Solutions - Quantum Mechanics August 2020 First Examination, Exams of Quantum Mechanics

Download Solutions - Quantum Mechanics August 2020 First Examination and more Exams Quantum Mechanics in PDF only on Docsity! Qualifying Exam Quantum Mechanics – SOLUTIONS August 2020 Solve one of the two problems in Part A, and one of the two problems in Part B. Each problem is worth 50 points. Part A Problem 1. Let us consider an electron whose squared orbital angular momentum L2 is measured to be 6h̄2. 1.a) [5 points] For each one of the two bases described above, list all the possible states for the electron which are compatible with this measurement. We have s = 1 2 and l = 2 (since l(l + 1) = 6). Since l + s ≤ j ≤ |l − s|, the allowed values for j will be j = 5 2 , j = 3 2 . As a consequence, there will be ten possible states in the basis {|jmj〉}: j = 5 2 with mj = 5 2 , 3 2 , 1 2 ,−1 2 ,−3 2 ,−5 2 (six states) and j = 3 2 with mj = 3 2 , 1 2 ,−1 2 ,−3 2 (four states). In the basis {|mlms〉}, we can have all possible combinations of ms = −1 2 , 1 2 with ml = 2, 1, 0,−1,−2, which is again ten states. 1.b) [15 points] Find the expression of the following basis states {|jmj〉} in terms of the appropriate elements of the basis {|mlms〉}. 1.b.1 : First state on the ladder, here is only one possible way of satisfying mj = ms + ml, therefore |j = 5 2 mj = 5 2 〉 = |ml = 2 ms = 1 2 〉 1.b.2 : We should apply the lowering operator on both sides to obtain |j = 5 2 mj = 3 2 〉 = √ 4 5 |ml = 1 ms = 1 2 〉+ √ 1 5 |ml = 2 ms = −1 2 〉 and then apply the orthogonality condition to get |j = 3 2 mj = 3 2 〉 = √ 1 5 |ml = 1 ms = 1 2 〉 − √ 4 5 |ml = 2 ms = −1 2 〉 1.b.3 : Not a possible state. We cannot have j = 1 2 if s = 1 2 and l = 2. 1 Solutions - Quantum Mechanics August 2020 First Examination Ph.D. Program in Physics - The Graduate Center CUNY 1.b.4 : We can continue to apply the lowering operator until we reach the minimum of mj, or start from the bottom of the ladder (which is faster). The only combination of ms and ml that allows for mj = −5 2 is |j = 5 2 mj = −5 2 〉 = |ml = −2 ms = −1 2 〉 1.c) [15 points] Now assume the electron to be in the state with j = 3 2 , and mj = 1 2 (and the value of l is the same as in the previous questions). If one measures the z-components of the electron orbital angular momentum and spin, what are the possible values and their probabilities? In our problem, there are only two configurations of ms and ml that are compatible with |j = 3 2 , and mj = 1 2 , namely the states |ml = 1 ms = −1 2 〉 and |ml = 0 ms = 1 2 〉. To obtain the probabilities, we start from the expression that we obtained in question 1.b.2), and we apply once more the lowering operator, to obtain: |j = 3 2 mj = 1 2 〉 = √ 2 5 |ml = 0 ms = 1 2 〉 − √ 3 5 |ml = 1 ms = −1 2 〉 which allows to read the probabilities 2 5 and 3 5 , respectively for the pair of values (ml = 0,ms = 1 2 ) and (ml = 1,ms = −1 2 ). 1.d) [15 points] Let us now assume that the electron is in the state with ml = 1 and ms = −1 2 (again, l did not change). What are the possible values of j and their probabilities? Here we have the “inverse” problem as 1.c). The state |ml = 1ms = −1 2 〉 must be a linear combination of the states |j = 3 2 mj = 1 2 〉 (which we know already) and |j = 5 2 mj = 1 2 〉 (which we do not have yet). Let us first find |j = 5 2 mj = 1 2 〉. We can obtain it by applying the lowering operator to |j = 5 2 mj = 3 2 〉 or simply by using the orthogonality with |j = 3 2 mj = 1 2 〉. In either way, we get: |j = 5 2 mj = 1 2 〉 = √ 3 5 |ml = 0 ms = 1 2 〉+ √ 2 5 |ml = 1 ms = −1 2 〉 We can solve now for |ml = 1 ms = −1 2 〉, to get |ml = 1 ms = −1 2 〉 = √ 2 5 |j = 5 2 mj = 1 2 〉 − √ 3 5 |j = 3 2 mj = 1 2 〉 . Therefor, we will have j = 5 2 with probability 2 5 and j = 3 2 with probability 3 5 . 2 Solutions - Quantum Mechanics August 2020 First Examination Ph.D. Program in Physics - The Graduate Center CUNY Formulas for Part A Schrödinger Equation: ih̄∂ψ(x,t) ∂t = Ĥψ(x, t) , Ĥ = p̂2 2m + V (x) , ĤψE(x) = EψE(x) , j(x) = Re { −ih̄ m ψ∗(x)∂ψ(x) ∂x } Angular Momentum Operators: [Ĵx, Ĵy] = ih̄Ĵz (or [Ĵi, Ĵj ] = ih̄εijkĴk) , [Ĵ2, Ĵi] = 0 , Ĵ± = Ĵx ± iĴy , Ĵ±|j m〉 = h̄ √ j(j + 1)−m(m± 1) |j m± 1〉 5 Solutions - Quantum Mechanics August 2020 First Examination Ph.D. Program in Physics - The Graduate Center CUNY Problems for QMII, 2020 CUNY Graduate Center Li Ge P.1 The WKB approximation is valuable in a wide range of systems that resemble the Schrödinger equation. Consider the Helmholtz equation in a one-dimensional optical cavity:[ d2 dx2 + ε(x)k2 ] ψ(x) = 0. (1) Here ε(x) is the real-valued dielectric function and its frequency dependence is neglected. k = ω/c is the wave vector in free-space, where ω is the circular frequency and c is the speed of light in vacuum. (I) Assume for now that k is real. Establish one connection between Eq. (1) and the time-independent Schrödinger equation. Point out what quatities play the roles of the mass, the potential and the energy. Note that there are multiple acceptable answers and take ~ = 1. Answer 1: The energy E is zero. −ε(x)k2 is the potential (Note: the minus sign is required). m = 0.5. Answer 2: The energy E is zero. −ε(x) is the potential. m = 0.5k2. Answer 3: The energy E is k2. (1− ε(x))k2 is the potential. m = 0.5. . . . Here we consider a cavity of a finite size, i.e., x ∈ [−L/2, L/2], and the refractive index n(x) = √ ε(x) is position-dependent inside the cavity but is a constant (ne > 0) outside. Now we focus on the states that are defined by a solution of Eq. (1) with the incoming boundary condition ψ(x, t) ∝ e−inek|x| in |x| > L/2. They correspond to the zeros of the scattering matrix. (II) Write the boundary condition in terms of ψ′/ψ at x = ±L/2 where ψ′ = dψ/dx. Answer: ψ′/ψ = −inek at x = L/2 and inek at x = −L/2. (III) Because of the incoming boundary condition, the system is no longer Hermitian and k is complex in general. Can you recover the WKB result learned in class using ψ(x) = A(x)eiφ(x)? Here A(x), φ(x) are real. Answer: ψ′ = (A′ + iAφ′)eiφ, ′′ = (A′′ + 2iA′φ′ + iAφ′′ −A(φ′)2)eikφ = −εk2Aeikφ By dropping A′′, we find i(2A′φ′ +Aφ′′) = A((φ′)2 − εk2). Denoting k2 = f2 + ig (f, g ∈ R), we separate the real and imaginary part of the equation above: 2A′φ′ +Aφ′′ = −Aεg, 0 = (φ′)2 − εf2. The second equation above gives φ′ = ± √ εf2 ≡ ±n(x)f , similar to what we learned in class. However, the first equation needs more attention. We rewrite it as (ln y)′ = y′ y = −gφ ′ f2 , where y = A2φ′L is dimensionless. We then find y ∝ e−gφ/f2 and A(x) ∝ 1√ n(x) e−gφ/2f 2 → (x) ∝ 1√ n(x) ei(1+ig/2f 2)φ = 1√ n(x) e±i(f+ig/2f) ∫ dxn(x) Note that if we denote k = kr + iki and assume 0 < ki  kr, we then find f ≈ kr and g = 2krki. Therefore, the factor in the exponent is approximatedly k. If a student is unable to derive the result above, he has the P.1 continued on next page. . . Page 1 of 4 Solutions - Quantum Mechanics August 2020 First Examination Ph.D. Program in Physics - The Graduate Center CUNY Problems for QMII, 2020 CUNY Graduate Center Li Ge option to start with a simpler ansatz in the next question. (IV) How about using ψ(x) = A(x)eikφ(x)? If it works, apply the WKB approximation using A′′ ≈ 0. Note that the result depends on whether Re[k] vanishes. Answer: ψ′ = (A′ + ikAφ′)eikφ, ′′ = (A′′ + 2ikA′φ′ + ikAφ′′ −A(kφ′)2)eikφ = −εk2Aeikφ By dropping A′′, we find i(2A′φ′ +Aφ′′) = Ak((φ′)2 − ε). If k were real, then the derivation that follows would be exactly the same as in the Schrödinger equation. The imaginary part of k makes the result slightly different. Denoting again k = kr + iki (kr,i > 0), we separate the real and imaginary part of the equation above: 2A′φ′ +Aφ′′ = Aki((φ ′)2 − ε), 0 = kr((φ ′)2 − ε). If kr 6= 0, the second equation tells us that φ′ = ± √ ε = ±n(x), and the first equation indicates Aφ′ = const. The wave function is then given by ψ(x) ∝ 1√ n(x) eik ∫ n(x)dx, which is again the same as in the Schrödinger equation. However, there exists zeros of the scattering matrix with kr = 0 for some ε(x), and we find (φ′)2 6= ε. As a result, the right hand side of 2A′φ′+Aφ′′ = Aki((φ ′)2−ε) is not zero either. In other words, in this case the phase and amplitude of the wave function are not simply related. (V) Assume ε(x) = ε(−x), write down the wave function in the WKB approximation as sine and cosine functions. What is the argument of these sinusoidal functions? Answer: ψ+(x) = 1√ n(x) cos ( k ∫ x 0 n(x)dx ) and ψ−(x) = 1√ n(x) sin ( k ∫ x 0 n(x)dx ) , one for even-parity and the other for odd-parity. Check: ψ−(−x) = 1√ n(x) sin ( k ∫ −x 0 n(x)dx ) = 1√ n(x) sin ( −k ∫ x 0 n(−x)dx ) = −ψ−(x). We have used x→ −x in the second step and n(x) = n(−x) in the last step. Similarly, we find ψ+(−x) = ψ+(x). (VI) Using the boundary condition at either x = L/2 or −L/2, derive the analytical expressions satisfied by k = kr + iki (kr,i > 0), one for even-parity and the other for odd-parity. Assuming ki  kr, analyze these express to show that they lead to the same form k ≈ 1 n̄L [ q + i ln 2q |α| ] , (2) when |α|  q. Specify the expressions for α, q, and the average refractive index n̄ inside the cavity. Answer: Using the boundary condition at x = L/2, we find ψ′− ψ− ∣∣∣∣ x=L/2 = √ ne[k cos(z/2)− n′ 2n2 e sin(z/2)] 1√ ne sin(z/2) = −inek, or tan(z/2) = z −α+ iz , P.1 continued on next page. . . Page 2 of 4 Solutions - Quantum Mechanics August 2020 First Examination Ph.D. Program in Physics - The Graduate Center CUNY Jan.2021-Pb.1 Monday, January 4, 2021 10:03 AM QM Fall 2020 Page 1 Solutions - Quantum Mechanics January 2021 First Examination Ph.D. Program in Physics - The Graduate Center CUNY QM Fall 2020 Page 2 Solutions - Quantum Mechanics January 2021 First Examination Ph.D. Program in Physics - The Graduate Center CUNY QM Fall 2020 Page 3 Solutions - Quantum Mechanics January 2021 First Examination Ph.D. Program in Physics - The Graduate Center CUNY QM Fall 2020 Page 3 Solutions - Quantum Mechanics January 2021 First Examination Ph.D. Program in Physics - The Graduate Center CUNY QM Fall 2020 Page 4 Solutions - Quantum Mechanics January 2021 First Examination Ph.D. Program in Physics - The Graduate Center CUNY ^ŽůƵƚŝŽŶƐĨŽƌWĂƌƚ tĞĚŶĞƐĚĂLJ͕:ĂŶƵĂƌLJϲ͕ϮϬϮϭ ϭϮ͗ϭϭD Solutions - Quantum Mechanics January 2021 First Examination Ph.D. Program in Physics - The Graduate Center CUNY Qualifying Exam Quantum Mechanics June 2021 Problem 1. Consider two observables  and B̂ in a three-dimensional Hilbert space. In the basis: |1〉 = 1 0 0  , |2〉 = 0 1 0  , |3〉 = 0 0 1  , the observables  and B̂ are represented, respectively, by the matrices A→ a 1 0 0 0 2 0 0 0 6  , B → b −1 0 0 0 0 2 0 2 0  where b << a. a) [5 points] Show that the observables  and B̂ are not compatible. Check the commutator [Â, B̂]. Since  and B̂ do not commute, the observables are NOT compatible. in particular,  and B̂ do not admit a common set of eigenstates. b) [10 points] Compute the possible outcomes and the corresponding probabilities of separate (independent) measurements of  and B̂ in the state |χ〉 = √ 1 2 (|2〉 − |3〉). Â: in general, we have three possible outcomes in the eigenvalues a1 = 1a, a2 = 2a, a3 = 6a, which correspond to the three eigenstates |1〉, |2〉, and |3〉. The probabilities are given by Pi = |〈χ|i〉|2, where i = 1, 2, 3. Since |χ〉 = √ 1 2 (|2〉 − |3〉), we will measure a2 = 2a with a probability of 50% and a3 = 6a with a probability of 50%. B̂: the possible outcomes are given by the eigenvalues. Let us firts compute them together with the corresponding eigenstates |B1〉, |B2〉, and |B3〉. The probabilities will be given by Pi = |〈χ|Bi〉|2, where i = 1, 2, 3. The first eigenvalue is b1 = −1b, and |B1〉 = |1〉. By diagonalizing the sub-matrix ( 0 2 2 0 ) , we find that the additional eigenvalues are b2 = −2b and b3 = +2b, with respective eigenvectors |B2〉 = √ 1 2 (|2〉 − |3〉) and |B3〉 √ 1 2 (|2〉+ |3〉). Since |χ〉 = |B2〉, we will measure b2 = −2b with a probability of 100%. c) [10 points] Compute the possible outcomes and the corresponding probabilities of a mea- surements of B̂ that follows a measurement of Â, if the system is initially in the state |χ〉. What about a measurement of  that follows a measurements of B̂? After the measurement of Â, the state of the system will be in |2〉 or |3〉 with a probability of 50% each. Noting that |2〉 = √ 1 2 (|B2〉+ |B3〉) and |3〉 = √ 1 2 (|B2〉 − |B3〉), we can conclude 1 Quantum Mechanics - Solutions First Examination - June 2021 Ph.D. Program in Physics The Graduate Center CUNY that the measurement of B̂ will have a 50% probability of b2 = −2b a 50% probability of b3 = +2b. Note that, if we were to perform the measurement of B̂ first, it would give a b2 = −2b with 100% probability and turn the state of the system in — B2〉. However, since |χ〉 = |B2〉, the subsequent measurement of  would give the same outcomes as in part b) of the problem. Let us now construct the Hamiltonian Ĥ =  + B̂. Since b << a, we can use perturbation theory to study Ĥ. d) [10 points] After writing down the eigenstates and eigenvalues of Ĥ0 = Â, compute the first- and second-order corrections to the energy levels due to the correction Ĥp = B̂. Leading Order. (just for defining notation). In the basis defined by |1〉 = 1 0 0  , |2〉 = 0 1 0  , |3〉 = 0 0 1  , we have E0 1 = 1a , E0 2 = 2a , E0 3 = 6a. First Order. We use E1 n = 〈n|Ĥp|n〉, to obtain E1 1 = 〈1|Ĥp|1〉 = −1b, E1 2 = 〈2|Ĥp|2〉 = 0, E1 3 = 〈3|Ĥp|3〉 = 0. Second Order. Using E2 n = ∑ k 6=n |〈k|Ĥp|n〉|2 E0 n−E0 k , we get E2 1 = 0, E2 2 = − b2 a , E1 3 = + b2 a . Therefore, up to second order E1 = a− b+ . . . = a [1−X +O(X3)] E2 = 2a− b2 a + . . . = a [2−X2 +O(X3)] E3 = 6a+ b2 a + . . . = a [6 +X2 +O(X3)] where X = b/a. e) [10 points] Compute the energy eigenstates of Ĥ up to first order in perturbation theory. |p11〉 = 0 |p12〉 = ∑ k 6=2 〈k|Ĥp|2〉 E0 2−E0 k |k〉 = 〈3|Ĥp|2〉 E0 2−E0 3 |3〉 = 2b −4a |3〉 = − b 2a |3〉 |p13〉 = ∑ k 6=3 〈k|Ĥp|3〉 E0 3−E0 k |k〉 = 〈2|Ĥp|3〉 E0 3−E0 2 |2〉 = 2b 4a |2〉 = b 2a |2〉 Therefore, up to first order: |p1〉 = |1〉 |p2〉 = |2〉 − b 2a |3〉+ . . . |p3〉 = |3〉+ b 2a |2〉+ . . . f) [5 points] After finding the exact solutions, check that your results for the energy levels obtained in perturbation theory are correct. By diagonalizing the matrix Ĥ = a 1−X x 0 x 2 2X 0 2X 6  , 2 Quantum Mechanics - Solutions First Examination - June 2021 Ph.D. Program in Physics The Graduate Center CUNY we get the eigenvalues λ1 = (1−X) and λ2,3 = 4∓ 2 √ 1 + x2. By expanding λ2,3 for small X, i.e. √ 1 +X2 = 1 + 1 2 X2 +O(X4), we obtain: λ2 = 4− 2(1 + 1 2 X2) = 2−X2, and λ3 = 4 + 2(1 + 1 2 X2) = 6 +X2, in agreement with part a). Time-independent Perturbation Theory: E1 n = 〈n|Ĥp|n〉; Ei n = 〈p0n|Ĥp|pi−1n 〉; |p1n〉 = ∑ k 6=n 〈k|Ĥp|n〉 E0 n−E0 k |k〉 3 Quantum Mechanics - Solutions First Examination - June 2021 Ph.D. Program in Physics The Graduate Center CUNY Some useful definitions and formulas: Legendre polynomials Pl(z): (1− z2) d2 dz2 Pl(x)−2z dPl(z) dz + l(l +1)Pl(z) = 0, ∫ 1 −1 Pk(z)Pl(z)dz = 2δkl 2l +1 , P0(z) = 1, P1(z) = z, P2(z) = 1 2 (3z2−1), P3(z) = 1 2 (5z3−1), . . . . Spherical Bessel functions jl(ρ): eiρz = ∞ ∑ l=0 (2l +1)il jl(ρ)Pl(z), 1 ρ2 d dρ [ ρ 2 d jl(ρ) dρ ] + [ 1− l(l +1) ρ2 ] jl(ρ) = 0, jl(ρ) = (−ρ)l ( 1 ρ d dρ )l sinρ ρ −→ 2ll! (2l +1)! ρ l, as ρ → 0, jl(ρ)−→ 1 ρ cos ( ρ− πl 2 ) , as ρ → ∞. The partial wave expansion for positive-energy eigenfunctions. with~k = kẑ: ψk(r,θ)−→ eikr cosθ + f (θ) r eikr, as r→ ∞, f (θ) = 1 2ik ∞ ∑ l=0 (2l +1)(e2iδl −1)Pl(cosθ). 3 Quantum Mechanics - Solutions First Examination - June 2021 Ph.D. Program in Physics The Graduate Center CUNY Qualifying Exam Quantum Mechanics August 2021 Solutions Problem 1. Consider a spin-1 particle in the state: |ψ〉 = √ 1 3 ( |+〉+ |0〉+ |−〉 ) , where |+〉 ≡ |s = 1, sz = 1〉, |0〉 ≡ |s = 1, sz = 0〉, and |−〉 ≡ |s = 1, sz = −1〉 are the eigenstates of Ŝz. 1.a) (5 points) Show that the Ŝz operator can be written as Ŝz = h̄(|+〉〈+|) − h̄(|−〉〈−|). Write the expression for the operator Ŝ2 z . Textbook problem. The simplest way is to use the spectral decomposition  = ∑ i |ai〉λi〈ai|, where |ai〉 and λi are the eigenvectors and eigenvalues of the operator Â, respectively. By squaring Ŝz, we obtain Ŝ2 z = h̄2|+〉〈+|+ |−〉〈−|. In matrix form (if need be) this would be represented by Ŝz → h̄ 1 0 0 0 0 0 0 0 −1  , Ŝ2 z → h̄2 1 0 0 0 0 0 0 0 1  . 1.b) (15 points) Beginning with the state |ψ〉, consider measuring first Ŝ2 z , then measuring afterwards Ŝz. What are the possible measurement outcomes of Ŝ2 z and the corresponding probabilities? What are the possible outcomes of a measurement of Ŝz after Ŝ2 z has been measured and the corresponding probabilities? First measurement (Ŝ2 z ): The possible outcomes are given by the eigenvalues of Ŝ2 z , which are 0 (non degenerate) and h̄2 (degenerate of degree 2) respectively. We will measure S2 z = 0 with probability P0 = |〈ψ|0〉|2 = 1 3 . We will measure S2 z = h̄2 with probability P+ = |〈ψ|1〉|2 = 2 3 , where |1〉 = √ 1 2 ( |+〉 + |−〉 ) is the eigenstate of eigenvalue h̄2 and the state of the system after the measurement. Note that the from of |1〉 can be obtained directly (by observing the form of the operator S2 z and the initial state of the system) or by applying the projector P = |+〉〈+|+ |−〉〈−| to the initial state of the system and normalizing the outcome. Second measurement (Ŝz): If the outcome of the first measurement was S2 z = 0, the system is now in the state |0 >. We will then measure Sz = 0 with probability P0 = |〈0|0〉|2 = 1. If the outcome of the first measurement was S2 z = 1, the system is now in the state |1〉 = √ 1 2 ( |+〉+ |−〉 ) . We will then measure Sz = h̄ with probability P+ = |〈1|+〉|2 = 1 2 and Sz = −h̄ with probability P− = |〈1|−〉|2 = 1 2 . Assume the Hamiltonian of the system to be Ĥ = AŜ2 z +B(Ŝ2 x − Ŝ2 y) . 1 PhD Program in Physics - The Graduate Center CUNY Graduate Physics Qualifying Exam Solutions, CUNY Graduate Center, August, 2021 1.a. (10 points) The time derivatives of the components of the position vector are dr j dt = 1 ih̄ [H,r j] = 1 ih̄ [−ih̄c~α ·~∇,r j] =−cα j, or d~r dt =−c~α. (Note: this illustrates the phenomenon of Zitterbewegung, which is that the expectation value of the velocity of the massive electron has magnitude c. The electron’s path fluctuates, result- ing in an average velocity < c.) The time derivatives of the components of the momentum vector are d p j dt = 1 ih̄ [H, p j] = 1 ih̄ [ −ec~α ·~A(~r, t)+ eφ(~r, t) , −ih̄ ∂ ∂ r j ] =−ec ∂ ∂ r j ( ~α ·~A ) + e ∂φ ∂ r j , so the velocity operator is d~p dt =−ec~∇ ( ~α ·~A ) + e~∇φ . 1.b. (10 points) The time derivative of ~p− e~A = m~v (the force on the electron) is ~F = d dt (~p− e~A) = d~p dt − e ( 1 ih̄ [H,~A]+ ∂~A ∂ t ) =−ec~∇(~α ·~A)+ e~∇φ + ec(α ·~∇)~A− e ∂~A ∂ t . (Had you found this time derivative of ~p−e~A above, you would have received full credit for Part b., but we remark that using the formulas for the electric and magnetic fields respectively, ~E =−~∇φ + ∂~A ∂ t , ~B = ~∇×~A, it may be written as ~F =−e(~E + c~α ×~B), the Lorentz force of a particle with velocity~v = c~α . ) We now have d~L dt = d~r dt × (~p− e~A)+~r× d dt (~p− e~A) =−c~α × (~p− e~A)+~r×~F , 1 PhD Program in Physics - The Graduate Center CUNY Qualifying Exam Quantum Mechanics June 2022 Solutions Problem 1. Let’s consider a perturbation Hp to a one-dimensional harmonic oscillator. The Hamiltonian of the system can be written as Ĥ = Ĥ0 + Ĥp where Ĥ0 = p̂2 2m + 1 2 mω2x̂2 and Ĥp = βx̂4 Define the operators â and ↠as â = 1√ 2mωh̄ (p̂− imωx̂) ↠= 1√ 2mωh̄ (p̂+ imωx̂) 1.a) Using the properties of the ladder operators â and â†, find the expression for all the non-vanishing matrix elements 〈m|x̂2|n〉, where |n〉 and |m〉 are eigenstates of the unperturbed Hamiltonian Ĥ0 [15 points]. We should evaluate all non-zero matrix elements 〈m|x̂2|n〉, namely the cases m = −2, 0,+2. All other matrix elements are zero. Since x̂ = i √ h̄ 2mω ( ↠− â ) , then x̂2 = − h̄ 2mω ( ↠− â ) ( ↠− â ) . Using â|n〉 = √ n|n− 1〉 and â†|n〉 = √ n+ 1|n+ 1〉, we get: 〈n|x̂2|n〉 = h̄ 2mω (2n+ 1) 〈n|x̂2|n− 2〉 = − h̄ 2mω √ n(n− 1) 〈n|x̂2|n+ 2〉 = − h̄ 2mω √ (n+ 2)(n+ 1) 1.b) Evaluate the expectation value 〈n|x̂4|n〉 [15 points]. Hint: You can simplify this calculation by using the completeness relation for the energy eigenstates and the result of part 1.a. We can compute directly 〈n|x̂4|n〉, starting from the expression x̂4 = h̄2 4m2ω2 ( ↠− â )4 or use the hint provided in the text. We will follow the latter route. Using completeness, we can write: 〈n|x̂4|n〉 = ∑ m〈n|x̂2|m〉〈m|x̂2|n〉 = ∑ m ∣∣〈n|x̂2|m〉 ∣∣2 Using the expressions from part 1.b, we finally get: 〈n|x̂4|n〉 = h̄2 4m2ω2 [ (2n+ 1)2 + n(n− 1) + (n+ 1)(n+ 2) ] = h̄2 4m2ω2 [ (6n2 + 6n+ 3 ] . 1.c) Write the energy spectrum for the Hamiltonian Ĥ up to first order in perturbation theory [10 points]. The spectrum En will be given by En = E0 n +E1 n + . . . where E0 n = h̄ω ( n+ 1 2 ) (Harmonic oscillator) and E1 n = 〈n|Ĥp|n〉 = β〈n|x̂4|n〉 = βh̄2 4m2ω2 [ (6n2 + 6n+ 3 ] . Thus, En = h̄ω ( n+ 1 2 ) + βh̄2 4m2ω2 [ (6n2 + 6n+ 3 ] + . . . 1.d) (5 points) Evaluate the commutator [Ĥ0, Ĥp]. The evaluation the commutator [Ĥ0, Ĥp] comes down to the evaluation of [p̂2, x̂4] = −4ih(p̂x̂3 + x̂3p̂). 1 Solutions Quantum Mechanics Qualifying Examination - June 2022 PhD Program in Physics CUNY 1.e) (5 points) Given the result for the commutator in part 1.d), should we expect higher-order corrections to the energy spectrum of H? (Explain your reasoning). Since their [p̂2, x̂4] is not equal to zero, we do not have a complete basis in which the two operators Ĥ0 and Ĥp are diagonal at the same time. In particular, the basis of eigenstates of Ĥ0 is not a basis of eigenstates for the perturbation Ĥp (or the full Hamiltonian Ĥ. As a consequences, we can expect higher order corrections to the energy levels, which should be computed in perturbation theory. Some useful definitions and formulas: One-dimensional Harmonic oscillator: Ĥ0 = p̂2 2m + 1 2 mω2x̂2, â = 1√ 2mωh̄ (p̂− imωx̂), ↠= 1√ 2mωh̄ (p̂ + imωx̂), [â, â†] = 1, [Ĥ0, â †] = h̄ωâ†, â|n〉 = √ n|n− 1〉, â†|n〉 = √ n + 1|n + 1〉 Time-independent Perturbation Theory: E1 n = 〈n|Ĥp|n〉, Ei n = 〈p0 n|Ĥp|pi−1 n 〉, |p1 n〉 = ∑ k 6=n 〈k|Ĥp|n〉 E0 n−E0 k |k〉, E2 n = ∑ k 6=n |〈k|Ĥp|n〉|2 E0 n−E0 k . 2 Solutions Quantum Mechanics Qualifying Examination - June 2022 PhD Program in Physics CUNY Qualifying Exam Solutions, Quantum Mechanics, June 8, 2022 2. Consider the attractive central potential V (~r) = −Ω/r2 for a particle of mass m, with Ω > 0. 2.a. (20 points) Write the differential equation for χl(r) = Rl(r)r, where Rl is the radial part of the eigenfunction (see the mathematical formulas below). This equation should only contain the radial variable r, not the angular variables θ , φ . Note that Rl(r) must be finite at the origin r = 0, which gives a boundary condition on χl(r) at r = 0. Furthermore, Rl(r) may depend on additional quantum numbers (which are not essential to know for this exercise). Solution: The spherical harmonic Ylm(θ ,φ) is an eigenfunction of ~L2 with eigenvalue h̄2l(l + 1). Multiplying the eigenvalue equation by 2m/h̄2, after sub- stituting h̄2l(l +1) for~L2, we find χl(r) satisfies the radial equation d2χl(r) dr2 + [ ε−u(r)− l(l +1) r2 ] χl(r) = 0, For the central potential u(r) =−L(L+1)/r2, the radial equation is d2χl(r) dr2 + [ ε + L(L+1)− l(l +1) r2 ] χl(r) = 0, 2.b. (10 points) Next consider the scattering problem. Show that for the special value Ω = h̄2L(L+1) 2m , L being a positive integer, the phase shift of the l = L partial wave, is δL = Lπ/2. Do not forget the boundary condition at r = 0. Solution: Recall that ψ(r,θ) = ∞ ∑ l=0 (2l +1)il r χl(r)Pl(cosθ), where χl(r) satisfies the radial equation with ε = k2. d2χl(r) dr2 + [ k2 + L(L+1)− l(l +1) r2 ] χl(r) = 0, with u(r) = 2mV (r)/h̄2. For l = L, we have d2χL(r) dr2 + k2χL(r) = 0. This must sat- isfy the boundary condition that χL(0) = 0. Thus χL(r) = AL sinkr = AL sin(kr− Lπ 2 +δL) and δL = Lπ/2 (not zero, except for L = 0). 1 Solutions Quantum Mechanics Qualifying Examination - June 2022 PhD Program in Physics CUNY