1. At time t = 0, the wave function of a free particle moving in a one-dimension is given by, ψ(x,0) = N

Transcription

1 Physics 15 Solution Set Winter At time t = 0, the wave function of a free particle moving in a one-imension is given by, ψ(x,0) = N where N an k 0 are real positive constants. + e k /k 0 e ikx k, (1) Before attacking this problem, we shall first evaluate the integral that appears in eq. (1) an etermine the value of the constant N. Since k 0 > 0 the integral is convergent, an it follows that ψ(x,0) = N = N { 0 { ( )} 1 k exp k +ix +N k 0 ( ) 1 ( ) ] ix + ix = Nk 0 k 0 k 0 1+k. 0 x The normalization constant N is etermine by emaning that Hence, 4 N k 0 0 { ( )} } 1 k exp k ix k 0 ψ(x,0) x = 1. () x = 1. (3) (1+k0x ) The integral in eq. (3) is easily compute if you remember that x I k 0 (1+k0x ) = tan 1 (k 0 x) = π. (4) If we multiply eq. (4) by k 0 an efine q 1/k0, then eq. (4) becomes, k 0 I = x q +x = π q. Finally, taking the erivative of the above result with respect to q yiels q (k 0I) = x (q +x ) = k4 0 after making use of q 1/ = 1/k 0. We therefore en up with, x (1+k0 x ) = π. k 0 1 x (1+k0x ) = π q = 3/ 1 πk3 0,

2 Inserting this last result into eq. (3) yiels, π N k 0 = 1. (5) Since N is a real positive constant by assumption (one is always free to rephase the wave function such that N > 0), it follows that k0 1 ψ(x,0) = π 1+k0x. (6) REMARKS: 1. A goo check of the results above is to make use of imensional analysis. If x has imensions of length, then k 0 has imensions of inverse length. We can inicate this by writing x] = L an k] = L 1. Then, eq. (3) implies that N] = L 1/ an ψ(x,0)] = L 1/. In aition, the parameter q = 1/k 0 has imension q] = L. Using these results, one can verify that all the equations above are imensionally correct. Such a check helps immensely in avoiing algebraic mistakes.. Actually, there is a much quicker way to erive eq. (5). Using the results of problem 3(b) of Problem Set 1, it follows that if f(x) = 1 + π a(k)eikx k, then + f(x) x = + a(k) k. Applying this result to the evaluation of eq. (), we can ientify a(k) = N πe k /k 0 an f(x) = ψ(x,0). Hence, π N e k /k 0 k = 1. The above integral is elementary, an we immeiately recover eq. (5). (a) What is the probability that a measurement of the momentum performe at time t = 0 will yiel a result between p 1 an p 1? How oes the probability change if the measurement is performe instea at time t? Explain your result. Applying the postulates of quantum mechanics, the probability that a measurement of the momentum performe at time t = 0 will yiel a result between p 1 an p 1, enote by P(p 1,0) below, is given by P(p 1,0) = p1 p 1 p ψ p = p1 p 1 φ(p,0) p, (7) where the momentum-space wave function at t = 0 is the Fourier transform of ψ(x,0), φ(p,0) = 1 π e ipx/ ψ(x,0)x,

3 Inverting the Fourier transform yiels, ψ(x,0) = 1 π e ipx/ φ(p,0)x. If we write p = k an compare with eq. (1), we can ientify π φ(p,0) = Ne p /p 0, where we have efine p 0 k 0. Hence, using N = (πp 0 / ) 1/ cf. eq. (5)], it follows that φ(p,0) = 1 p0 e p /p 0. (8) Finally, inserting this result into eq. (7) yiels P(p 1,0) = 1 p 0 p1 p 1 e p /p 0 p = p 0 p1 0 e p/p 0 p = 1 e p 1/p 0. Next, we evaluate P(p 1,t). The wave function evolves accoring to, ψ(t) = U(t) ψ(0), where the time evolution operator is given by U(t) = exp ( iht/ ). For a free particle moving in one imension, the Hamiltonian is H = P /(m). Hence, { } U(t) = exp ip t. (9) m We procee to evaluate the momentum space wave function, φ(p,t) = p ψ(t) = p U(t) p p ψ(0) p = p U(t) p φ(p,0)p, (10) after inserting a complete set of momentum eigenstates. Using eq. (9) an P p = p p, where p p = δ(p p ), { } } p U(t) p = p exp ip t p = exp { i p δ(p p ). m m Plugging this result into eq. (10), the integration over p is now trivial ue to the elta function. Hence, } φ(p,t) = exp { i p φ(p,0). (11) m It therefore follows that φ(p,t) = φ(p,0). That is, φ(p,t) is inepenent of time. This result immeiately implies that the probability P(p, t) is also inepenent of time. 3

4 This result is expecte since for a free particle, the energy eigenstates are also momentum eigenstates. Since the Hamiltonian, H = P /(m) is time-inepenent, it follows that the energy (an consequently the momentum) eigenstates are stationary states. In particular, the probability of fining a particle with a given range of momenta must also be timeinepenent. (b) What is the form of the (position space) wave packet at time t = 0? Calculate the prouct X P at time t = 0. Describe qualitatively the subsequent evolution of the wave packet. The form of the wave packet was explicitly obtaine in eq. (6). Using this result as well as the corresponing result for the momentum space wave function obtaine in eq. (8), we compute the following expectation values. X = P = x ψ(x,0) x = k 0 π xx (1+k 0x ) = 0, p φ(p,0) p = 1 pe p /p 0 p = 0, p 0 X = x ψ(x,0) x = k 0 π x x (1+k 0 x ) = 1 k 0, P = p φ(p,0) p = 1 p e p /p 0 p = 1 p p 0, 0 (1) where p 0 k 0. It then follows that X = X X ] 1/ = 1 k 0, (13) P = P P ] 1/ = k 0. (14) Hence, X P = 1, which is consistent with the Heisenberg uncertainty principle, X P 1. Finally, as time evolves, the wave packet in coorinate space spreas out. In principle, one can emonstrate this behavior by computing, ψ(x,t) = 1 π φ(p,t)e ipx/ p = 1 π p0 { ( )} i exp px p t e p /p 0 p. m Unfortunately, the above integral cannot be evaluate in terms of elementary functions alone one woul nee to employ the error function erf(x)]. However, if one simply wants 4

5 to emonstrate the qualitative behavior of the spreaing wave packet, one can make use of the uncertainty relation obtaine in eq. (..30) on p. 85 of Sakurai an Napolitano, X(0) X(t) t m. Using X(0) = 1/k 0 obtaine in eq. (13), it follows that X(t) k 0 t/(m), which inicates that the with of the wave packet increases (at least as fast as) linearly in the time t.. (a) Consier a quantum mechanical ensemble characterize by a ensity matrix ρ. Suppose that the system is governe by a Hamiltonian H (which may be time epenent). Show that the time evolution of ρ (in the Schröinger picture) is given by: ρ t = i H,ρ]. The ensity operator for a quantum ensemble mae up of N representatives is efine by ρ = i p i ψ i ψ i, where the state ψ i appears n i times in the ensemble an p i n i /N. That is, p i is the probability that an arbitrarily chosen element of the ensemble is in the state ψ i. Thetimeevolutionofthestate ψ i (intheschröingerpicture)isgivenbytheschröinger equation, i t ψ i = H ψ i. (15) Uner the assumption that the Hamiltonian is a self-ajoint operator, the ajoint of eq. (15) is given by, i t ψ i = ψ i H. (16) Noting that the probabilities p i o not epen on the time t, it follows that ρ t = ( ) ( )] p i t ψ i ψ i + ψ i t ψ i i = 1 i p i H ψi ψ i ψ i ψ i H ] = i ] H, ρ. It follows that ρ t + i H,ρ] = 0, which is calle the von Neumann equation. i 5

6 (b) Let U(t,t 0 ) be the time evolution operator. Fin a general expression for ρ(t) in terms of ρ(t 0 ) an U. Starting from ρ(t) = i p i ψ i (t) ψ i (t), ρ(t 0 ) = i p i ψ i (t 0 ) ψ i (t 0 ), with ψ i (t) = U(t,t 0 ) ψ(t 0 ), it immeiately follows that ρ(t) = U(t,t 0 )ρ(t 0 )U (t,t 0 ). (17) REMARKS: Note that U(t,t 0 ) satisfies the ifferential equation i t U(t,t 0) = H(t)U(t,t 0 ), (18) which is equivalent to the Schröinger equation given by eq. (15). Taking the ajoint of this equation, uner the assumption that the Hamiltonian is a self-ajoint operator, i t U (t,t 0 ) = U (t,t 0 )H(t). (19) Taking the time erivative of eq. (17) an employing eqs. (18) an (19), one again erives the von Neumann equation obtaine in part (a). (c) Prove that Trρ is time-inepenent. Hence, show that a pure state cannot evolve into a mixe state. Using eq. (17), Trρ (t) = Tr { U(t,t 0 )ρ(t 0 )U (t,t 0 ) ] U(t,t 0 )ρ(t 0 )U (t,t 0 ) ]} = Tr { U(t,t 0 )ρ (t 0 )U (t,t 0 ) } = Trρ (t 0 ). In the penultimate step above, we cyclically permute the arguments of the trace (which oes not change its value) an put U U = I after noting that U is unitary. It follows that Trρ is time-inepenent. We now recall that a pure state satisfies Trρ = 1, whereas a mixe state satisfies Trρ < 1. We can therefore conclue that pure states always evolve into pure states, whereas mixe states always evolve into mixe states. 6

7 3. In a one-imensional problem, consier a particle of potential energy V(x) = f x, where f > 0. (a) Write Ehrenfest s theorem for the mean values of the position X an the momentum P of the particle. Integrate these equations an compare with the classical motion. Ehrenfest s theorem states that t X = H, P H t P =. (0) X For this problem, the Hamiltonian is given by H = P m fx, where f > 0. Using H/ P = P/m an H/ X = fi (where I is the ientity operator), eq. (0) yiels, t X = 1 m P, P = f. (1) t In obtaining the secon result above, we use the fact that the expectation value of the ientity operator I with respect to normalize state is equal to 1. Integrating eq. (1) then yiels t P = P 0 +ft, X = X 0 +P 0 m + ft m, () where P 0 P t=0 an X 0 X t=0. As expecte, eq. () is ientical in form to the corresponing classical mechanics results. (b) Show that the root-mean-square eviation P oes not vary with time. The root-mean square eviation satisfies the following relation, ( P) = P P. (3) To evaluate P, we use the fact that H /t = 0, which is a consequence of the fact that H has no explicit time epenence. 1 It then follows that 1 P +f X = 0. mt t 1 Recall that the generalize Ehrenfest theorem is given by t Ω = i Ω Ω, H] +, t for any operator Ω. In the case of Ω = H, we have H /t = H/ t (since H commutes with itself). For a time-inepenent Hamiltonian, we have H/ t = 0. It then follows that H /t = 0. 7

8 Using eq. (1) for X /t, we immeiately obtain, P = f P. (4) t Next, we make use of eq. (1) for P /t to conclue that Hence, the time erivative of eq. (3) yiels t P = P P = f P. (5) t t ( P) = f P +f P = 0, after using eqs. (4) an (5). Hence, P oes not vary in time. (c) Write the Schröinger equation in the p-representation an euce a relation between t p ψ(t) an p p ψ(t) Solve the equation thus obtaine an give a physical interpretation. The Schröinger equation cf. eq. (15)] for this problem is, i ( ) P t ψ(t) = m fx ψ(t). (6) In the p-representation, i ( ) P t p ψ(t) = p m fx ψ(t). (7) Using the ajoint of P p = p p recalling that P is self-ajoint an hence its eigenvalues are real), an recalling that p X ψ(t) = i p ψ(t), (8) p it then follows from eq. (7) that i φ(p,t) t ( p = m i f ) φ(p,t), (9) p where φ(p, t) p ψ(t) is the wave function in the p-representation. Due to eq. (8), one says that the operator X is represente by the ifferential operator i / p in the p-representation. 8

9 Using eq. (9), t φ(p,t) = φ (p,t) φ(p,t) t ( ip = φ (p,t) = f +φ(p,t) φ (p,t) t ) φ(p,t)+φ(p,t) m +f p ( φ (p,t) φ(p,t) +φ(p,t) φ (p,t) p p Denoting P(p,t) φ(p,t), the above equation reas, ( ip m f p ) = f p φ(p,t). ) φ (p,t) P(p, t) t = f P(p,t) p. (30) To solve eq. (30), it is convenient to change variables by introucing Then r p ft, P p = P r + P s, Inserting these results into eq. (30) yiels s p+ft. P t = f P r +f P s. P s = 0. That is P(r,s) is a function of r alone. This means that P(p,t) = f(p ft), where f is an arbitrary function. In particular, it follows that P(p,t) = P(p ft,0). The interpretation of this result is that the probability istribution in p moves accoring to the classical equation of motion, p = p 0 +ft. () Write the Schröinger equation in the x-representation. What are the energy eigenfunctions? The Schröinger equation in the x-representation is obtaine from eq. (6) by multiplying from the left by x. The wave function in the x-representation is ψ(x,t) = x ψ(t). After writing ψ(x,t) = ψ(x)e iet/, we obtain the time-inepenent Schröinger equation in the x-representation, ( ) mx fx ψ(x) = Eψ(x), 9

10 which we can rewrite as It is convenient to introuce a new variable, ( ψ mf x + x+ me ) ψ = 0. (31) ( mf y By changing variables from x to y, eq. (31) simplifies to ) /3 ( mf x+ me ). (3) ψ +yψ = 0. (33) y Consulting, e.g., N.N. Lebeev, Special Functions an Their Applications (Dover Publications, Inc., New York, 197), pp , we see that the solutions to eq. (33) are the Airy functions, ψ(y) = c 1 Ai( y)+c Bi( y), (34) where c 1 an c are arbitrary constants are etermine by the bounary conitions relevant for the problem. (e) Is the energy spectrum continuous or iscrete? What is the behavior of the energy eigenfunctions as x? If the potential were replace by V(x) = f x, how woul your answer change? To see whether the energy spectrum is iscrete or continuous, let us examine the asymptotic behavior of the Airy functions. On p. 138 of Lebeev (op. cit.), the following results are given: 1 π x 1/4 exp ( x3/), for x, 3 Ai(x) 1 ( x) 1/4 cos ( π 3 ( x)3/ 1π), for x, 4 1 x 1/4 exp ( x3/), for x, π 3 Bi(x) 1 ( x) 1/4 sin ( π 3 ( x)3/ 1π), for x. 4 10

11 Noting that f > 0, the potential V(x) = fx is exhibite below. V(x) E x Since V(x) > E as x, we eman that lim x ψ(x) = 0. Thus, we must choose c = 0 in the solution given in eq. (34). We conclue that ψ(y) = cai( y), where y is given in terms of x by eq. (3) an the constant c is etermine by an appropriate normalization conition. Note that as x, we expect oscillatory behavior of the wave function (in analogy to the wave function ψ(x) e ikx of a free particle). This is inee the case as is evient from the asymptotic form for Ai( x) as x given above. Note that no conition on E nees to be impose to satisfy the bounary conitions (i.e., an exponentially vanishing wave function as x an an oscillatory wave function as x ). Thus, theenergy spectrumiscontinuous (anthereareno bounstatesolutions). In contrast, consier the potential V(x) = f x, with f > 0. The previous figure is now moifie as follows. V(x) E x In this case, V(x) > E as x ±. The spectrum must consist entirely of boun states, an the energy spectrum is iscrete. To solve for the boun state energies, we must solve ( ψ x + mf ) me x + ψ = 0. (35) 11

12 separately for x > 0 an x < 0. Following our previous analysis, we obtain, { cai( y), for x < 0, ψ(x) = c Ai(y ), for x > 0, (36) where y is given in terms of x by eq. (3) an ( mf y ) /3 ( mf x me ). Notethatthewave functiongiven byeq. (36)exponentially approaches zero asx ±, as require by the boun state solutions to this problem. To etermine the relation between the constants c an c, we require that both ψ(x) an ψ/x are continuous functions at x = 0. These two conitions can only be satisfie for a iscrete set of energy eigenvalues E with a fixe value of c /c. There remains one overall unetermine coefficient (which epens on the boun state energy E), which is etermine by requiring that the norm of the boun state wave functions are unity. 4. Consier a particle in three imensions whose Hamiltonian is given by: H = P m +V( X). (37) By calculating the commutator, X P, H], erive the quantum Virial Theorem, X P P = X V. (38) t m To ientify eq. (38) with the quantum mechanical analog of the Virial Theorem, it is essential that the left-han sie of eq. (38) vanish. Uner what conition oes this happen? We shall make use of the generalize Ehrenfest theorem, t Ω = i Ω Ω, H] +, (39) t Using Ω = X P, it follows that For H given by eq. (37), X P = i X P, H ]. (40) t X P, H ] = X P, P m +V( X) ] = X P, 1 P ] + X P, V( X) ]. (41) m

13 We evaluate separately the two commutators on the right han sie of eq. (41) First, X P, P m ] 1 ] 1 ] = Xi P i, P j P j = Xi, P j P j Pi = 1 ] {P j Xi, P j Pi + } X i, P j ]P j P i, m m m where there is an implicit sum over the pairs of repeate inices. Using X i, P j ] = i δij an X i, X j ] = Pi, P j ] = 0, it follows that Secon, we evaluate X P, P m ] = i m P. (4) X P, V( X) ] = X i Pi, V( X) ]. (43) In light of problem 5(b) on Problem Set 1, P, F(X) ] = i F X. The extension to three imensions is straightforwar, Hence, using eq. (43) we obtain Pi, F(X) ] = i F X i. X P, V( X) ] = X i Pi, V( X) ] = i X i V X i = i X V. (44) Combining eqs. (4) an (44), we en up with X P, H ] = X P, P m +V( X) ] = i P m X V ]. Thus, eq. (40) yiels X P = t P m X V. (45) To ientify eq. (45) with the quantum mechanical analog of the Virial Theorem, it is essential that the left-han sie of eq. (45) vanish. This will happen if the expectation values are taken with respect to a stationary state. In this case, X P ψ(t) X P ψ(t) = ψ(0) e iht/ X P e iht/ ψ(0) t = ψ(0) e iet/ X P e iet/ ψ(0) = ψ(0) X P ψ(0) = X P. 0 13

14 That is, X P is time-inepenent when taken with respect to a stationary state. In this case, X P = 0, (46) t an we arrive at the quantum mechanical virial theorem, P = X V. (47) m REMARKS: A technical remark is in orer. The classical mechanical virial theorem is erive uner the assumption that all the particle orbits are boune. In the quantum mechanical context, this conition translates into the requirement that the stationary states use in computing the expectation values above are boun states. That is, the states use to evaluate expectation values must be iscrete energy levels rather than continuous energy levels. To see where this conition arises in the above analysis, we note that if Ω is timeinepenent, then eq. (39) yiels, t Ω = i Ω, H]. With respect to an energy eigenstate, Ω, H] = E ΩH HΩ E = EE Ω E EE Ω E = 0, (48) after using H E = E E an the corresponing ajoint relation. But this last computation is correct as long as E Ω E is finite. Note that for iscrete energy levels, we have E E = δ EE. Thus, we expect that E Ω E is also finite, in which case eq. (48) is vali. In contrast, for energy eigenvalues in the continuum corresponing to scattering states, we have E E = δ(e E ). That is, for continuum energy levels, E E =. Likewise E Ω E is also infinite, so we cannot conclue as in eq. (48) that Ω, H] = 0. Hence, eq. (46) is not expecte to be vali for expectation values with respect to stationary scattering states (in the continuum). That is, the quantum mechanical virial theorem given by eq. (47) applies only for the vacuum expectation values with respect to stationary boun states. 5. In this problem, you are aske to erive the Feynman-Hellmann Theorem. (a) If the Hamiltonian H(λ) epens on a real parameter λ, i.e., H(λ) ψ = E(λ) ψ, then show that: E λ = ψ H λ ψ. 14

15 Consier E(λ) ψ H(λ) ψ. For a normalize state, ψ ψ = 1. Then, E λ = ψ λ ψ H(λ) ψ = λ H(λ) ψ + ψ H(λ) ψ + ψ H λ λ ψ { } ψ = E(λ) λ + ψ ψ ψ + ψ H λ λ ψ. (49) Note that if one ifferentiates the normalization conition, ψ ψ = 1 with respect to λ, we immeiately obtain, 0 = ψ λ ψ ψ = λ + ψ ψ ψ. λ Using this last result in eq. (49), we en up with E λ = ψ H λ ψ, (50) which is the Feynman-Hellman theorem. (b) Consier the one-imensional problem with the Hamiltonian: H = mx +V(x), where V is inepenent of the parameter m. Suppose one fins that this Hamiltonian possesses a particular energy eigenstate with energy eigenvalue E. Describe the behavior of E as m ecreases. If we rewrite the Hamiltonian in terms of the momentum operator P an the position operator X, then H = P m +V(X). Since V(x) is inepenent of m, it follows that Hence, using eq. (50), E λ = ψ H m ψ H m = P m. = 1 m ψ P ψ < 0, since the operator P is non-negative. 3 More explicitly, ψ H m ψ = ψ m x < 0. We conclue that E/ m < 0. That is, as m ecreases, the boun state energies increase. 3 If we efine φ P ψ where P is self-ajoint (P = P ), then ψ P ψ = ψ P P ψ = φ φ > 0 since the norm of any non-zero state vector is positive. 15

16 6. Consier the operator: a = mωx +ip m ω, (51) where X an P are the position an momentum operators, respectively. (a) Let z be an eigenvector of a with eigenvalue z. This state is calle a coherent state. Compute x z. Using this result, show that z z 0. Why oes the lack of orthogonality of these states not violate any of our quantum mechanics postulates? The eigenvalue equation for the operator a is ( ) mωx +ip a z = z = z z. m ω Consier the eigenvalue equation with respect to the x-basis. Multiplying on the left by x then yiels, ( ) x mωx +ip z = z z. m ω To evaluate this equation, we make use of x X z = xx z, x P z = i x x z. It then follows that ( mω ) ( 1/ x+ ) x z = zx z. mωx This is a ifferential equation fro x z. Denoting ψ z (x) x z, we procee to solve x + mωx ( ) 1/ mω z] ψ z (x) = 0. The solution to this equation is x z ψ z (x) = C z exp mω x + ( ) 1/ mω zx], (5) where C z is a normalization constant (which will epen on the eigenvalue z). We can check to see whether states of ifferent values of z are orthogonal by computing z z. After inserting a compete set of position eigenstates,, z z = Cz C z exp = C zc z exp mω x + mω x + ( ) 1/ mω zx] exp ( ) 1/ mω (z +z )x] mω x + ( ) 1/ mω z x] ( ) 1/ π = Cz C z exp 1 mω (z +z ) ]. (53) 16

17 For z z, it is clear that z z 0. Hence, the eigenstates of the operator a are not mutually orthogonal. This oes not violate any of the quantum mechanics postulates since a is not a self-ajoint (nor is it an hermitian) operator. Thus, its eigenvalues nee not be real an its eigenstates nee not be mutually orthogonal. (b) Consier the operator a in the context of the one-imensional harmonic oscillator. Compute n z, where n is the nth energy eigenstate. (Assume that z is normalize to unity.) Given a coherent state z, fin the most probable value of n (an corresponing energy E). Repeate application of a n = n+1 n+1 yiels n = (a ) n n! 0. (54) Taking the ajoint of eq. (54), it then follows that n z = 0 a n z = zn 0 z. (55) n! n! Since { n } are complete set of states, we can use the completeness relation to obtain z n z = n n z = 0 z n. (56) n! n=0 Using eq. (56), we can compute the norm of z, n=0 n=0 z z = 0 z z n = 0 z e z. n! We shall normalize z to unity, i.e. we set z z = 1. It then follows that 0 z = e z /, (57) after (conventionally) setting an overall phase factor to 1. Inserting this last result back into eq. (55) then yiels, n z = zn n! e z /. Given the state z, the probability of measuring a value n is given by n z = z n! e z. The most probable value of n epens on z. For z 1, we see that n = 0 is the most probable value. As z becomes larger, so oes the most probable value of n. For z 1 we can make use of Stirling s approximation, n! πnn n e n. 17

18 Thus, for z 1, we approximately have, n z 1 ( ) z n e n z. πn n We woul like to maximize the function, { ( ) 1 z n } f(n) = ln n e n z = n z +nln z (n+ 1 n )lnn. Taking a erivative an setting f (n) = 0 yiels ln( z /n) = 1/(n). For z 1, it is clear that n z is an approximate solution to f (n) = 0. Moreover, f (n) 1/n < 0 at the extremum, which inicates that n z is a maximum. We conclue that the maximum value of n z occurs for n z, That is, for large z, the most probable energy is given by E ω z. (c) Prove that the normalize coherent state can be written as, z = e z / e za 0. Combining eqs. (56) an (57), it follows that z = e z / n=0 z n n! n = e z / n=0 z n n! (a ) n 0, (58) after making use of eq. (54) to obtain the final result above. The sum over n can now be performe. The en result is z = e z / e za 0. () Prove that the coherent state is a state of minimum uncertainty, i.e., X P = /. We first compute the expectation values of X, X P an P with respect to the coherent state z. From eq. (51) an its ajoint, we can solve for the operators X an P, X = ( ) 1/ ] ( ) 1/ a+a, P = i a a ], (59) m 0 ω m 0 ω where we have use the fact that X an P are self-ajoint operators. It then follows that X = m 0 ω (a+a ) = a +(a ) +aa +1 ], (60) m 0 ω P = m 0 ω (a a) = a +(a ) aa 1 ]. (61) m 0 ω Employing the above results an making use of, a z = z z, z a = z z, (6) 18

19 it follows that ( ) 1/ ( X z X z = ) z +z, mω X z X z = (z +z ) +1 ], mω ( ) 1/ h ω ( P z P z = i z z ), Using the above results, we obtain P z P z = 1 m ω (z z) 1 ]. ( X) = X X ] = mω, ( P) = P P ] = 1 m ω. Hence, ( X) ( P) = 1 4. Taking the positive square root yiels X P = 1, corresponing to a state of minimum uncertainty. (e) Consier coherent states in which the parameter z is a real positive number much larger than 1. Evaluate the expectation value of the quantum Hamiltonian with respect to a coherent state with z 1. In what way is this state a goo approximation to the classical limit of the harmonic oscillator? Recall that the quantum Hamiltonian of the harmonic oscillator can be written in terms of the raising an lowering operators as H = ω ( a a+ 1 ), where the ientity operator multiplying the factor of 1 has been suppresse in the notation above. Taking the expectation value of H with respect to the coherent state z anmaking use of eq. (6), In the case of z 1, we have H z H z = ω ( z a a z + 1 ) = ω z + 1 ]. (63) REMARK: One can also compute H starting from H = P m + 1 mω X, Using the results of part (), it follows that H z H z = 1 m H ω z, for z 1. (64) P + 1 m ω X = 1 4 ω ((z z) 1 ] ω (z +z ) +1 ] = ω z + 1 ], thereby reproucing the result of eq. (63). 19

20 In orer to appreciate why z 1 correspons to the classical limit, we nee to examine the time epenence of the coherent states. Suppose that we examine the state z cf. eq. (58)] at time t = 0. Then at time t, z z,t = 0 = e z / z,t = e z / n=0 n=0 z n n! n. z n n! e ient/ n, where E n = 1 ω(n+ 1 ) are the eigenvalues of the harmonic oscillator Hamiltonian. That is, z,t = e z / e iωt/ (ze iωt ) n n = e iωt/ ze iωt. (65) n! n=0 The wave function corresponing to the coherent state in the x-representation is ψ z (x,t) x z,t = e iωt/ x ze iωt. (66) The wave function x z was previously obtaine in eq. (5). We can now evaluate C z by requiring that z z = 1. Using eq. (53), it follows that C z = ( mω ) 1/4exp 1 4 π (z +z ) ], where we have (conventionally) set the arbitrary phase of C z to unity. Inserting this back into eq. (5) yiels ( mω ) 1/4exp x z = mω ( ) ] 1/ mω π x + zx 1 4 (z +z ). (67) Replacing z with ze iωt in eq. (67) an using the resulting expression in eq. (66) yiels ( mω ) 1/4exp ψ z (x,t) = e iωt/ mω ( ) ] 1/ mω π x + xze iωt 1 4 (ze iωt +z e iωt ). We now evaluate the square magnitue of ψ(z, t) which yiels the probability ensity of the coherent state wave packet. Remarkably, it takes on a very simple form, where ψ z (x,t) = x c ( mω ) 1/exp mω π (x x c) ], (68) ( ) 1/ ( ze iωt +z e iωt). (69) mω 0

21 We see that the center of the wave packet, x = x c, oscillates with the frequency of the harmonic oscillator. But the shape of the wave packet is time-inepenent! In particular, this wave packet oes not sprea. Inee, we can easily repeat the calculation of part () by taking expectation values with respect to z,t. The en result is the same, namely X P = 1, inepenently of the time t. Thus, the coherent state wave packet is a minimum uncertainty wave packet for all times t. Using the results of part (), we can easily work out the expectation values of X an P with respect to z,t, ( ) 1/ X z,t X z,t = Re(ze iωt ), (70) mω P z,t P z,t = ( m ω ) 1/ Im(ze iωt ). (71) It is convenient to write z = z e iδ. The above equations then take the form, where X = Acos(ωt δ), P = mωasin(ωt δ), (7) A ( ) 1/ z. (73) mω As expecte from Ehrenfest s theorem, the mean position an momentum of the coherent state wave packet follows precisely the position an momentum of the classical harmonic oscillator as a function of time. 4 Moreover, for z 1, we have H = ω z cf. eq. (64)], which can be rewritten as H 1 mω A, for z 1, which is the expecte behavior of the classical harmonic oscillator (where quantum fluctuations are negligible). 5 Hence, the coherent state wave packet follows the classical motion of the harmonic oscillator without changing its form. Moreover as suggeste above, the relative size of the quantum fluctuations vanish in the limit of z. For example, using eq. (65) an the results of part (), we easily compute X = A cos (ωt δ)+ mω, It then follows that 6 ( X) = X X ] = mω, P = m ω A sin (ωt δ)+ 1 mω. ( P) = P P ] = 1 mω. 4 This shoul be compare with the expectation values of X an P with respect to the energy eigenstates n. Inee, in contrast to eqs. (70) an (71), n X n = n P n = 0 since X an P are parity-o operators an the n are states of efinite parity. Moreover, these expectation values are time-inepenent since the n are stationary states. 5 Fortheclassicalvariables, H = P /(m)+ 1 mω X = 1 mω A cos (ωt δ)+sin (ωt δ) ] = 1 mω A. 6 As asserte below eq. (69), one immeiately obtains X P = 1, inepenently of the time t. 1

22 Comparing X to the amplitue A of harmonic motion given by eq. (73)], in the limit of z 1. Likewise, X A = 1 z 1, P mωa = 1 z 1. Hence, the quantum fluctuations are suppresse in the limit of z 1. A similar conclusion can be obtaine by computing the root-mean square of the Hamiltonian. H = z H z = z ( ) a a+ 1 z = ω z ( ) a a aa+a a+ 1 4 z = ω z 4 + z + 1 4] = H ] + ω z, after making use of a, a ] = 1 an employing eq. (63). Hence, it follows that an H = H H ] = ω z, H H 1 z 1, in the limit of z 1. Inconclusion, acoherent statewith z 1isavery gooapproximationtotheclassical limit of the harmonic oscillator.