Understanding the Uncertainty Principle in Quantum Mechanics, Study notes of Physics

Download Understanding the Uncertainty Principle in Quantum Mechanics and more Study notes Physics in PDF only on Docsity! 5 Probability and Uncertainty Principle 5.1 Probability Interpretation of ψ Wave-particle “duality”: particles are supposed to be found where we expect from interference pattern that ψ is large. e.g.: Davisson-Germer experiment: Max Born: probability that particle described by ψ(r, t) found at time t in volume elt. δV at r is δP = |ψ(r, t)|2δV (1) Basic requirements for probability function: 1. Positive semidefinite. δP ≥ 0 since ψψ∗ ≥ 0 always. 2. Normalized. ∫ dP = 1. Particle must be somewhere! Wave func- tion is therefore normalized to 1:∫ d3r|ψ(r, t)|2 = 1 (2) If ∫ d3r|ψ|2 is not 0 or ∞, can always normalize given eigensolution ψ of S.’s eqn. by multiplying with const., since ih̄∂ψ/∂t = Hψ is linear. 1 3. Conserved. Conservation of probability: if ψ normalized at t, must stay normalized at all times: 0 = d dt ∫ ψ∗ψ(r, t)d3r = ∫ (∂ψ∗ ∂t ψ + ψ∗ ∂ψ ∂t ) (3) Is this consistent with S.-equation? The motion of ψ is determined by ih̄∂ψ∂t = Hψ, while the motion of ψ ∗ is determined by the complex conj. eqn. −ih̄∂ψ∗∂t = Hψ∗. Note H = H∗ since potential V (r) in H = − h̄22m∇2 + V (r) is real. Therefore t-derivative of |ψ|2 is ∂ ∂t |ψ(r, t)|2 = ψ∗∂ψ ∂t + ψ ∂ψ∗ ∂t = ih̄ 2m ψ∗∇2ψ − ih̄ 2m ψ∇2ψ∗ (4) Now note we can write ψ∗∇2ψ − (∇2ψ∗)ψ = ∇ · (ψ∗∇ψ − (∇ψ∗)ψ) (5) so we get ∂ ∂t |ψ|2 +∇ · [− ih̄ 2m [ψ∗∇ψ − ((∇ψ∗)ψ)] = 0, (6) which just looks like a quantum-mechanical continuity equation: ∂ρ ∂t +∇ · j = 0, (7) where ρ = |ψ|2 = prob. density (8) j = − ih̄ 2m [ψ∗∇ψ − ((∇ψ∗)ψ)] = prob. current (9) or prob. flux density 2 just Gedanken experiments, but now doable!) by assuming there is a probability amplitude ψ1 for particle to go through slit 1, ψ2 to go through slit 2. With slit 2 closed, easy to see P1 = |ψ1|2 gives the smooth intensity profile observed, by extension P2 = |ψ2|2 when slit 1 is closed. But in order to explain interference pattern, deduce probability profile with both open is P12 = |ψ1 + ψ2|2 = |ψ1|2 + |ψ2|2 + ψ1ψ∗2 + ψ∗1ψ2. (14) Last terms provide interference effects. This is silly—why not just check to see which slit particle goes through? If we do (see figure) interference pattern disappears due to collapse of wave function, i.e. ψ is no longer ψ1 + ψ2, but just, e.g. ψ1. We must regard particle, if we do not measure it, as being a superpo- sition of a particle which went through slit 1 and one which goes through slit 2. 5.2 Particle in a box Let’s talk about the use of ψ as a probability function with more concrete example. We talked about the allowed normal modes of a cavity already, but let’s revisit the problem from the point of view of Herr Schrödinger. 5 One way to write the Hamiltonian of a cavity or “box” in 1D is to simply say there is an infinite potential if the particle is outside the box: V (x) =    ∞ x < −a/2 0 − a/2 ≤ x ≤ a/2 ∞ x > a/2 (15) Outside the box, there is zero probablity amplitude to find the particle, ψ = 0. Inside the box, however, there is no potential, so Schrödinger’s equation has the form for a free particle, (h̄2∇2/2m)ψ = Eψ. We know that the solutions are waves, but we also recall that they must satisfy the correct boundary conditions, in this case that ψ go to zero at the ends of the box (what we called standing wave boundary conditions). The first (lowest energy) standing wave has no nodes, and has a maximum at x = 0, so this must be cos πx/a. The next has a node at zero, so it must be sin 2πx/a. Then we get cos 3πx/a, and sin 4πx/a, and so on. So the solutions may be written therefore ψn = { An cos kx k = nπ/a, n = 1, 3, 5, ... An sin kx k = nπ/a, n = 2, 4, 6... (16) (Don’t be confused if you remember dimly that we used sin instead cos before (we used a coordinate system before where 0 was the left side of the cavity, now 0 is the middle! See, e.g. Griffiths 2.28 and make sure you understand the difference.). Check to make sure that for n = 1, 2, 3... you get the usual standing waves. We should use our normalization condition ∫ dx|ψn|2 = 1 condition to determine each of the An. For 6 example, determine A0 by A20 ∫ a/2 −a/2 cos2 πx/a dx = A20 · a 2 , (17) so that A0 = √ 2/a. One final remark about the eigenfunctions. Note that each successive function has a different symmetry with respect to reflection about the x axis, or parity. ψ0, ψ2, etc. are even functions of x, whereas ψ1, ψ3, etc. are odd functions of x. There are no functions which are neither even nor odd. This is a special case of a more general theorem which says if H displays a certain symmetry (in this case invariance under x → −x), the eigenfunctions are eigenfunctions of a differential operator which implements this symmetry. In other words, the ψn(x) are eigenfunctions of parity, too. We’ll revisit this concept later. Inserting any of the wavefunctions into the S-eqn., the energy levels are found to be just En = h̄2k2 2m = h̄2n2π2 2ma2 (18) Let’s use this example to discuss some basic concepts of quantum- mechanical measurement. Once we know ψ(x), we claim to know the probability |ψ(x)|2 of finding the particle at x if we make a measurement. But what if we make many measurements–what will be the average value, or expectation value of such a series of measurements? Fortunately since we know the probability function, we know the answer immediately, 〈x〉 = ∫ xP (x)dx = ∫ ∞ −∞ x|ψ(x)|2dx, (19) i.e. each possible value of the thing measured, here x, is weighted by the probability of finding it. If the particle is in the ground state ψ1(x) (usually we call ψ0 the ground state, but here the lowest standing wave 7 f (x) = ∑ n e2πinx/a ∫ a 0 dy a f (y)e−2πiny/a (31) = ∫ a 0 f (y)dy ( 1 a ∑ n e2πin(x−y)/a ) ︸ ︷︷ ︸ (32) δ(x− y) or ∫ a 0 f (y)δ(x− y)dy = f (x) (33) True for any well behaved fctn. f (x). Note mathematically δ(x) itself not function but distribution. For case f (x) = 1, get normalization condition ∫ δ(x)dx = 1 (34) Can think of δ(x) as limit of sequence of functions δm(x) = m∑ n=−m 1 a e−2πinx/a (35) which look like (m=0,3,6–Maple) In sequence width of main peak given by a/m, height by 2m/a. Note for large m oscillations outside of central peak die out. So true δ-fctn. infinitely sharp, but with area 1. • 1D Fourier integral transform. Want to do similar things with function f (x) which isn’t periodic but which vanishes suff. rapidly at∞. Crudely, replace f (x) with another function which agrees with it over a large interval (−a/2, a/2), make periodic. 10 -2 0 2 4 6 8 10 12 -3 -2 -1 1 2 3 x Now define kn = 2πn/a, afn = g(kn) (36) Using this notation, Eqs. (23) and (28) become f (x) = ∑ kn 1 a g(kn)e ikx, (37) g(kn) = ∫ a/2 −a/2 dx f (x) e−iknx (38) (since new f (x) periodic, any interval of length a is ok. Now let a get very large =⇒ in sum over kn, The difference between two 11 neighboring k’s gets small, ∆kn = 2π a , so replace sum by integral as usual: f (x) = ∫ ∞ −∞ dk 2π g(k) eikx, (39) g(k) = ∫ ∞ −∞ dx f (x) e−ikx (40) Again substituting Eq. for f into one for g, get relation f (x) = ∫ ∞ −∞ dy f (y) ( 1 2π ∫ ∞ −∞ dkeik(x−y) ) ︸ ︷︷ ︸ (41) ≡ δ(x− y) (42) i.e. integral representation for Dirac delta-fctn: δ(x) = 1 2π ∫ ∞ −∞ dk eikx (43) 5.3.2 Momentum measurement Let’s consider a free particle first, so we know classically its motion will be uniform. Time of flight measurement ok: suppose at t = 0 we measure & find it to be in a small volume around r = 0. At later time t remeasure, found at r, so velocity is v = r/t, momentum mr/t. Note there was some uncertainty in our determination of initial position, but if we make t large this becomes unimportant. In 1D S.-eqn. reads ih̄ ∂ψ ∂t = − h̄ 2 2m ∂2ψ ∂x2 (44) Plane wave (free-particle) solution is ψ ∼ ei(px/h̄−p2t/2mh̄), (45) and as in the wave packet example, gen. soln. is lin. comb. of such waves: ψ(x) = ∫ dp√ 2πh̄ f (p) ei(px/h̄−p 2t/2mh̄) (46) 12 Integration contour for ∫ eiz 2 dz integration. and from Eqs. (41-42) this prob. is properly normalized ∫ dp|f (p)|2 = 1. 5.3.3 Heisenberg uncertainty principle At t = 0, ψ(x) = ∫ dp√ 2πh̄ f (p) eipx, (57) and as usual assume f (p) peaks at some p0, width ∆p. • Case I. πh̄ 2x À ∆p Then Re eipx/h̄ varies slowly over the range of p where f (p) appreciable =⇒ eipx/h̄ ' eip0x/h̄ in this range, so ψ(x) = ∫ dp f (p) eipx/h̄ ' eip0x/h̄ ∫ dp f (p) (58) and |ψ(x)|2 nearly ind. of x. 15 • Case II. πh̄ 2x ¿ ∆p Here eipx/h̄ oscillates rapidly over ∆p =⇒ ψ(x) = ∫ dp f (p) eipx/h̄ ' 0. (59) Then ψ(x) must look like “bump” of width ∼ h̄/∆p: Was true for particular choice of f (p), what about in general? Could have f (p) varying rapidly within “spread” ∆p: At x = 0 ψ large since ∫ dpf (p) 6= 0. As x increases, faster oscillations eipx/h̄ don’t kill ψ as above, since they can match up with fast oscillations of f (p) within envelope. ψ(x) only gets small when oscillations of eipx/h̄ significantly faster than those in f (p), i.e. when ∆x À h̄/δp (À h̄/∆p!). For wiggly f (p), ∆x bigger than for “bump” f (p) =⇒ ∆x we found 16 Dashed line is “envelope” of f(p), of width ∆p, solid line f(p) itself. Scale of fast oscillations within envelope is δp before is rough lower bound for the position uncertainty, ∆x>∼h̄/∆p, or, as Heisenberg put it: ∆x∆p>∼h̄ (Heisenberg Uncertainty Principle) (60) 17