes sense in the classical picture: Imagine a ball rolling in a (parabolic) bowl. The more energy this ball has, the further away from the bottom of the bowl it can go. That’s it: If n is small, then it doesn’t really matter that H ′ eventually diverges for sufficiently large x because the wavefunction of that state is essentially zero there anyway. However, if n is too large, then the wavefunction might extend far enough into the region where H ′ starts becoming too big and then perturbation theory breaks down. So, you see that in evaluating how reasonable perturbation theory will be, one must examine not just what H ′ is doing, but what H ′ and the zeroth order wavefunctions are doing together. Example 2 (Finite-sized nucleus): You might think that all you need to ensure is that the zeroth order wavefunctions basically vanish wherever H ′ is not small. This is not strictly true. It is possible that H ′ might become big or indeed diverge, but only in a very small region. In this case, it might be permissible for the wavefunction to not vanish in the region where H ′ is large. For example, consider the correction to the ground state of a hydrogen-like atom (all electrons ionized except one) due to the finite size of the nucleus assuming, for simplicity, that the charge of the nucleus, Ze, is evenly distributed throughout the volume of a sphere of radius R. This assumes that the charge of the electron does not significantly affect the distribution of nuclear charge, which is reasonable given that the proton-proton electrostatic forces and the strong nuclear forces are much much stronger that the proton-electron electrostatic forces. Furthermore, the electron is to be in the ground state, which is spherically symmetric. Using Gauss’ law, you can work out the potential energy of the electron in the presence of the electric field produced by the nuclear charge: V (r) = { −Ze 2 R ( 3 2 − 1 2 r2 R2 ) , 0 ≤ r ≤ R, −Ze 2 r , r ≥ R. (1.1.11) This is plotted below as a solid line. The dotted line is what we would have if the nucleus were a point. 1 2 3 r R -1 -2 -3 VR Ze 2 H ′ is zero for r ≥ R but quickly goes to −∞ as r → 0. What saves us is that we know that the radius of the nucleus, R, is much smaller than the Bohr radius, a. The ground state wavefunction is ψ100(r) = 1√ πa3 e−r/a, a = ~2 Zme2 . (1.1.12) 48 CHAPTER 1. DISCUSSION Admittedly, we expect R to increase with Z, while we see a decreases with Z. The hard-sphere packing model roughly gives R ∼ Z1/3. Therefore, the ratio R/a increases with Z roughly at the rate Z4/3. Thus, you might think that this approximation breaks down for sufficiently large Z. However, Z only goes up to order 100 or so (we only have so many elements) and this is not enough to make up the large difference between R ∼ 1 to 10 fm and aB = 53, 000 fm. If you work it out, the first order correction to the ground state energy, E1, where the subscript is the value of the principal quantum number n, is approximately given by E1 ≈ E(0) 1 ( 1− 4 5 R2 a2 ) . (1.1.13) Since R/a << 1, this is indeed a small correction. Furthermore, one can check that the following condition holds: |H ′mn| << |E(0) n − E(0) m |, (1.1.14) which is a necessary condition for the validity of the second order approximation. Perturbations to the potential energy do not have to “be” small: Oh surely you’re joking now! Well sort of. Consider the following example: Treat the Helium atom as two separate copies of the hydrogen-like Helium atom neglecting the Coulomb repulsion between the two electrons. Doing so gives a first ionization energy of ∼ 109 eV. Then, you can treat the Coulomb repulsion between the two electrons in first-order perturbation theory and find the first ionization energy to be ∼ 79 eV. That’s a big difference! But it’s not that bad of an approximation to the actual experimental value of ∼ 70 eV! Strictly speaking, perturbation theory cannot be trusted in this situation. You could have said that right from the start. The zeroth order result is 109 eV whereas the experimental result is 70 eV. Well, that’s not approximately correct so the starting point is clearly a bad one. Indeed, none of the higher order corrections gets you any closer to the correct result. It turns out that the reason why perturbation theory does an unreasonably good job in this particular case is that it is actually essentially equivalent to the variational procedure with the same starting point (in this case). 1.3. DISCUSSION 3 51 1.3 Discussion 3 Comments on Homework 2 Exercise in variational principle and WKB approximation: In the y-z plane is a uniformly charged infinite sheet of surface charge density σ. A point charge of mass m and charge −q, with q > 0, is confined to move along the x axis only and can freely pass through the charged sheet. (a) Estimate the ground state energy of the point charge using the trial wave function ψ(x) ∝ exp [ − 1 2 ( x a )2] . (b) Estimate the ground state energy of the point charge using the WKB approximation. SOLUTION: (a) Recall that the electric field produced by a uniformly charged infinite plane is constant, E = σ 2ε0 , pointing away from the plane (if σ > 0; towards the plane if σ < 0), i.e. E = σ 2ε0 x |x| x̂. The electric potential, Φ, is related to the electric field via E = −∇Φ, and so Φ = −E|x|. The potential energy of the point charge is V (x) = −qΦ = qE|x|. (1.3.1) Therefore, the Hamiltonian for the point charge is H = − ~2 2m d2 dx2 + qE|x|. (1.3.2) Before we proceed, note that the combination qE has units of force, or energy per length. Meanwhile, ~ has units of energy times time and the only other dimensionful quantity we have is m, a mass. Therefore, the combination (qE)2~2/m has units of energy cubed. Therefore, all of the energy eigenvalues of this system must take the form En = cn [ (~qE)2 m ]1/3 , (1.3.3) where cn is some number which depends on the nth state. Now, let us normalize the wavefunction. Set ψ(x) = A exp [ − 1 2 ( x a )2] and determine A: 1 = A2 ∫ ∞ −∞ dx e−(x/a)2 = A2a √ π. (1.3.4) Therefore, our normalized trial wavefunction is ψ(x) = (a2π)−1/4 exp [ − 1 2 ( x a )2] . (1.3.5) Let us calculate the expectation value of the potential energy in this wavefunction: 〈V 〉 = ∫ ∞ −∞ dxψ∗(x)V (x)ψ(x) = 2qE a √ π ∫ ∞ 0 dxx e−(x/a)2 = aqE√ π . (1.3.6) Now, we will evaluate the expectation value of the kinetic energy. The first derivative may be written as dψ dx = − 1 π1/4 1 a3/2 ( x a ) e− 1 2 ( xa )2 . (1.3.7) 52 CHAPTER 1. DISCUSSION The second derivative may be written as d2ψ dx2 = 1 π1/4 1 a3/2 [( x a )2 − 1 ] e− 1 2 ( xa )2 . (1.3.8) Therefore, the expectation value of the kinetic energy is 〈T 〉 = − ~2 2m 1√ π 1 a3 ∫ ∞ −∞ dx [( x a )2 − 1 ] e−(x/a)2 = − 1√ π ~2 2ma2 (√ π 4 − √ π ) = 3~2 8ma2 . (1.3.9) Therefore, the expectation value of the Hamiltonian is E(α) = 〈H〉 = 3~2 8ma2 + aqE√ π . (1.3.10) Now, we must minimize this as a function of a: 0 = dE dα = − 3~2 4ma3 + qE√ π =⇒ a = ( 3 √ π 4 ~2 mqE )1/3 . (1.3.11) Plugging this back in yields our estimate for the ground state energy: E0 = [ 34 25π (~qE)2 m ]1/3 ≈ 0.931 [ (~qE)2 m ]1/3 . (1.3.12) (b) For a state with energy E, the classical turning points are given by E = V (x). This gives two turning points (one negative, x1, and one positive, x2): x1 = − E qE , x2 = E qE . (1.3.13) Let region I be x < x1, region II be x1 < x < x2, and region III be x > x2. The solution in region II is ψII(x) = 1√ p(x) sin [ 1 ~ ∫ x x1 p(x′) dx′ − π 4 + δ ] , (1.3.14) where δ is some number and where p(x) = √ 2m(E − V (x)) = √ 2m(E − qE|x|). (1.3.15) Note that the solution could have been written in terms of sines and cosines. However, in this way of writing the solution, we know from lecture what the solution in region I looks like as x→ −∞: ψI(x) x→−∞−−−−−→ − cos δ√ |p(x)| exp [ 1 ~ ∫ x1 x |p(x′)| dx′ ] . (1.3.16) This diverges unless δ = ( n+ 1 2 ) π. (1.3.17) Note that n is any integer here, including negative integers. 1.3. DISCUSSION 3 53 We can write (1.3.14) as ψII(x) = 1√ p(x) sin [ −1 ~ ∫ x2 x p(x′) dx′ + 1 ~ ∫ x2 x1 p(x′) dx′ − π 4 + δ ] = − 1√ p(x) sin [ 1 ~ ∫ x2 x p(x′) dx′ − π 4 + π 2 − δ − 1 ~ ∫ x2 x1 p(x′) dx′ ] = − 1√ p(x) sin [ 1 ~ ∫ x2 x p(x′) dx′ − π 4 + δ′ ] , (1.3.18) where δ′ = π 2 − δ − 1 ~ ∫ x2 x1 p(x) dx. (1.3.19) In the region III, as x→∞, the solution becomes ψIII(x) x→∞−−−−→ cos δ′√ |p(x)| exp [ 1 ~ ∫ x2 x|p(x′)| dx′ ] , (1.3.20) which diverges unless δ′ = ( n′ + 1 2 ) π. (1.3.21) Combining Eqns. (1.3.17), (1.3.19) and (1.3.21) yields 1 ~ ∫ x2 x1 p(x′) dx′ = − ( n+ n′ + 1 2 ) π. (1.3.22) Since n and n′ are just arbitrary integers, we could also just as well write this condition as 1 ~ ∫ x2 x1 p(x) dx = ( n+ 1 2 ) π, (1.3.23) where n is an integer, now constrained to be non-negative since the integral obviously cannot be negative! The integral on the LHS is 1 ~ ∫ x2 x1 p(x) dx = 1 ~ ∫ E/qE −E/qE dx √ 2m(E − qE|x|) = 2 ~ √ 2mqE ∫ E/qE 0 dx ( E qE − x )1/2 = 2 ~ √ 2mqE [ −2 3 ( E qE − x )3/2]E/qE 0 = 4 √ 2 3 √ mE3/2 ~qE . (1.3.24) Setting this equal to ( n+ 1 2 ) π and solving for E, which we label En, gives En = ( 3 ( n+ 1 2 ) π 4 √ 2 )2/3[ (~qE)2 m ]1/3 . (1.3.25) For the ground state, we get E0 = (3π)2/3 27/3 [ (~qE)2 m ]1/3 ≈ 0.885 [ (~qE)2 m ]1/3 . (1.3.26) Note: Since the potential energy is an even function of x, the energy eigenstates of the system may be chosen to be completely even or completely odd functions of x. Since the ground state is even, we can make a variational estimate for the first excited energy by choosing a trial wavefunction, which is odd. The obvious choice is ψ(x) ∼ x exp [ − 1 2 ( x a )2] . Doing so, we find E1 = 2E0. For WKB, E1 = 32/3E0 ≈ 2.1E0. 56 CHAPTER 1. DISCUSSION in this region at all, then the integrals along the two contours are in fact the same. Well, the function e−z 2 is analytic on the complex plane. Therefore, we are free to shift the contour of integration up to the real axis. That is, we can simply drop the imaginary components of the limits of integration: c (1) 2 (u) = − i 2 √ 2 α e−α 2 ∫ u α −∞ dz e−z 2 . (1.4.20) Now, we may take the u→∞ limit: c (1) 2 (∞) = − i 2 √ 2 α e−α 2 ∫ ∞ −∞ dz e−z 2 = − i 2 √ π 2 α e−α 2 . (1.4.21) Remember that c2 = λc (1) 2 +O(λ2) =⇒ c2(∞) = − i 2 √ π 2 λα e−α 2 . (1.4.22) Then, the probability to be in an excited state is Pexcited = ∞∑ n=1 |cn(∞)|2 = |c2(∞)|2 = π 8 λ2α2 e−2α2 . (1.4.23) Adiabatic approximation: At any instant in time, the Hamiltonian is an harmonic oscillator one just with a frequency which depends on the time. The instantaneous eigenstates of H (u) are |nu〉 with eigenvalue En(u) = ( 1 + λe−(u/α)2 )( n+ 1 2 ) . (1.4.24) Let us expand the state in these instantaneous eigenstates: |ψ〉 = ∞∑ s=0 cs(u) |su〉 exp [ −i ∫ u −∞ du′ Es(u ′) ] . (1.4.25) Plugging this into the Schrödinger equation gives i ∞∑ s=0 ( ċs |su〉+ cs |ṡu〉 − iEscs |su〉 ) exp [ −i ∫ u −∞ du′ Es(u ′) ] = ∞∑ s=0 csEs |su〉 exp [ −i ∫ u −∞ du′ Es(u ′) ] . (1.4.26) The last term on the LHS cancels the RHS. Then, taking the inner product with 〈nu| gives ċn(u) = − ∑ s6=n cs〈nu|ṡu〉 exp [ i(n− s) ∫ u −∞ du′ ( 1 + λe−(u′/α)2 )] , (1.4.27) where we have already assumed that the s = n term has be gauged away. Taking the u-derivative of the instantaneous time-independent Schrödinger equation gives ˙H |su〉+ H |ṡu〉 = Ės |su〉+ Es |ṡu〉 . (1.4.28) Take the inner product with 〈nu|: 〈nu| ˙H |su〉+ 〈nu|H |ṡu〉 = Ės〈nu|su〉+ Es〈nu|ṡu〉 〈nu| ˙H |su〉+ En〈nu|ṡu〉 = Ėnδn,s + Es〈nu|ṡu〉. (1.4.29) This gives 〈nu|ṡu〉 = 〈nu| ˙H |su〉 − Ėnδn,s Es − En . (1.4.30) From this expression one can see why it is important to get rid of the s = n term on the RHS of Eqn. (1.4.27): if s = n then the denominator of the above equation goes to zero! It is conceivable, of course, that the s = n term in the numerator happens to vanish as well and thus protect us from a singularity at s = n. However, in our present case, we will find that it does not. 1.4. DISCUSSION 4 57 The derivative of the Hamiltonian is ˙H = −λ u α2 e−(u/α)2 ξ2. (1.4.31) Therefore, 〈nu| ˙H |su〉 = −λ u α2 e−(u/α)2〈nu|ξ2|su〉. (1.4.32) ξ is not the appropriate position operator at time u. Instead, define the following: ξu = √ mω(u) ~ x = ( 1 + λ e−(u/α)2 )1/2 ξ. (1.4.33) Then, 〈nu| ˙H |su〉 = −λ u α2 e−(u/α)2 1 + λ e−(u/α)2 〈nu|ξ2 u|su〉 = − λu 2α2 e−(u/α)2 1 + λ e−(u/α)2 (√ n(n− 1) δs,n−2 + (2n+ 1)δs,n + √ (n+ 1)(n+ 2) δs,n+2 ) (1.4.34) Therefore, for s 6= n, s 6= n : 〈nu|ṡu〉 = − λu 2α2 e−(u/α)2( 1 + λ e−(u/α)2 )2( √ n(n− 1) s− n δs,n−2 + √ (n+ 1)(n+ 2) s− n δs,n+2 ) = λu 4α2 e−(u/α)2( 1 + λ e−(u/α)2 )2(√n(n− 1) δs,n−2 − √ (n+ 1)(n+ 2) δs,n+2 ) . (1.4.35) Plugging this into Eqn. (1.4.27) gives ċn = − λu 4α2 e−(u/α)2( 1 + λ e−(u/α)2 )2{√n(n− 1) cn−2 exp [ 2i ∫ u −∞ du′ ( 1 + λ e−(u′/α)2 )] − √ (n+ 1)(n+ 2) cn+2 exp [ −2i ∫ u −∞ du′ ( 1 + λ e−(u′/α)2 )]} . (1.4.36) This is quite a complicated set of coupled differential equations. However, as usual, we expand the coefficients in powers of λ and consider only the λ0 and λ1 equation. This will vastly simplify the complicated dependence of the RHS on λ. As usual, c(0) n = δn,0. (1.4.37) The RHS of Eqn. (1.4.36) is at least of order λ1 because of the λ sitting out front. Since we will only care about this equation to order λ1, we can actually set λ = 0 everywhere else on the RHS (other than the λ sitting out front). Thank goodness! The only coefficient involved at order λ is c (1) 2 and its equation reads ċ (1) 2 = − 1 2 √ 2 u α2 e−(u/α)2 exp [ 2i ∫ u −∞ du′ ] = − 1 2 √ 2 u α2 e−(u/α)2+2iu e2i∞. (1.4.38) Thus, c (1) 2 (u) = − 1 2 √ 2 e2i∞ ∫ u −∞ du′ u′ α2 e−(u′/α)2+2iu′ . (1.4.39) In terms of the variable, z, defined in Eqn. (1.4.18), and taking the u→∞ limit, this reads c (1) 2 (∞) = − 1 2 √ 2 e−α 2 e2i∞ ∫ ∞ −∞ dz (z + iα) e−z 2 = − i 2 √ π 2 α e−α 2 e2i∞. (1.4.40) c2 is simply λ times this. Notice that e2i∞ drops out of the square magnitude, thankfully! Thus, Pexcited = ∞∑ n=1 |cn(∞)|2 = |c2(∞)|2 = π 8 λ2α2 e−2α2 . (1.4.41) 58 CHAPTER 1. DISCUSSION 1.5 Discussion 5 Comments on Homework 4 Hund’s rules: Consider the carbon atom with atomic number 6. The electron configuration is 1s22s22p2. The filled subshells form a spherically symmetric fixed electron cloud in which the outermost electrons sit. We do not have to worry about the electrons in the filled subshells, just the outermost ones. So, we only have to consider the two outermost electrons. These are two spin-1/2 particles, so the total spin can be S = 0 or S = 1. They are also both have ` = 1 (p-subshell), so the total orbital angular momentum can be L = 0, L = 1 or L = 2. However, not all possible combinations (L, S) are possible because the two electrons are fermions and thus the total wavefunction (orbital × spin) must be antisymmetric under the interchange of the two electrons. For any number of electrons, the state with the highest S and MS or highest L and ML is always going to be symmetric because this is the state in which each electron is separately in the highest possible s and ms or ` and m`, respectively. Since the operation of switching the two electrons commutes with the lowering operator for total spin, S−, and orbital angular momentum, L−, it follows that the entire set of states with the highest possible value of S or L will be symmetric. For two electrons, this implies that the next highest value of S or L is antisymmetric. The next one down is symmetric, then the next is antisymmetric, and so on. This continues all the way down to the lowest values of S and L. Therefore, for the two-electron system of the outer shell electrons in the carbon atom, S = 1 is symmetric while S = 0 is antisymmetric. Meanwhile, L = 2 and L = 0 are symmetric while S = 1 is antisymmetric. Therefore, the only possible values for (L, S) are (L, S) = (0, 0) or (1, 1) or (2, 0). The corresponding possible values for J are J = 0 or {0, 1, 2} or 2. Therefore, the Russel-Saunders terms for carbon are 1S0 or 3P0,1,2 or 1D2. But which one is the ground state? To answer this question, we turn to Hund’s rules applied in the following order: 1. Maximize S. 2. Of the states with maximum S, maximize L. 3. If the outermost subshell is ≤ half-filled, then minimize J ; otherwise, maximize J . For carbon, the first rule limits the ground state to one of the three 3P0,1,2. There are no other possible values of L, but if there had been, then the maximum one would have been the ground state. The 2p subshell can host six electrons. Therefore, carbon’s is less than half-filled and so we must minimize J , namely J = 0. Therefore, the ground state is 3P0. We can motivate these rules: 1. This is sometimes referred to as the bus seat rule. Each ML state can host two electrons (spin up and down). However, electrons will populate different ML states before they pair up in one ML state. It used to be thought that this was so that these electrons would be further apart on average thus minimizing electron- electron repulsion. However, computer simulations indicate that the main reason is that this reduces the efficacy of screening of the nucleus and thus these outermost electrons “see” more of the nucleus than they would if they paired off. This is fine. But, in order to maximize S, the electrons in different ML states must have their spin pointing parallel with each other. Why is that? This is due to the magnetic dipole-dipole interaction between the two spins. These interactions prefer parallel magnetic dipoles as opposed to opposite ones. 2. Maximizing L tends to minimize the electron-electron repulsion energy. The planetary orbits classical model suffices to intuit this: maximizing L means having the electrons orbit in the same direction in which case they tend to spend less time in the vicinity of each other. If they were orbiting in opposite directions, then they would pass each other quite frequently. 3. This rule consider the spin-orbit coupling, which goes like J(J + 1)− L(L + 1)− S(S + 1) with a coefficient that changes from positive to negative when we go from ≤ half-filled subshell to > half-filled. Since L(L+ 1) and S(S + 1) are subtracted, it seems we would want to maximize L and S for ≤ half-filled and minimize them for > half-filled. However, the effects of the previous two rules are far greater than spin-orbit coupling (usually). 1.5. DISCUSSION 5 61 Note that we could have jumped directly to the states φ± just using symmetry. Ethylene is symmetric under reflection over perpendicular bisecting plane of the line segment between the two carbon atoms. This means that this operation commutes with the Hamiltonian and the states can be chosen to form eigenstates of this operation. This operation, being a reflection has only two eigenvalues: ±1. The symmetric combination, φ+, is the eigenstate of this reflection with eigenvalue +1 and the antisymmetric combination, φ−, has eigenvalues −1. As another example, consider the allyl free radical H H H H H C C C This is a free radical because there are two dangling covalent bonds. Eqn. (1.5.9) reads∣∣∣∣∣∣ α− E β 0 β α− E β 0 β α− E ∣∣∣∣∣∣ = (α− E) [ (α− E)2 − 2β2 ] = 0. (1.5.15) The solutions are E1 = α+ β √ 2, E2 = α, E3 = α− β √ 2. (1.5.16) One finds that the corresponding eigenstates are φ1 = 1 2 p1 + 1√ 2 p2 + 1 2 p3, φ2 = 1√ 2 ( p1 − p3 ) , φ3 = 1 2 p1 − 1√ 2 p2 + 1 2 p3. (1.5.17) Below are plots of these states from lowest energy (left) to highest energy (right). It is clear that φ1 is the state that exhibits bonding between the three carbon atoms. Calculation of Eqn. 10.123 BJ p.503, LCAO for H+ 2 : Let |A〉 and |B〉 stand for the atomic orbital 1s centered on the A hydrogen atom and B hydrogen atom, respectively. The non-normalized gerade state is |χg〉 = |A〉+ |B〉 . (1.5.18) The square amplitude of this state is 〈χg|χg〉 = 2 + 〈A|B〉+ 〈B|A〉 = 2 ( 1 + 〈A|B〉 ) , (1.5.19) where we have used the fact that these hydrogen states have real wavefunctions. The electronic Hamiltonian reads H = − ~2 2m ∇2 r − e2 4πε0 ( 1 rA + 1 rB − 1 R ) . (1.5.20) Note that the first two terms serve as the Hydrogen Hamiltonian just for the A hydrogen atom. Therefore, HAA = 〈A|H|A〉 = E1s + e2 4πε0R − e2 4πε0 〈A|r−1 B |A〉. (1.5.21) Here, E1s = − e2 8πε0a0 = −13.6 eV is the ground state energy of atomic hydrogen. By symmetry under switching A and B, we must have HBB = HAA. Likewise, HBA = HAB and this off-diagonal element is given by HAB = 〈A|H|B〉 = 〈 A ∣∣∣∣[− ~2 2m ∇2 r − e2 4πε0 ( 1 rA + 1 rB − 1 R )]∣∣∣∣B〉 = 〈 A ∣∣∣∣[E1s + e2 4πε0R − e2 4πε0rA ]∣∣∣∣B〉 = ( E1s + e2 4πε0R ) 〈A|B〉 − e2 4πε0 〈A|r−1 A |B〉. (1.5.22) 62 CHAPTER 1. DISCUSSION Therefore, Eg(R) = 〈χg|H|χg〉 〈χg|χg〉 = 2(HAA +HAB) 2 ( 1 + 〈A|B〉 ) = E1s + e2 4πε0R − e2 4πε0 〈A|r−1 B |A〉+ 〈A|r−1 A |B〉 1 + 〈A|B〉 . (1.5.23) Let us try to calculate 〈A|B〉: 〈A|B〉 = ∫ d3r 〈A|r〉〈r|B〉 = 1 πa3 0 ∫ d3r e−(rA+rB)/a0 . (1.5.24) As vectors, rA = r + 1 2R, rB = r− 1 2R. (1.5.25) Therefore, rA = √ r2 + rR cos θ + (R/2)2, rB = √ r2 − rR cos θ + (R/2)2, (1.5.26) where θ is the angle between r and R. The exponential in the integrand in Eqn. (1.5.24) suggests shifting to the variables µ = rA + rB R , ν = rA − rB R . (1.5.27) We will need the Jacobian for the change of variables. The derivatives of rA and rB are ∂rA ∂r = r + (R/2) cos θ rA , (1.5.28a) ∂rB ∂r = r − (R/2) cos θ rB , (1.5.28b) ∂rA ∂θ = −r(R/2) sin θ rA , (1.5.28c) ∂rB ∂θ = r(R/2) sin θ rB . (1.5.28d) Therefore, ∂µ ∂r = (rA + rB)r − (rA − rB)(R/2) cos θ rArBR , (1.5.29a) ∂µ ∂θ = (rA − rB)r sin θ 2rArB , (1.5.29b) ∂ν ∂r = −(rA − rB)r + (rA + rB)(R/2) cos θ rArBR , (1.5.29c) ∂ν ∂θ = − (rA + rB)r sin θ 2rArB . (1.5.29d) Therefore, the Jacobian is J = ∣∣∣∣∂rµ ∂θµ ∂rν ∂θν ∣∣∣∣ = − ( (rA + rB)r rArB )2 sin θ 2R + (r2 A − r2 B)r sin θ cos θ (2rArB)2 + ( (rA − rB)r rArB )2 sin θ 2R − (r2 A − r2 B)r sin θ cos θ (2rArB)2 = −2r2 sin θ rArBR . (1.5.30) 1.5. DISCUSSION 5 63 Note that µ2 − ν2 = 4rArB R2 . (1.5.31) Therefore, J = − 8r2 sin θ (µ2 − ν2)R3 . (1.5.32) Note that µ2 ≥ ν2 Therefore, It follows that dµ dν dφ = |J | dr dθ dφ = 8r2 sin θ (µ2 − ν2)R3 dr dθ dφ, (1.5.33) or d3r = r2 sin θ dr dθ dφ = R3 8 (µ2 − ν2)dµ dν dφ. (1.5.34) Now, we just need the limits of integration. The minimum possible value for rA + rB is obviously R. The maximum is ∞. Therefore, 1 ≤ µ ≤ ∞. The minimum rA − rB is −R and the maximum is R. Therefore, −1 ≤ ν ≤ 1. Therefore, Eqn. (1.5.24) may be written as 〈A|B〉 = R3 8πa3 0 ∫ ∞ 1 dµ ∫ 1 −1 dν ∫ 2π 0 dφ (µ2 − ν2) e−(R/a0)µ = R3 2a3 0 ∫ ∞ 1 dµ ( µ2 − 1 3 ) e−(R/a0)µ = − R 2 2a2 0 µ2 e−(R/a0)µ ∣∣∣∣∞ 1 + R2 a2 0 ∫ ∞ 1 dµµ e−(R/a0)µ + R2 6a2 0 e−(R/a0)µ ∣∣∣∣∞ 1 = R2 2a2 0 e−R/a0 − R a0 µ e−(R/a0)µ ∣∣∣∣∞ 1 + R a0 ∫ ∞ 1 dµ e−(R/a0)µ − R2 6a2 0 e−R/a0 = R2 3a2 0 e−R/a0 + R a0 e−R/a0 + e−R/a0 . (1.5.35) Define x = R/a0. (1.5.36) Then, 〈A|B〉 = ( 1 + x+ 1 3x 2 ) e−x. (1.5.37) Note that this nicely reproduces the complicated denominator in Eqn. 10.123a BJ p.503. Next we calculate 〈A|r−1 B |A〉 = 1 πa3 0 ∫ d3r e−2rA/a0 rB . (1.5.38) Note that rA = R 2 (µ+ ν), rB = R 2 (µ− ν). (1.5.39) Therefore, 〈A|r−1 B |A〉 = R3 8πa3 0 ∫ ∞ 1 dµ ∫ 1 −1 dν ∫ 2π 0 dφ (µ2 − ν2) e−R(µ+ν)/a0 R(µ− ν)/2 = R2 2a3 0 ∫ ∞ 1 dµ ∫ 1 −1 dν (µ+ ν) e−(R/a0)(µ+ν) = R2 2a3 0 [ a0 R ( −µ e−(R/a0)µ ∣∣∣∣∞ 1 + ∫ ∞ 1 dµ e−(R/a0)µ )( −a0 R e−(R/a0)ν ∣∣∣∣1 −1 ) + ( −a0 R e−(R/a0)µ ∣∣∣∣∞ 1 ) a R0 ( −νe−(R/a0)ν ∣∣∣∣1 −1 + ∫ 1 −1 dν e−(R/a0)ν )] = 1 2a0 [( 1 + a0 R ) e−R/a0 ( eR/a0 − e−R/a0 ) + e−R/a0 ( −eR/a0 − e−R/a0 + a0 R eR/a0 − a0 R e−R/a0 )] (1.5.40) 66 CHAPTER 1. DISCUSSION 1.7 Discussion 7 Classical Hall Effect: In 1879, Edwin Hall discovered what would later be called the Hall Effect. Suppose we have a box-shaped conductor. We pass a current along the length of the conductor and apply a magnetic field along the width. Hall discovered that an electric potential difference builds up along the height of the conductor as well as a current in that direction if leads are attached along that direction that allow the mobile charges to move out of the bulk conductor. This behavior is explained by the Lorentz force on a charge moving in a magnetic field. Suppose a charge, e, is moving in the x-direction with speed V (i.e., q is part of the current that is flowing in the x-direction), and let the magnetic field point in the z-direction, Bz. Then, the Lorentz force on the charge is F = ev×B = evb(−ŷ). (1.7.1) That is, in addition to moving along the length of the conductor, the charge is forced to the bottom face of the conductor by the magnetic field. This creates a buildup of charges on the bottom face and a deficit on the top and thus builds up a potential difference between the top and bottom faces. Of course, the stronger is this buildup the more new charges are repelled by the charges on the bottom face. At some point, the system reaches equilibrium and the new charges flowing in the x-direction are no longer deflected up or down. The voltage difference in the y-direction, called the Hall voltage VH , will clearly increase linearly with the magnetic field since equilibrium is reached when the magnetic and electric forces in the y-direction cancel out. Each for is linearly proportional to the magnetic and electric fields, respectively, and the voltage is linearly proportional to the electric field. Therefore, VH ∝ Bz. The Hall resistance, RH , is defined as RH = VH/I, where I is the current that we are feeding into the conductor in the x-direction. (Caution: many texts use RH for the Hall coefficient, which is related, but not the same as the Hall resistance). RH is a calculable quantity derived from VH , which is a measurable quantity, and I, which is a controlled quantity, and thus measurable. The take-away message is that we should expect that RH be linearly proportional to the magnetic field, RH ∝ B . (1.7.2) In fact, one use of the Hall effect is as a device for measuring magnetic fields very accurately. If we know I and the dimensions of the conductor, then we can work out the current density, j. If we know the charge carrier density, n, then we can work out the speed of the charges in that current via j = nev. Given v and the Hall voltage that we measure using a sensitive voltmeter, we can work backwards to determine the magnetic field since it is simply proportional to the voltage. Integer Quantum Hall Effect: In 1980, Klaus von Klitzing et. al. discovered that, at very large magnetic fields or extremely low temperatures, RH was not linearly proportional to B. Instead, it exhibits plateaus and goes up in steps. This is shown in the diagram below, taken from the 1985 Nobel prize in physics lecture (the award was given to von Klitzing that year). The graph rising in steps is ρxy, which is the Hall resistance RH . 1.7. DISCUSSION 7 67 Amazingly, it was found that the value of RH at these plateaus were given by RH = h νe2 , (1.7.3) where ν is a positive integer. Furthermore, the longitudinal resistance, RL, which is the usual resistance along the direction of current, vanishes at the Hall plateaus and peaks sharply at the tradition steps. Note that the standard resistance of the conductor does not vanish. We are not making the usual approximation that the resistance of a wire is basically zero. That is because its resistance is usually much smaller than anything else in the circuit. Here, this conductor clearly has a nonzero resistance at zero magnetic field. Thus, having vanishing longitudinal resistance at certain values of the magnetic field is quite remarkable. It is as if the system acts like a superconductor in the longitudinal direction but for a completely different resin; there are no Cooper pairs condensing here! Why the steps in Hall resistance?: You can guess that the quantization of the Hall resistance is probably related to the quantization of the states of the system into Landau levels. For this system, it is convenient to use the gauge which is translationally invariant in the y-direction. This is because we want to describe the system at a Hall plateau where the current is purely in the y-direction. In Homework 7 problem 2, you will solve the system in the gauge which is translationally invariant in the x-direction, but other than that, the results are the same. The take-away message is that the state of the system is a plane wave of wavevector k in the y-direction and a harmonic oscillator state in the x-direction centered at x = ~/eB (note: the homework problem sets c = 1; you can work out how to reintroduce c by dimensional analysis). Of course, the system is actually finite. Impose periodic boundary conditions in the y-direction. Impose Dirichlet boundary conditions in the x-direction (infinite potential outside the sample). This forces the center of the harmonic oscillator in the x-direction, ~k/eB, to lie between x = 0 and x = Lx, where Lx is the length of the sample in the x-direction: 0 ≤ k ≤ eBLx/~ or 0 ≤ s ≤ eBLxLy/h. The density of states, which is the number of staes per unit area, is simply nB = eB h = B Φ0 , (1.7.4) where Φ0 = h/e is the fundamental magnetic flux quantum. That is, this is the degeneracy per unit area of each Landau level. If n is the density of charges in the system, then ν = n nB = nh eB (1.7.5) is the number of distinct Landau levels that are occupied. The assumption here is taht the system is cold enough so that low Landau levels are fully occupied before any higher Landau level is occupied. Furthermore, the charges are assumed to be fermions so that no two charges can occupy the same state; of course, they can have different spin, which we have not been counting. Now, if you were to solve the force equation, mẍ = eẋ×B + eE, with E = Ex̂ and B = Bẑ, then you find x = X + r cos ( eBt m ) , y = Y − r sin ( eBt m ) − E B t, (1.7.6) where X, Y and r are constants. Notice that there is the expected rotational motion. However, on top of that, there is constant speed drift in the y-direction; this is the Hall current! The current density is j = nev = neE B . (1.7.7) Therefore, the conductivity is σ = j E = ne B , (1.7.8) and the resistivity is ρ = 1 σ = B ne = h νe2 . (1.7.9) As mentioned earlier, the Hall resistivity and resistances are the same. Indeed, this coincides with Eqn. (1.7.3). However, we have not explained why ν must be an integer. Indeed, as we raise the magnetic field, B, the Landau 68 CHAPTER 1. DISCUSSION levels separate and go up in energy. Eventually, one will be pushed up beyond the Fermi energy. Then, all of the sudden, there are fewer Landau levels to carry current and the resistance goes up. That is where the steps in the Hall resistance are supposed to come from. However, as B increases, the degeneracy of each Landau level also increases. So all that really happens is that the lower Landau levels get occupied more and more. There is basi- cally stil the same number of current-carrying charges; they occupy fewer, but more degenerate, Landau levels. In truth, what actually happens is that the levels are not perfectly discrete and separated. There will always be some spread due to impurities in the system and interactions among the levels. There are states in between the sharp Landau levels, but they are localized and unable to carry bulk current (e.g., states orbiting impurities). The current is really only carried by what are called edge states. Basically, hardly any current is carreid through the bulk and essentially all of ti moves along the edges of the sample. Only the extended Landau levels can go all the way to the edge and contribute edge states. Therefore, there really is only one current-carrying channel per occupied Landau level. Thsu, when a level is pushed up above the Fermi energy by increasing the magnetic fiel sufficiently, then the number of current-carrying channels is decreased by one and the Hall resistance goes up a step. Shubnikov-de-Haas oscilations: The Fermi energy is in fact not constant throughout the sample. There are small variations from point to point and there is a macroscopic difference from the left to the right end, which we apply as a voltage to run the current in the first place. As for the longitudinal resistance, only the states near the Fermi energy can be expected to travel across the sample. If these states are localized, then they will not be able to move towards the regions of lower local Fermi energy because the characteristic length scale for the variation of the Fermi energy will be too large compared to the localization length scale of the state. Thus, they cannot move to a lower energy state in order to dissipate energy. That is, the longitudinal resistance vanishes. However, right when a Landau level crosses the local Fermi energy, these extended staes can flow along the length of the sample. Thus, rigth at these points, we get a sudden peak in longitudinal resistance. This may be confusing: when we have localized states that cannot carry current, we get zero resistance, and when all of a sudden the system can carry longitudinal current, the resistance goes up? Isn’t this the wrong way round? The answer is no. Resistance is all about whether or not the system can dissipate energy away, usually in the form of heat. It usually means that a higher resistance for a given voltage means a lower current. But that’s the usual consequence, it is not the microscopic definition. Indeed, it does not hold in this case. The fractional occupation, f , of the states above the highest Landau level below the Fermi energy, EF , is simply given by f ∼ EF ~ω − [ EF ~ω ] , (1.7.10) where [x] is the integer part of x, and ω = eB/m = 2ωL, where ωL is the Larmore frequency. The exact form is not important. What really matters is that this is periodic in 1/B (since f ∼ 1/B). Therefore, we expect the longitudinal resistance to be periodic in 1/B. These are called Shubnikov-de-Haas oscillations. You can see them in the low-B ringing of the graph of ρxx in the previous plots. 1.8. DISCUSSION 8 71 All types of resonant behaviors are observed in all sorts of different systems. For example, the Higgs was discovered as a very tiny peak over a very large background in the scattering of two protons at the LHC with a total center of momentum mass of about 125 GeV (which is thus interpreted as the Higgs mass). Meanwhile, in the scattering of electrons by noble gases, Ramsauer and Townsend observed a resonance of the mostly-dipping kind (the third pair of diagrams in the above set). 72 CHAPTER 1. DISCUSSION 1.9 Discussion 9 Møller Scattering: Møller scattering is the scattering of two electrons. Let us suppose that the electrons interact just via a Coulomb potential: V (r) = α~c r , (1.9.1) where α = e2 4πε0~c is the usual fine-structure constant. The exact scattering amplitude for this potential is f(θ) = − γ 2k sin2 θ 2 Γ(1 + iγ) Γ(1− iγ) e−iγ log sin2 θ 2 , (1.9.2) where γ = 2mα~c/~2 = 2αmc2/~c. If you want to see how this is derived, see section 7.9 of Weinberg’s Lecture on Quantum Mechanics. The upshot is that the solutions involve hypergeometric functions and their asymptotic forms as r →∞, so it’s nice to know, but it’s not very instructive. In lecture, the factor inolving the Euler gamma function was repeatedly referred to as a phase. That means that its magnitude is 1. We can see this from the integral definition of the gamma function: Γ(t) = ∫ ∞ 0 dxxt−1 e−x =⇒ Γ(1± iγ) = ∫ ∞ 0 dxx±iγ e−x. (1.9.3) Therefore, Γ(1− iγ) = [Γ(1 + iγ)]∗ =⇒ ∣∣∣∣Γ(1 + iγ) Γ(1− iγ) ∣∣∣∣ = 1. (1.9.4) Since this factor is just a phase that is independent of the angles, we will simply drop it henceforth. The other phase, which does involve the angle cannot be dropped, however, because we will need to add amplitudes at different angles before squaring and that phase will make a difference! Since the two electrons are identical fermions, we must combine their wavefunctions in a way that is totally antisymmetric under the interchange of the two electrons. The wavefunction is a product of a purely spatial piece and a purely spin-dependent piece: Φ(r1, r2;σ1, σ2) = Ψ(r1, r2)X(σ1, σ2). (1.9.5) We have two options: (a) Ψ is symmetric and X is antisymmetric; and (b) Ψ is antisymmetric and X is symmetric. There is one state of type (a). The total spin of this state is S = 0. Φ+(1, 2) = ψ1(1)ψ2(2) + ψ1(2)ψ2(1)√ 2 × α(1)β(2)− α(2)β(1)√ 2 . (1.9.6) There are three states of type (b). The total spin of these states is S = 1. Φ−(1, 2) = ψ1(1)ψ2(2)− ψ1(2)ψ2(1)√ 2 ×  α(1)α(2), MS = 1, 1√ 2 [ α(1)β(2) + α(2)β(1) ] , MS = 0, β(1)β(2), MS = −1. (1.9.7) In the state of type (a), the total scattering amplitude is the symmetric one. In the states of type (b), it is the antisymmetric one. Note that we have dropped the Euler gamma functions as previously discussed. f±(θ) = f(θ)± f(π − θ) = − γ 2k ( e−iγ log sin2 θ 2 sin2 θ 2 ± e−iγ log cos2 θ 2 cos2 θ 2 ) . (1.9.8) The differential scattering cross-sections are( dσ dΩ ) ± = |f±|2 = ( γ 2k )2[ 1 sin4 θ 2 + 1 cos4 θ 2 ± 2 cos ( 2γ log tan θ 2 ) sin2 θ 2 cos2 θ 2 ] . (1.9.9) 1.9. DISCUSSION 9 73 If the incoming electrons are completely unpolarized, then they have a probability of 1 4 to be in the state Φ+ and 3 4 to be in any of the states Φ−. In this case, the total unpolarized differential scattering cross-section is given by the weighted average: dσ dΩ = 1 4 ( dσ dΩ ) + + 3 4 ( dσ dΩ ) − = ( γ 2k )2[ 1 sin4 θ 2 + 1 cos4 θ 2 − 2 cos ( 2γ log tan θ 2 ) sin2 θ 2 cos2 θ 2 ] . (1.9.10) Below is a graph of this. We see the customary divergence in the forward direction at θ = 0. However, since these are identical particles and one is moving forward and the other backward, the graph should be even around θ = π/2 and so there should be a divergence in the backward scattering as well, at θ = π. Furthermore, we observe the osillations characteristic of interference between identical particles, that would be absent of distinguishable ones. 76 CHAPTER 1. DISCUSSION We can solve for the a’s:∫ d3x e−ipxφ(x) = ∫ d3x e−ipx ∫ d̄ 3k√ 2Ek [ ake ikx + a†ke −ikx] = ∫ d3k√ 2Ek ∫ d̄ 3x [ ake i(k−p)x + a†ke −i(k+p)x ] = ∫ d3k√ 2Ek [ ake i(Ep−Ek)tδ(k− p) + a†ke i(Ep+Ek)tδ(k + p) ] = 1√ 2Ep [ ap + a†−pe 2iEpt ] . (1.10.22) Similarly, we find all the following relations: ap + a†−pe 2iEpt = √ 2Ep ∫ d3x e−ipx φ(x), (1.10.23a) ap − a†−pe2iEpt = i √ 2 Ep ∫ d3x e−ipx π(x), (1.10.23b) a−pe −2iEpt + a†p = √ 2Ep ∫ d3x eipx φ(x), (1.10.23c) a−pe −2iEpt − a†p = i √ 2 Ep ∫ d3x eipx (x), . (1.10.23d) Therefore, ap = i ∫ d3x√ 2Ep e−ipx [ π(x)− iEpφ(x) ] , (1.10.24a) a†p = −i ∫ d3x√ 2Ep eipx [ π(x) + iEpφ(x) ] . (1.10.24b) So far, the a’s have been arbitrary functions. Now, let us impose canonical quantization. In quantum mechanics, this amounts to promoting x and p to operators and imposing the canonical commutation relation [x, p] = i. Here, we promote φ and π to operators and impose the equal-time commutation relations [φ(t,x), φ(t,y)] = [π(t,x), π(t,y)] = 0, [φ(t,x), π(t,y)] = iδ(x− y). (1.10.25) These imply [ap, a † k] = ∫ d3x d3y 2 √ EpEk ei(ky−px) ( [π(x), π(y)] + EpEk[φ(x), φ(y)] + iEk[π(x), φ(y)]− iEp[φ(x), π(y)] ) = ∫ d3x d3y 2 √ EpEk ei(ky−px)(Ek + Ep)δ(x− y) = Ek + Ep 2 √ EpEk ∫ d3x ei(k−p)x = Ek + Ep 2 √ EpEk ei(Ep−Ek)t (2π)3δ(p− k) = (2π)3δ(p− k). (1.10.26) Note that the humongous pre-factor in the penultimate line is simply equal to 1 when p = k, which is imposed by the delta function. You can check that the commutators of two a’s and two a†’s vanish. Therefore, we can interpret these things as creation and annihilation operators! Chapter 2 Homework x aq 78 CHAPTER 2. HOMEWORK 2.1 Homework 1 2.1.1 BJ 8.2 p.427 Derive Eqn. (8.18) p.378 for the third order correction E (3) n : E(3) n = 〈ψ(1) n |H ′ − E(1) n |ψ(1) n 〉 − 2E(2) n 〈ψ(0) n |ψ(1) n 〉. (2.1.1) SOLUTION: Outline: • Since λ << 1, Taylor expand the Schrödinger equation in powers of λ. • Isolate the pieces proportional to one and the same power of λ to form equations at various orders of λ. • Multiply the O(λ3) equation by 〈ψ(0) n | to isolate E (3) n . The resulting equation, call it E , involves |ψ(2) n 〉. • Multiply the lower order equations (or their adjoints) by whatever is necessary to produce terms that will cancel the terms in the equation E that involve |ψ(2) n 〉. At this point, I don’t yet know what exactly what to multiply with, but I know that the lower order equations are all I have to work with. Expand the time-independent Schrödinger equations (tiSEq) in powers of λ as usual: (H0 + λH ′) ( |ψ(0) n 〉+ λ|ψ(1) n 〉+ λ2|ψ(2) n 〉+ λ3|ψ(3) n 〉+ · · · ) = ( E(0) n + λE(1) n + λ2E(2) n + λ3E(3) n + · · · ) × ( |ψ(0) n 〉+ λ|ψ(1) n 〉+ λ2|ψ(2) n 〉+ λ3|ψ(3) n 〉+ · · · ) . The O(λ0) equation does not tell us anything new: H0|ψ(0) n 〉 = E(0) n |ψ(0) n 〉. (3.1.2a) The O(λ1) equation reads H0|ψ(1) n 〉+H ′|ψ(0) n 〉 = E(0) n |ψ(1) n 〉+ E(1) n |ψ(0) n 〉. (3.1.2b) The O(λ2) equation reads H0|ψ(2) n 〉+H ′|ψ(1) n 〉 = E(0) n |ψ(2) n 〉+ E(1) n |ψ(1) n 〉+ E(2) n |ψ(0) n 〉. (3.1.2c) The O(λ3) equation reads H0|ψ(3) n 〉+H ′|ψ(2) n 〉 = E(0) n |ψ(3) n 〉+ E(1) n |ψ(2) n 〉+ E(2) n |ψ(1) n 〉+ E(3) n |ψ(0) n 〉. (3.1.2d) Hermiticity of H implies 〈ψ(0) n |H0 = E(0) n 〈ψ(0) n | (2.1.3) Therefore, the first terms on either side of the inner product of Eqns. (3.1.2a) to (3.1.2d) with 〈ψ(0) n | cancel each other. Using the orthonormality condition 〈ψ(0) m |ψ(0) n 〉 = δmn, the last term on the right hand side (RHS) of the inner product of the O(λj) equation with 〈ψ(0) n | is E (j) n . Therefore, Eqn. (3.1.2d) becomes 〈ψ(0) n |H ′|ψ(2) n 〉 = E(1) n 〈ψ(0) n |ψ(2) n 〉+ E(2) n 〈ψ(0) n |ψ(1) n 〉+ E(3) n . Reorganizing this equation gives something similar to, but not quite, what we want: E(3) n = 〈ψ(0) n |H ′ − E(1) n |ψ(2) n 〉 − E(2) n 〈ψ(0) n |ψ(1) n 〉. (2.1.4) While this is correct, it is not as useful as Eqn. (2.1.1) because it involves ψ (2) n , whereas Eqn. (2.1.1) involves at most ψ (1) n . Clearly, we would like to replace the first term on the RHS with something else. We have gotten all the mileage we can get out of Eqn. (3.1.2d). We could take the inner product with respect to a different state other than 〈ψ(0) n |, but that would not isolate E (3) n and hence would not be helpful. We could consider higher order 2.1. HOMEWORK 1 81 Therefore, (x3)kn = ( ~ 2mω )3/2(√ (n+ 1)(n+ 2)(n+ 3) δk,n+3 + 3(n+ 1)3/2δk,n+1 + 3n3/2δk,n−1 + √ n(n− 1)(n− 2) δk,n−3 ) . Finally, x4 |n〉 = ( ~ 2mω )3/2(√ (n+ 1)(n+ 2)(n+ 3)x |n+ 3〉+ 3(n+ 1)3/2x |n+ 1〉 + 3n3/2x |n− 1〉+ √ n(n− 1)(n− 2)x |n− 3〉 ) = ( ~ 2mω )2(√ (n+ 1)(n+ 2)(n+ 3)(n+ 4) |n+ 4〉 + [√ (n+ 1)(n+ 2) (n+ 3) + 3(n+ 1)3/2 √ n+ 2 ] |n+ 2〉+ 3 [ (n+ 1)2 + n2 ] |n〉 + [ 3n3/2 √ m− 1 + √ n(n− 1) (n− 2) ] |n− 2〉+ √ n(n− 1)(n− 2)(n− 3) |n− 4〉 ) = ( ~ 2mω )2(√ (n+ 1)(n+ 2)(n+ 3)(n+ 4) |n+ 4〉 + 2 √ (n+ 1)(n+ 2) (2n+ 3) |n+ 2〉+ 3(2n2 + 2n+ 1) |m〉 + 2 √ n(n− 1) (2n− 1) |m− 2〉+ √ n(n− 1)(n− 2)(n− 3) |n− 4〉 ) . (2.1.21) Therefore, (x4)kn = ( ~ 2mω )2(√ (n+ 1)(n+ 2)(n+ 3)(n+ 4) δk,n+4 + 2 √ (n+ 1)(n+ 2) (2n+ 3)δk,n+2 + 3(2n2 + 2n+ 1)δk,n + 2 √ n(n− 1) (2n− 1)δk,n−2 + √ n(n− 1)(n− 2)(n− 3) δk,n−4 ) . 2.1.3 BJ 8.8 p.427 Derive Eqn. (8.56) for E (2) n in the perturbed harmonic oscillator potential, H = H0 + ax3 + bx4: E(2) n = −1 8 a2 ~ω ( ~ mω )3 (30n2 + 30n+ 11)− 1 8 b2 ~ω ( ~ mω )4 (34n3 + 51n2 + 59n+ 21). (2.1.22) SOLUTION: The expression for E (2) n reads E(2) n = ∑ k 6=n |a(x3)kn + b(x4)kn|2 E (0) n − E(0) k (2.1.23) It is clear that there are no cross-terms, i.e. terms ∼ ab, since there is no value of k for which (x3)kn and (x4)kn are simultaneously non-vanishing. This is corroborated by the final answer, (2.1.22), since it contains no ab terms, only a2 and b2 terms. Note that E (0) n − E(0) k = ~ω(n− k). Plugging in the results of the previous problem yields E(2) n = a2 ~ω ( ~ 2mω )3( (n+ 1)(n+ 2)(n+ 3) n− (n+ 3) + 9(n+ 1)3 n− (n+ 1) + 9n3 n− (n− 1) + n(n− 1)(n− 2) n− (n− 3) ) + b2 ~ω ( ~ 2mω )4( (n+ 1)(n+ 2)(n+ 3)(n+ 4) n− (n+ 4) + 4(n+ 1)(n+ 2)(2n+ 3)2 n− (n+ 2) + 4n(n− 1)(2n− 1)2 n− (n− 2) + n(n− 1)(n− 2)(n− 3) n− (n− 4) ) (2.1.24) The rest is algebra. This simplifies to (2.1.22). 82 CHAPTER 2. HOMEWORK 2.1.4 Homework 1 Comments • Problem sets go through drafts just like essays do. What you turn in should not be the first draft. • When I was an undergrad, it was mandatory to write a brief outline of ones solution to a problem (no algebra) at the beginning of the solution. This is very important: it forces you to identify the main concepts involved in the problem rather than focusing on the algebraic steps. It forces you to put your thoughts in some logical order and think about why you are doing something other than that it works. Why would you think of using some equation or another? What is the motivation? It is not enough to simply say that it will solve the problem. How do you know that your plan will solve the problem before you actually enact it? Of course, you can exercise a degree of discretion: for example, I did not feel an outline for problem 3 was necessary since it was pretty much all algebra. (See Using qualitative problem-solving strategies to highlight the role of conceptual knowledge in solving problems by Leonard, Dufresne and Mestre in Am. J. Phys. 64 (12), December 1996). • N.S. stands for “non-sequitur”: Please do not just write a whole bunch of equations some of which do not follow from the previous ones with no comment. It is extremely difficult to follow your work in this case. • Note that 〈ψ|H|χ〉 = ( 〈χ|H†|ψ〉 )∗ , where the asterisk denotes complex conjugation. Therefore, if H is Hermitian, then 〈ψ|H|χ〉 = ( 〈χ|H|ψ〉 )∗ . It is not necessarily true that 〈ψ|H|χ〉 = 〈χ|H|ψ〉. • Hermiticity of H does not imply that 〈ψ|H|χ〉 is real; it only implies that the eigenvalues of H are real, namely 〈ψ|H|ψ〉 is real if |ψ〉 is an eigenstate of H. As an example, consider the momentum operator, p, for a one-dimensional harmonic oscillator. Write p = iα(a+ − a−), where α = ( m~ω 2 )1/2 . Then, the matrix element, 〈n+ 1|p|n〉 is iα √ n+ 1, which is imaginary, not real. For a second example, consider the y-component of the spin operator. • The comments about the reality of 〈ψ(0) n |ψ(1) n 〉 in problem 1 are sufficiently crucial to the solution that it is in the body of the solution. Please make sure you understand those comments. 2.2. HOMEWORK 2 83 2.2 Homework 2 2.2.1 BJ 8.6 p.427 Consider a one-dimensional linear harmonic oscillator perturbed by a Gaussian perturbation H ′ = λ e−ax 2 . Cal- culate the first-order correction to the ground-state energy and the energy of the first excited state. SOLUTION: Outline: • States of the one-dimensional harmonic oscillator are non-degenerate, so we can use non-degenerate per- turbation theory. • The wavefunctions of the ground state and first excited state are known: the ground state is a Gaussian and the first excited state is a Gaussian multiplied by x. • The first order energy correction is just the expectation value of the perturbation in the zeroth order state. • We will clearly need some Gaussian integrals: ∫∞ −∞ dx e−x 2 = √ π and ∫∞ −∞ dx, x2 e−x 2 = √ π/2. The states of a one-dimensional harmonic oscillator are non-degenerate. Hence, we can use non-degenerate perturbation theory. Let ψ (0) n (x) be the wave function of the nth state of the unperturbed one-dimensional harmonic oscillator. You can probably jump immediately to the expression E(1) n = ∫ ∞ −∞ dx ( ψ(0) n (x) )∗ λ e−ax 2 ψ(0) n (x). (2.2.1) However, let us see how to get here from the bra-ket notation. First, we must introduce a new basis of states, called the position basis, |x〉. This is an eigenstate of the position operator with eigenvalue x. Let us reintroduce the hat notation: x̂ |x〉 = x |x〉 . (2.2.2) This is the state whose wavefunction is a Dirac delta function at x. This basis is orthonormal: 〈x′|x〉 = δ(x− x′) (2.2.3) and complete: ∫ ∞ −∞ dx |x〉 〈x| = 1, (2.2.4) where 1 is the identity operator. Therefore, |ψ〉 = ∫ ∞ −∞ dx |x〉 〈x|ψ〉. (2.2.5) Since 〈x|ψ〉 is the coefficient of |x〉 in the rewriting of |ψ〉 in the position basis, we recognize this as the wavefunction associated with the state |ψ〉: ψ(x) = 〈x|ψ〉. (2.2.6) Then, the first order correction equation reads E(1) n = 〈ψ(0) n |H ′|ψ(0) n 〉. We can insert two copies of the identity in the form of (2.2.3) between the states and H ′: E(1) n = ∫ ∞ −∞ dx′ ∫ ∞ −∞ dx 〈ψ(0) n |x′〉︸ ︷︷ ︸ (ψ (0) n (x′))∗ 〈x′|H ′|x〉 〈x|ψ(0) n 〉︸ ︷︷ ︸ ψ (0) n (x) . It is easy to act H ′ on |x〉 since |x〉 is an eigenstate of the position operator. Again, it is helpful to reintroduce the hats. Remember that H ′ is an operator and the x in it is also an operator. Thus, Ĥ ′ |x〉 = λe−ax̂ 2 |x〉 = λe−ax 2 |x〉 . (2.2.7) 86 CHAPTER 2. HOMEWORK As we do for the energies, we put superscripts (0) on these bras and kets, (0) 〈· · ·| and |· · ·〉(0) , because they are the states of the unperturbed Hamiltonian. As in Eqn. (2.1.14), from homework 1, we have x = ( ~ 2mω )1/2 (a+ + a−), y = ( ~ 2mω )1/2 (b+ + b−). (2.2.25) Therefore, H ′ = λ~ 2mω (a+ + a−)(b+ + b−). (2.2.26) Therefore, H ′ can raise or lower both nx and ny, or it can raise one of them and lower the other. Therefore, the expectation value of H ′ within any one state is automatically zero. Since the ground state is non-degenerate, this immediately implies that it does not get perturbed: E (1) 0,0 = (0)〈0, 0|H ′|0, 0〉(0) = 0 . (2.2.27) It also implies that (0)〈1, 0|H ′|1, 0〉(0) = (0)〈0, 1|H ′|0, 1〉(0) = 0. (2.2.28) However, we find (0)〈1, 0|H ′|0, 1〉(0) = (0)〈0, 1|H ′|1, 0〉(0) = λ~ 2mω . (2.2.29) Therefore, in the two-dimensional subspace of the first excited states, the expectation values of H ′ forms the 2× 2 matrix H ′ = λ~ 2mω ( 0 1 1 0 ) . (2.2.30) This is easy to diagonalize. Simply rewrite it in the basis |1〉± = 1√ 2 ( |1, 0〉(0) ± |0, 1〉(0)) . (2.2.31) In this basis, H ′ reads H ′ = λ~ 2mω ( 1 0 0 −1 ) . (2.2.32) Therefore, the two degenerate first excited states are split: E (1) 1,± = ± λ~ 2mω . (2.2.33) 2.2.4 Homework 2 Comments • On the whole, your homework 2 solutions were clearer and easier to follow than your homework 1 solutions. Thank you. • When we write H = H0 +λH ′, the λ factor is just meant to keep track of the order of some other parameter in H ′, which is presumed to be small in some sense. I say “in some sense” because the parameter in H ′ that is supposed to be small will be dimensionful and so it can’t just be small; it must be small compared to something else. In any case, this is why we set λ = 1 at the end of the perturbation calculation: λ is just a proxy for that other small parameter, which is the one we actually care about. In other words, what we really care about is H = H0 +H ′, not H = H0 + λH ′. However, in actual practical problems, we often include the small parameter λ into H ′ and often λ is dimensionful. For example, when we say H ′ = λ e−ax 2 , then λ has units of energy, or H ′ = λxy, in which case λ has units of energy per area (energy density in two dimensions). If you set λ = 1 in these cases, then you are admitting that you are measuring energy in units of λ in the first case, and energy density in units of λ in the second case. That is fine as long as you are aware of it, but it is rather unnatural. Why would you measure things in units of the perturbation? It seems more natural to measure things in terms of unperturbed quantities. 2.2. HOMEWORK 2 87 For example, in the last problem ~ω is a natural unit to use for energy, and ~/mω is a natural unit to use for area because these are intrinsic to the unperturbed system. Then, a good unit for energy per area would be mω2. In this way, it actually makes sense to write Eqn. (2.2.33) as E (1) 1,± = ±1 2 ( λ mω2 ) ~ω. (2.2.34) This is the exact same expression, but it makes clear that you are measuring energy in units of ~ω and it also clarifies what it means for λ to be small: λ is a two-dimensional energy density and it should be small compared to mω2. Furthermore, this dimensional analysis allows us to conclude that the jth order correction to an energy will always take the form E(j) = c(j) ( λ mω2 )j ~ω, (2.2.35) where c(j) is a constant, which may, of course, be different for different energy levels. For the first excited states, c(1) happens to be ± 1 2 . This is a very powerful statement: all the calculation in the third problem goes into determining this numerical prefactor. The rest of the answer you could have stated without doing any calculation! For the first problem, we have two parameters: λ and a. It is natural to remove their dimensions by defining λ̃ = λ/~ω and ã = a~/mω. Now, we have a similar expression as in Eqn. (2.2.35) except that λ is now measured in units of ~ω and c(j) can be an arbitrary function of ã: E(j) = c(j)(ã)λ̃j~ω. (2.2.36) So, all the calculation in that problem goes into determining this function of ã. For example, for the ground state, c(1)(ã) = (1 + ã)−1/2. • For the third problem, you can essentially guess the appropriate states, (2.2.31), in which to evaluate the first-order perturbation to the first excited state just by graphing things. You can even tell which one will have a slightly higher energy than the unperturbed first excited energy and which one will have a slightly lower energy. Figure 2.1 shows the countour plots of the first excited states |1, 0〉 and |0, 1〉. We know that this basis is not well-suited for the perturbation, H ′, because these states are not symmetric under the interchange x ↔ y or x ↔ −y (reflection over the diagonal axes). States that are symmetric or anti-symmetric under these reflections would be preferable because H ′ has this symmetry. The states in Figure 2.1 are symmetric under reflections over the central horizontal and vertical axes, but H ′ does not have this symmetry, as can be seen in Figure 2.2. (a) (nx, ny) = (1, 0) (b) (nx, ny) = (0, 1) Figure 2.1: Contour plot of the first excited state wavefunctions. Darker colors bulge into the page and lighter ones bulge out of the page. 88 CHAPTER 2. HOMEWORK Figure 2.2: Contour plot of H ′ = λxy. Darker colors dip below the page and lighter ones rise above the page. (a) Symmetric state. (b) Anti-symmetric state. Figure 2.3: Contour plot of the symmetric and anti-symmetric combinations of the first excited states. The symmetric and anti-symmetric combinations of the states |1, 0〉 and |0, 1〉 given in Eqn. (2.2.31) are plotted in Figure 2.3 and we see that these are symmetric or anti-symmetric under reflection over the diagonal axes. We can tell that the anti-symmetric state will have a lower energy and the symmetric state will have a higher energy than the unperturbed first excited energy. This is because the peaks and valleys of the anti-symmetric state lie along the x = −y axis, which is where the perturbing energy, H ′, is negative. Similarly, the peaks and valleys of the symmetric state lie along the x = y axis, which is where H ′ is positive. 2.3. HOMEWORK 3 91 • Use matching conditions and impose normalizability of the wavefunction to derive the Wilson-Sommerfield quantization condition. • Solve for E. Let E be a number such that −V0 < E < 0 so that it might be possible for E to be the energy of a bound state. The classical turning points, when V (x) = E are x1 = − ( 1 + E V0 ) a, x2 = ( 1 + E V0 ) a, (2.3.16) where x1 is a left-handed turning point and x2 is a right-handed one. Let region I be the region to the left of x1, region II be between x1 and x2, and region III to be to the right of x2. Of course, in reality, we should avoid a small region around the turning points themselves. Suppose that the wavefunction in region II is ψII(x) = 1√ p(x) sin [ 1 ~ ∫ x x1 p(x′) dx′ − π 4 + δ ] , (2.3.17) where p(x) = √ 2m(E − V (x)) (2.3.18) and δ is some phase. Then, the matching conditions imply that the asymptotic form of the solution in region I far away from the turning point x1 is ψI(x) x→−∞−−−−−→ − cos δ√ |p(x)| exp [ 1 ~ ∫ x1 x |p(x′)| dx′ ] . (2.3.19) Normalizability of the wavefunction requires that ψI(x) x→−∞−−−−−→ 0, which implies cos δ = 0 =⇒ δ = ( s+ 1 2 ) π, (2.3.20) where s is an arbitrary integer. Rewrite Eqn. (2.3.17) as ψII(x) = − 1√ p(x) sin [ 1 ~ ∫ x2 x p(x′) dx′ − π 4 + δ′ ] , (2.3.21) where δ′ = π 2 − δ − 1 ~ ∫ x2 x1 p(x) dx (2.3.22) The matching conditions now imply that the asymptotic solution in the region III far away from x2 is ψIII(x) x→∞−−−−→ cos δ′√ |p(x)| exp [ 1 ~ ∫ x x2 |p(x′)|dx′ ] , (2.3.23) which again implies that cos δ′ = 0 =⇒ δ′ = ( s′ + 1 2 ) π. (2.3.24) Hence, 1 ~ ∫ x2 x1 p(x) dx = π 2 − δ − δ′ = − ( s+ s′ + 1 2 ) π = ( n+ 1 2 ) π, (2.3.25) where n is again some integer, related to s and s′ via n = −(s+ s′ + 1). (2.3.26) In this form, we know that n = 0, 1, 2, . . . because the integral on the left must be non-negative. 92 CHAPTER 2. HOMEWORK Let us calculate the integral: 1 ~ ∫ x2 x1 p(x) dx = √ 2m ~ ∫ (1+ E V0 )a −(1+ E V0 )a dx √ E + V0 ( 1− |x|a ) = 2 √ 2mV0 ~ ∫ (1+ E V0 )a 0 dx √ E V0 + 1− x a . (2.3.27) Define the rescaled integration variable ξ = x/a. (2.3.28) Then, 1 ~ ∫ x2 x1 p(x) dx = 2a √ 2mV0 ~ ∫ 1+ E V0 0 dx √ 1 + E V0 − ξ = 2a √ 2mV0 ~ [ −2 3 ( 1 + E V0 − ξ )3/2]1+ E V0 0 = 4a √ 2mV0 3~ ( 1 + E V0 )3/2 . (2.3.29) We set this equal to ( n+ 1 2 ) π and solve for E, the solution of which is called En: En = [( 9 ( n+ 1 2 )2 π2~2 32mV0a2 )1/3 − 1 ] V0 = [(( n+ 1 2 ) 3π 40 )2/3 − 1 ] V0. (2.3.30) Now, we just plug in some values for n. Eventually, En given above will be positive and hence not a bound state energy. This happens for n ≥ 4. Therefore, there are four bound states with energies approximately given by E0 = −0.76V0 E1 = −0.50V0 E2 = −0.30V0 E3 = −0.12V0. (2.3.31) 2.3.3 Homework 3 Comments • Most people did not compare the estimate for the hydrogen ground state energy in problem 1 to the actual hydrogen ground state. If you had plugged in numbers, you would have noticed if your answer is sensible (or even has the correct units). • Similarly, plugging in numbers in the second problem allows you to determine how many bound states there are as well as their energies. Once again, one would easily rule out unreasonable answers such as energies below −V0 or above 0, or answers that do not have units of energy. 2.4. HOMEWORK 4 93 2.4 Homework 4 2.4.1 BJ 9.2 p.466 A particle of charge q, in simple harmonic motion along the x-axis, is acted on by a time-dependent homogeneous electric field E(t) = E0e−(t/τ)2 , where E0 and τ are constants. If the oscillators is in its ground state at t = −∞, find the probability that it will be found in an excited state as t→∞. SOLUTION: Outline: • Determine the electric potential energy of the charge in the presence of the electric field as V = qΦ, where Φ is the electric potential related to the electric field via E = −∇Φ. • Expand the state |ψ〉 of the system in terms of the eigenstates of the unperturbed harmonic oscillator, |n〉. • Plug this expansion into the Schrödinger equation to get coupled first order differential equations for the expansion coefficients, cn. • Expand these coefficients, cn, in powers of the appropriate small constant prefactor in the perturbing Hamiltonian. The initial condition gives the zeroth order term in this expansion: c (0) n = δn,0. Solve for each term in the expansion order by order; we will only need to go to first order. The full time-dependent Hamiltonian is H(t) = − ~2 2m d2 dx2 + 1 2 mω2x2︸ ︷︷ ︸ H0 +−qE0xe−(t/τ)2︸ ︷︷ ︸ H′(t) . (2.4.1) The eigenstates of H0 are |n〉 with energy E (0) n = ( n + 1 2 ) ~ω. These form a complete orthonormal basis and so we can expand the solution of the time-dependent Hamiltonian above in them: |ψ〉 = ∞∑ n=0 cn(t) e−iωnt |n〉 , (2.4.2) where ωn = En/~ = ( n+ 1 2 ) ω. The time-dependent Schrödinger equation reads E . i~ ∂ ∂t |ψ〉 = H(t) |ψ〉 = (H0 +H ′(t)) |ψ〉 . Get rid of the dimensions of time and space by defining the variables u ≡ ωt, ξ = √ mω ~ x. (2.4.3) For convenience, also define the dimensionless quantities ε = qE0√ 2m~ω3 , α = ωτ. (2.4.4) These are the two dimensionless versions of the two free parameters in H ′, namely the combination qE0 and the time constant, τ . The √ 2 in the denominator of ε is for future convenience. In terms of these dimensionless variables, H ′ reads H ′(t) ~ω = − √ 2 εξe−(u/α)2 . (2.4.5)