8.04 Spring 2013 March 12, 2013 Problem 1. (10 points) The Probability Current

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1 Prolem Set 5 Solutions 8.04 Spring 03 March, 03 Prolem. (0 points) The Proaility Current We wish to prove that dp a = J(a, t) J(, t). () dt Since P a (t) is the proaility of finding the particle in the range a < x < at time t it is mathematically equal to Its time derivative is therefore given y P a (t) = ψ(x, t) dx = ψ (x, t)ψ(x, t)dx. () a a dp a d ψ ψ = ψ (x, t)ψ(x, t)dx = [ψ (x, t)ψ(x, t)] dx = ψ + ψ dx, (3) dt dt t t t a a a where we were allowed to take the time derivative inside the integral ecause the integral is in time whereas the integral is over space. Note, though, that the total derivative ecame a partial derivative when we took it inside the integral, ecause whereas a ψ (x, t)ψ(x, t)dx is a function of t only (since x has already een integrated over), ψ is in general a function of oth x and t, and we only want to take its time derivative. We now make use of the fact that regardless of what the wavefunction ψ(x, t) happens to e, it must oey the Schrödinger equation: ψ(x, t) ψ(x, t) i = + V (x)ψ(x, t) (4a) dt m x ψ (x, t) ψ (x, t) i = + V (x)ψ (x, t) (4) dt m x We can thus eliminate the partial derivatives in Equation 3: dp a ψ ψ = ψ + V ψ + V ψ ψ dx (5a) dt i a m x m x ψ ψ = ψ ψ + (ψ V ψ V ψ ψ) dx (5) i m x x a i ψ ψ = ψ ψ dx, (5c) m a x x where in the last equality we used the fact that ψ V ψ = V ψ ψ ecause in position space the potential energy operator commutes with oth ψ and ψ. To proceed, we integrate the two remaining terms y parts. Recall that g f f dx = fg g dx. (6) a x a a x

2 This means ( ) dp ψ ψ [ ( ) ( )] a i i ψ ψ = ψ ψ dx = ψ ψ dx (7a) dt m a x x m a x x x x [( ) i ψ ψ ( ) ψ ψ ] ψ ψ ψ = ψ dx + dx (7) m x a x a a x x a x x [( ) i ψ ψ ( ) ψ ] = ψ (7c) m x a x a [( ) i ψ ψ ( ) ψ ψ ] = ψ ψ ψ ψ, (7d) m x x x=a x x x= so if we let ( ) i ψ ψ J(x, t) ψ ψ, (8) m x x then dp a = J(a, t) J(, t), (9) dt which is what we were trying to prove. From this equation, we can see that the units of J(x, t) must e (time), ecause P a is a proaility and is therefore a pure numer, and t has units of time.

3 Prolem Set 5 Solutions 8.04 Spring 03 March, 03 Prolem. (0 points) Visual Oservation of a Quantum Harmonic Oscillator (a) (5 points) The energy of a classical harmonic oscillator is given y E = mω 0 A, (0) where ω 0 is the angular frequency, m is the mass, and A is the amplitude of the oscillation. The quantum harmonic oscillator, on the other hand, has energy ( ) E n = ω 0 n +. () Equating these expressions and rearranging gives mω 0 A mω 0 πmν 0 n = A = A, () ω 0 where we have converted to using the frequency ν 0 = ω 0 /π instead of the angular frequency ω 0, and have assumed that n is so large that n + n (we can check our result against this assumption later). Plugging in m = 0 g, ν 0 = 0 3 Hz, and A = 0 3 cm, we get n 3 0. (3) From this result we see that we were justified in neglecting the / term in n +. () (5 points) If this oscillator were in its ground state, its energy would e E 0 = ω 0 = π ν 0 =. 0 ev, (4) where once again ν 0 = 0 3 Hz. This is smaller than the average thermal energy of air molecules (5 mev) y a factor of E air E ev = 0 0. (5). 0 ev (c) (5 points) We can perform the same manipulations as we did in part (a), except this time we cannot neglect the / term in the oscillator s energy, since we are dealing with the ground state: E 0 = ω 0 = mω 0 A A = =. (6) mω 0 πmν 0 There is really no difference etween a classical harmonic oscillator and a quantum one the Universe is governed y quantum mechanics, so in principle all oscillators are quantum mechanical. The only difference is whether or not the quantum effects are ovious. 3

4 Note that this is width of the Gaussian ground state wavefunction. Numerically, this comes out to e A = nm, which is much smaller than the wavelength of visile light: λ 400 nm = 0 5. (7) A nm (d) (5 points) I would not recommend award of a grant to carry out this research. From part (a), we can see that the experimenter s proposed system corresponds to a high quantum numer state, which means the experiment is unlikely to really proe quantum mechanical effects. At est, it would e ale to investigate the approach of the quantum system to the classical limit, ut even this is unlikely, given the results of () and (c). In part (), we saw that air molecules possess energies that are much larger than the spacings etween energy levels of the quantum harmonic oscillator. Thus, unless the experimenter can perform his/her experiment in a perfect vacuum (something that noody can achieve), the air molecules will likely interact with the system and cause the oscillator to make transitions to higher energy levels. Finally, the calculation in part (c) shows that using visile light to proe the system is not practical. Even if one neglects the collapse of the wavefunction caused y measuring the system with light, the fact that the relevant spatial scales ( nm) are so much smaller the wavelength of the light means that the light will essentially ignore features on those scales. This is why, for instance, mirrors for telescopes that operate in the visile wavelength and have to e grounded to such high precision, whereas those for microwave telescopes are aout as smooth as a nice piece of cardoard. 4

5 Prolem Set 5 Solutions 8.04 Spring 03 March, 03 Prolem 3. (30 points) Harmonic Oscillators Oscillate Harmonically (a) (4 points) Our wavefunction is initially ψ(x, 0) = [φ 0 (x) + iφ (x)], (8) where φ 0 and φ are the normalized eigenstates for the ground and first excited states of the harmonic oscillator respectively. The eigenstates evolve in time in the usual fashion (phase factor with angular frequency equal to E n / ), so the principle of superposition tells us that ψ(x, t) = e ie 0t/ φ 0 (x) + ie ie t/ φ (x) = e iω0t/ φ 0 (x) + ie i3ω0t/ φ (x), (9) where in the last step we used the fact that the n th excited state of a harmonic oscillator has energy E n = ω 0 (n + ). The proaility distriution ψ(x, t) is given y ψ(x, t) ψ ψ = e iω 0t/ φ 0 (x) + ie i3ω 0t/ φ (x) e iω 0t/ φ 0 (x) + ie i3ω 0t/ φ (x) (0a) iω = e 0 t/ φ 0(x) ie i3ω0t/ φ (x) e iω0t/ φ 0 (x) + ie i3ω0t/ φ (x) (0) = φ φ 0 + ie iω0t φ 0 (x) + φ (x) ie iω0t φ 0φ (0c) iω 0 t = φ e iω 0t 0 (x) + φ (x) + iφ 0 φ e (0d) = φ 0 (x) + φ (x) + φ 0 φ sin ω 0 t, (0e) where in the second last step we took advantage of the fact that energy eigenstates can e taken to e real (see Prolem Set 3), so φ 0 = φ 0 and φ = φ. () (0 points) The expectation value (xˆ) is defined in the usual fashion: [ ] (xˆ) = ψ xψdx ˆ = x ψ(x, t) dx = x φ 0 (x) dx + x φ (x) dx + sin ω 0 t xφ 0 φ dx. () To proceed, we need to know something aout the form of φ n, the n th excited energy eigenstate of the harmonic oscillator. As discussed in lecture, the general form of φ n is given y φ n (x) = N n H n (x/a)e x a, () 5

6 where H n is the n th Hermite polynomial and a /mω 0. Since the Hermite polynomials are either even or odd, the eigenstates are also either even and odd, which means that φ 0 (x) and φ (x) are oth even. In turn, x φ 0 (x) and x φ (x) are odd, so x φ 0 (x) dx = 0 and x φ (x) dx = 0, (3) since the integrals are over symmetric intervals. This leaves the last term, for which we will need the precise forms of the energy eigenstates: mω /4 0 x mω /4 0 x x φ 0 (x) = e a and φ (x) = e a (4) π π a This gives mω 0 x (xˆ) = sin ω 0 t xφ 0 φ dx = a sin ω 0 t e ( x a) dx (5a) π a mω 0 u mω 0 = a sin ω 0 t u e du = a sin ω 0 t = sin ω o t. (5) π mω 0 The reader is encouraged to t check that this expression has the right units. The ampli tude of the oscillation is mω 0, and the angular frequency of the oscillation is ω 0. It is instructive to see how to calculate (xˆ) also via the operator method. Taking the result from Prolem 4 (e), we can express ˆx in terms of â, â : xˆ = (â + â ). (6) mω 0 Thus (xˆ) = dxψ (â + â )ψ = mω 0 iω 0 t = φ 0 (â + â )φ 0 + i e φ 0 (â + â )φ + mω 0 i iω0 t e φ )φ 0 + ] (â + â φ (â + â )φ. (7) Now, still from Prolem 4 we know that, for any energy eigenstate φ n, φ nˆ aφ n = φ n a ˆ φ n = φ n aφ ˆ n = φ n ˆ φ a n = 0, (8) and thus we have [ ] i iω 0 t i iω 0 t (xˆ) = e φ 0ˆ aφ e a φ ˆ φ 0 = sin ω 0 t. (9) mω 0 mω 0 Of course, we know from Prolem Set 3 that we can always take the energy eigenstates to e even or odd, ut it s nice to see it explicitly in Equation as well. 6

7 (c) (0 points) For (p) we have (pˆ) = ψ pψdx ˆ = ψ ψdx (30a) i x [ iω 0 ] [ ] = e t/ φ 0(x) ie i3ω0t/ φ (x) e iω0t/ φ0(x) + ie i3ω0t/ φ (x) dx(30) i [ ] = φ 0 φ 0 + φ φ ie iω0t φ φ 0 + ie iω0t φ 0 φ dx, (30c) i where we have once again used the fact that the energy eigenfunctions can e taken to e real. Let us examine this term-y-term. Since the derivative of an even function is odd (and vice versa 3 ), terms like φ n φ n must e odd overall. We thus have φ 0 φ 0 dx = 0 and φ φ dx = 0. (3) Plugging in the forms of the energy eigenfunctions gives [ ] (pˆ) = ie iω0t φ φ 0 + ie iω0t φ 0 φ dx (3a) i m ω 0 x x x x e x a iω 0 t iω 0 t iω 0 t = e e a e e a + e dx (3) π a3 a3 a m ω 0 iω 0 t u iω 0 t u iω 0 t u = e u e e u e + e e du (3c) π [ m ω ] 0 iω 0 t π iω 0 t π iω 0 t = e e + e π (3d) π m ω 0 = i sin ω 0 t π + (cos ω 0 t i sin ω 0 t) π (3e) π m ω 0 = cos ω 0 t. (3f) Again, let s otain the expression for (pˆ) also through the operator method. From Prolem 4 (e), we know that m ω 0 pˆ = i (â â ), (33) and hence, m ω 0 (pˆ) = i dxψ (â â )ψ. (34) 3 If you re not sure aout this, try proving it from the definition of the derivative: f ' (x) = f (x+δx) f(x) lim Δx 0 Δx. It may also help to compute the derivatives of the energy eigenstates of the harmonic oscillator explicitly. Finally, try drawing a sketch! A picture is worth 0 3 words, especially in physics. 7

8 Now, expanding like in Equation (7) and keeping into account Equation (8), we otain [ ] m ω 0 i iω 0 t i iω 0 t m ω 0 (pˆ) = i e φ 0ˆ aφ + e φ ˆ a φ 0 = cos ω 0 t. (35) (d) (6 points) Any wavefunction can e expanded in terms of the energy eigenstates of the system: ie ψ(x, t) = c nt/ n φ n (x)e, (36) n where E n is the energy of the n th excited state. With the harmonic oscillator, we have ( ) E n = ω 0 n +, (37) so the wavefunction takes the form ψ(x, t) = e iω 0t/ n c n φ n (x)e inω 0t. (38) Each term in the sum oscillates with period T n = π nω 0, i.e. some integer fraction of iω 0 t/ the classical period T = π/ω 0. For example, aside from the overall the e in front, the first term is constant, the second term oscillates with period T, the third term oscillates with period T/, and so on. Thus, after a time T all the phases will e aligned once again, except for phase in the prefactor, so that Taking the norm squared of oth sides gives ψ(x, t + T ) = e iω 0T/ ψ(x, t). (39) ψ(x, t + T ) = ψ(x, t), (40) which is a mathematical way of saying that the proaility distriution returns to its original shape after period T = π/ω 0. The proaility distriution is periodic ecause the difference etween angular frequencies (E n E m )/ are rational multiples of a common value. With the harmonic oscillator in particular, this arises ecause the energy levels are equally spaced apart (with ΔE = ω 0 ). Consider instead a system whose first three energy levels are E 0, E 0, E 0, with associated eigenstates φ 0, φ, φ. Then the state ( E 0 E 0 E i i i 0 ) ψ(x, t) = e t φ 0 + e t φ 0 + e t φ 3 is not periodic in time, or more precisely there is no value of T for which φ(x, t) = e iα ψ(x, t + T ). (4) 8

9 To see this, let s rewrite ψ as i E 0 t e ( ( )E 0 ) i t i φ 0 + e E 0 ψ(x, t) = φ t 0 + e φ, 3 so that (4) is equivalent to e i( ) E 0 T =, together with e i E 0 T =, which is equivalent to ( )T = πn, E0 together with T = πm E0 for some integers n and m. Dividing one equation y the other we get n =, m which has no solution. 9

10 Prolem Set 5 Solutions 8.04 Spring 03 March, 03 Prolem 4. (40 points) Operators for the Harmonic Oscillator (a) (0 points) First rememer what it means when we take the Hermitian conjugate ( dagger ) of an operator. The operator O is defined y the following equation: where ψ and ψ are aritrary wavefunctions that satisfy that usual requirements of continuity, normalizaility, and so on. With this, the norm squared of our state is given y (φ n, φ n) = φ nφ ndx = [(â ) n φ 0 ] (â ) n φ 0 dx = [(â ) n φ 0 ] ââ [(â ) n φ 0 ]dx (4) Now, ââ can e rewritten in the following way: ∠a = aˆ ˆa a ˆ ˆ a + a ˆ â = [ˆ a, aˆ ] + a ˆ â = + a ˆ ˆ a, (43) where we have used the identity [â, â ] = (prove this y expressing â and â in terms of ˆp and ˆx). Inserting this into our expression, we have (φ n, φ n) = (φ n, φ n ) + [(â ) n φ 0 ] â â[(â ) n φ 0 ]dx (44a) = (φ n, φ n ) + φ n â âφ n dx. (44) Now, recall that the energy operator for a harmonic oscillator can e written as ( Ê = ω â â + ), (45) which means we can rewrite the last line in terms of the energy operator: (φ n, φ n) = ( φ n, φ n ) (φ n, φ n ) + φ n Êφ n dx. (46) ω This is nice ecause φ n is an energy eigenstate (it may not e normalized, ut it s still an energy eigenstate, ecause a constant times an eigenfunction is still an eigenfunction). This means that Ê acts on it in a very simple way, ecause it satisfies the energy eigenvalue equation: ( ) Ĥφ n = E n φ n where E n = n + ω. (47) 0

11 In our case, then, we have E n (φ n, φ n) = (φ n, φ n ) + (φ n, φ n ) (48a) [ ω = (φ n, φ n ) + (n ) + ] (φ n, φ n ) (48) = n(φ n, φ n ). (48c) We have thus related the norm square of φ n to the norm square of φ n. We can repeat the process until we reach the ground state: (φ n, φ n) = n(φ n, φ n ) = n(n )(φ n, φ n ) = = n(n )... (φ 0, φ 0) = n!, (49) where in the last step we used the fact that φ 0 = φ 0, so φ 0 is in fact the normalized ground state. The norm of ñ) is thus t (φ n, φ n) = n!. (50) () (6 points) The normalized energy eigenstates are given y φ n (â ) n φ 0. (5) n! First we test for the normality of the states: n! (φ n φ n ) = φ 0â n (â ) n φ 0 dx = =, (5) n! n! where we made use of the result we proved in (a). For orthogonality, we have (φ m φ n ) = φ 0â m (â ) n φ 0 dx, (53) m!n! where m = n. Now, after acting on φ 0 with the creation operator a ˆ n times, we end up with a state that is proportional to the n th excited state φ n. Our equation then calls for the annihilation operator â to act on the state m times. If m > n, we have more annihilation operators than we do creation operators, and at some point we get to âφ 0, which gives us zero. If m < n, we can say φ 0â m (â ) n φ 0 dx = (â n (â ) m φ 0 ) φ 0 dx, (54) m!n! m!n! and again we end up lowering a state more than we raise it, only this time it s the left copy of the wavefunction. We thus conclude that (φ m φ n ) = 0 unless m = n, in which case (φ m φ n ) =. The states φ n are therefore orthonormal.

12 (c) (6 points) We wish to show that aφ ˆ n = nφ n and a ˆ φ n = n + φ n+. (55) So far, we only know that âφ n is proportional to φ n. In other words, aφ ˆ n = Dφ n, (56) where D is some proportionality constant. Let us dot this equation with itself and perform manipulations similar to what we did in part (a): [âφ n ] [âφ n ]dx = D (φ n φ n ) (57a) φ nâ ˆ aφ n dx = D (φ n φ n ) (57) φ n Êφ n dx (φ n φ n ) = ω D (φ n φ n ) (57c) E n (φ n φ n ) (φ n φ n ) ω = D (φ n φ n ) (57d) n(φ n φ n ) = D (φ n φ n ) (57e) n = D, (57f) ( ) where we have used the fact that E n = n + ω and that φ n and φ n are normalized energy eigenstates of the harmonic oscillator. We thus have aφ ˆ n = nφ n. (58) Similarly, let â φ n = F φ n+, (59) where F is some proportionality constant. Performing analogous manipulations, we have: [â φ n ] [â φ n ]dx = F (φ n φ n ) (60a) φ nââ φ n dx = F (φ n φ n ) (60) φ n [â, â ] + â ˆ a φ n dx = F (φ n φ n ) (60c) (φ n φ n ) + φ nâ aφ ˆ ndx = F (φ n φ n ) (60d) (φ n φ n ) + n(φ n φ n ) = F (φ n φ n ) (60e) (n + )(φ n φ n ) = F (60f) n + = F, (60g) where we have used the fact that [â, â ] = and the result D φ ˆ nâ aφ n dx = n that we otained aove. We thus have â φ n = n + φ n+. (6)

13 (d) (6 points) The numer operator is defined as N ˆ = â ˆ a. (6) We now examine the commutators of this operator with the annihilation and creation operators: Similarly, [ N, ˆ a] ˆ = Nˆ a ˆ a ˆN ˆ = aˆ ˆa aˆ ˆ ˆa a = (ˆa aˆ aˆ ˆ a)ˆ a = [ˆ a, a ˆ ]â = ˆ a. (63) ˆ ˆ ˆ ˆâ ˆ [ N, ˆ a ˆ ] = N a ˆ a ˆ N = aˆ aˆ a a ˆ a = a ˆ (ˆa aˆ a ˆ a) = aˆ [ˆ a, a ˆ ] = a ˆ. (64) From lecture, we know that for the harmonic oscillator the energy operator E ˆ can e written as 4 ( E ˆ ˆ ˆ ) ( ) ˆ ˆ E = ω 0 â a + = ω 0 N + N =. (65) ω 0 This means that the eigenfunctions of N ˆ are the eigenfunctions of E, ˆ i.e. the energy eigenstates of the harmonic oscillator. We can see this explicitly y having N ˆ act on φ n : ( ) ˆ ˆ ω 0 Nφ n = Eφ n φ n = n + φ n φ n = nφ n, (66) ω 0 ω 0 since we know that the energy eigenvalues of the harmonic oscillator are given y E n = ω 0 n + ˆ. We can also see from this that the eigenvalues of N are simply the various values of n (i.e. the quantum numer corresponding to each energy eigenstate). (e) (6 points) Our original definitions of â and â were xˆ pˆ xˆ pˆ â + i and â i. (67) /mω 0 m ω 0 /mω 0 m ω 0 We can solve these to get expression for ˆx and ˆp: m ω 0 xˆ = (â + â ) and pˆ = i (â â ). (68) mω 0 Now, for an eigenstate φ n, we have (xˆ) = φ n xφ ˆ n dx = φ n(ˆ a + a ˆ )φ n dx = 0, (69) mω 0 ecause when â acts on φ n, the result is proportional to φ n+, which is orthogonal to φ n, and similarly for â. For the momentum, we have m ω 0 (pˆ) = φ n pφ ˆ n dx = i φ n(ˆ a a ˆ )φ n dx = 0, (70) using exactly the same reasoning. 4 If you re not sure where this comes from, express â and â in terms of ˆx and ˆp, and make use of the ˆ commutation relation [ˆx, pˆ] = i to express E of the harmonic oscillator in terms of â and â. 3

14 (f) (6 points) First we express ˆx and ˆp in terms of â and â : xˆ = (â + â ) (â + â ) = (â + â )(â + â ) (7a) mω 0 mω 0 mω 0 = (â + ââ + â â + (â ) ) = (â + (â ) + â â + [â, â ]) (7) mω 0 mω 0 = (â + (â ) + Nˆ + ). (7c) mω 0 Similarly, for ˆp, we have m ω 0 m ω 0 m ω 0 pˆ = i (â â ) i (â â ) = (â â )(â â ) (7a) mω 0 m ω 0 = (â ââ â â + (â ) ) = (â + (â ) â â [â, â ]) (7) mω 0 = (â + (â ) Nˆ ). (7c) Plugging these operators into (xˆ) and (pˆ) gives [ ] (xˆ) = φ n (â + (â ) + Nˆ + ) φ n dx (73a) mω 0 = φ n(nˆ + )φ n dx (73) mω 0 = (n + ), (73c) mω 0 where in going from the first line to the second line we used the same reasoning as we did in part (e) i.e. orthogonality of φ n and (â ) φ n φ n+ or â φ n φ n. In going from the second line to the third line we used our results from part (d). The expectation value of the momentum looks similar: [ ] m ω 0 (pˆ) = φ n (â + (â ) Nˆ ) φ n dx (74a) mω 0 = φ n(nˆ + )φ n dx (74) mω 0 = (n + ). (74c) Now, recall that the uncertainties are defined as Δx (xˆ) (xˆ) and Δp (pˆ) (pˆ). Since (xˆ) and (pˆ) are zero from part (e), we get which is our desired result. ΔxΔp = (xˆ)(pˆ) = (n + ), (75) 4

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