Quantum Dynamics. March 10, 2017

Transcription

1 Quantum Dynamics March 0, 07 As in classical mechanics, time is a parameter in quantum mechanics. It is distinct from space in the sense that, while we have Hermitian operators, X, for position and therefore epect a measurement of postion to yield any eigenvalue of X, there is no T operator for measurements of time. We cannot epect measurements to yield distributions of eigenvalues of time, perhaps because we are stuck in one particular time ourselves. In any case, while there are differences between time and space, there does still eists a unitary time translation operator, and its infinitesimal generator energy is an central observable. In classical physics we make a distinction between active transformations which transform the physical position and other variables, and passive transformations which leave the physical particles fied and transform the coordinates. The same distinction occurs in quantum mechanics. A given time-dependent wave function, ψ, t ψ t, has the same two interpretations. We may describe the system by an eplicitly time-dependent ket, ψ ψ, t taking the basis kets, a, whatever they are as fied. The state is then a time varying vector in the space spanned by the fied basis kets. ψ, t ψ, t Alternatively, we may let the state be a fied ket, while the basis kets evolve in time, a, t, ψ, t, t ψ The two descriptions are equivalent, though one or the other may be convenient for a given problem. In perturbation theory, it is even useful to mi the two pictures, giving part of the time dependence to the basis and the rest to the state. For the moment, we consider time-dependent states in a fied basis. In quantum mechanics, the terms active and passive are replaced, respectively by speaking of the Schrödinger picture and the Heisenberg picture. In the Schrödinger picture, the state vector becomes time dependent with the basis fied, coresponding to an active transformation in classical physics. The Heisenberg picture begins by letting the operators become time dependent, with the consequence that their eigenstates and therefore the basis, becomes time dependent. The time translation operator We define the time translation operator to be the mapping that takes any time-dependent state from some initial time, t 0, to a later time, t, Û t, t 0 ψ, t 0 ψ, t 0 ; t In the limit as t t 0, this must be the identity, Û t 0, t 0 ψ, t 0 ψ, t 0 ; t 0 ψ, t 0

2 We require time evolution to preserve probabilities, so by Wigner s theorem it must be either unitary or antiunitary, and there is no conflict in taking it to be unitary, In a fied basis a, we may epand any state as Û t, t 0 Û t, t 0 ψ, t a c a t a Notice that the time dependence allows the state to move from one eigenstate to another. Since the state is normalized at all times, we require ψ, t 0 ; t ψ, t 0 ; t c a t a c a t a a a c a t c a t a a a a c a t c a t δ aa a a a c a t Unitarity of the time translation operator requires this to hold at every time t. We may write Û t, t 0 as an eponential, Then, for an infinitesimal time change, t t 0 +, where unitarity of Û t, t 0 requires Ω Ω. The Schrödinger equation Û t, t 0 e i Ωt Û t 0 +, t 0 i Ω t 0 Consider three sequential times, t 0 < t < t. It must be the case that acting with on an arbitrary state with Û t 0, t gives the same result as acting with Û t 0, t followed by Û t, t : Û t 0, t ψ, t 0 Û t, t Û t 0, t ψ, t 0 since both of these must give ψ, t 0 ; t. Since the times are arbitrary, this gives an operator equality, Û t 0, t Û t, t Û t 0, t Now eamine the time evolution of the time translation operator, Û t, t 0, at some time t. Translating the time t by an additional small amount t, Û t + t, t 0 Û t, t Û t, t 0 iĥ t t Û t, t 0 Û t, t i t Ĥ t Û t, t 0

3 Rearranging, so taking the limit as t 0, Ĥ t Û t, t 0 i Û t + t, t 0 Û t, t 0 t Ĥ t Û t, t 0 i Û t, t 0 This is the Schrödinger equation for the time evolution operator. Finally, let this operator relation act on an initial state, ψ, t 0, to give the time evolution equation for that state, This gives the time-dependent Schrödinger equation, Ĥ t Û t, t 0 ψ, t 0 i Û t, t 0 ψ, t 0 Ĥ t ψ, t i ψ, t This is basis independent form of the familiar Schrödinger equation. Since the state is arbitrary, we recognize the generator of time translations as the Hamiltonian, Ω t i Ĥ 3 The full time evolution operator As we did the the translation operator, we may recover the form for a finite time translation trom the infinitesimal solution. We distinguish three cases:. Ĥ independent of time. Ĥ depends on time, but Ĥ t, Ĥ t ] 0 for any two times, t, t. 3. Ĥ t, Ĥ t ] 0 The first case may be solved by direction integration of the Schrödinger equation for Û t, t 0, With Ĥ t Ĥ constant, this immediately gives Ĥ t Û t, t 0 i Û t, t 0 Û t, t 0 e i Ĥt t0 We may solve the second case by iterating the infinitesimal transformation. Dividing the time interval t t 0 into n equal parts of length t t t0 n, we take the product of n infinitesimal time translations, where the k th time translation may be written to first order as i Ĥ t 0 + k t t e i Ĥt0+k t t 3

4 Then the finite transformation is Û t, t 0 lim i n Ĥ t 0 t i Ĥ t 0 + t t i Ĥ t 0 + t t... iĥ t t t n e i Ĥt0+k t t n k0 lim lim n ep i n Ĥ t 0 + k t t k0 t ep i Ĥ t t 0 Notice that as n the linear approimation becomes eact and the sum becomes an integral. Because all of the Ĥ t commute, we see directly that this form of Û t, t 0 satisfies the Schrödinger equation, Ĥ t Û t, t 0 i Û t, t 0 The general case is considerably more involved because we have to keep track of the order of the terms in the power series. The result, called the Dyson series, has the form Û t, t 0 lim i n Ĥ t 0 t i Ĥ t 0 + t t i Ĥ t 0 + t t... n0 i n t t Ĥ t Ĥ t... t 0 t 0 t T ep i Ĥ t t 0 t n t 0 Ĥ t n n The crucial step occurs in noticing that the n th term in the sum, may be etend each of the n integrals all the way from t 0 to the final time t and dividing by n!, i n t t t n n Ĥ t Ĥ t... Ĥ t n n i t Ĥ t n! t 0 t 0 t 0 with the caveat that in the resulting n th power on the right must be time ordered, n n! T i t Ĥ t t 0 t 0 meaning that the n factors of the Hamiltonian must be returned to the product form on the left with t < t <... < t n in increasing order. This notational trick allows us to write the result as an eponential, but the final eponential is simply shorthand for the whole series Almost all of the applications we consider will involve time-independent Hamiltonians. 4 Energy eigenkets Suppose we choose as a basis the eigenkets of an operator  which commutes with the time-independent Hamiltonian, Â, Ĥ] 0 4

5 This operator could be the Hamiltonian itself, or another observable, but in either case, the eigenkets, a, may be chosen to simultaneously be eigenkets of Ĥ, Ĥ a E a a Consider the time evolution of an arbitrary initial state, ψ, t 0. In terms of the  basis, ψ, t 0 a a a a ψ, t 0 c a a where c a a ψ, t 0. Then the time evolution is given by ψ, t 0 ; t ep iĥt ψ, t 0 c a ep iĥt a a c a ep i E nt a a Defining time dependent epansion coefficients, c a t c a e i Eat, the state becomes ψ, t 0 ; t a c a t a In general, this means that the probabilities for measuring the system to be in different states varies with time. The only eceptions are the probatilities for measuring a fied energy eigenstate. Concretely, the probability for measuring the state to have energy E a0 at time t is P E a0 a 0 ψ, t 0 ; t c a0 t c a0 e i Ea 0 t c a0 that is, the same as the original probability. However, the probability for finding a general superpostion changes. Let χ E + E be a superposition of eigenstates for energies E and E. The probability of measuring the system to be in the state χ is P χ χ ψ, t 0 ; t c t + c t c e i Et + c e i Et We pull out a common phase and set E E E. Then P χ e i Et e i Et c e i Et + c e i c e i Et + c e i c e i Et + c e i c + c + c c e i E Et + c c e i E Et c + c + Re c c e i E Et 5

6 which is time dependent unless the energy is degenerate with E 0. The problem of quantum dynamics is now reduced to finding a maimal set of operators which commute with the Hamiltonian. We may then label states using these and study the time evolution of arbitrary initial states. Eercise: Find the time evolution of a particle in an infinite square well, < L V 0 L < < L > L when its state at time t 0 is: π. ψ, t 0 L cos L π. ψ, t 0 L cos L + 3π L cos L 5 Epectation values of observables Once we have the state of a system epanded in energy eigenkets, we can find the time dependence of the epectation values for any observable, Ô, ψ, t 0 ; t Ô ψ, t 0; t a,a ψ, t 0 ; t a a Ô a a ψ, t 0 ; t a,a c a t c a t a Ô a a,a c a c ae i Ea E a t a Ô a As an eample, consider a -state system. The Hamiltonian for a spin- µ e mcs, in a magnetic field, B, is particle with magnetic moment Ĥ µ B e mcŝ B Let B Bk be constant, so the time-independent Hamiltonian is Ĥ eb mcŝz The energy eigenkets are then just the ± eigenkets of Ŝz with energies Ĥ ± E ± ± ± eb mc ± ± ω ± where ω eb mc. Setting t 0 0, the time evolution of a general normalized state, χ cos θ + + e iϕ sin θ 6

7 is χ, t Û t, 0 χ e i Ĥt χ e i Ĥt cos θ + + e iϕ sin θ e i E+t cos θ + + e i E t e iϕ sin θ e iωt cos θ + + e iωt e iωt e iϕ sin θ ] cos θ + + e iωt+ϕ sin θ The probability for measuring the system to be in the spin-up state is then ] + χ, t e iωt cos θ e iωt+ϕ sin θ + cos θ which is the same as the initial probability. However, if we look at the probability that the -component of spin is up, we find Ŝ, + χ, t + + ] e iωt cos θ + + e iωt+ϕ sin θ cos θ + e iωt+ϕ sin θ cos θ + e iωt+ϕ sin θ cos θ + e iωt+ϕ sin θ cos θ + sin θ + e iωt+ϕ + e iωt+ϕ sin θ cos θ 6 Time-energy uncertainty + sin θ cos ωt + ϕ Consider a system with a continuous energy spectrum and many particles. Let the state be epanded in energy eigenkets, ψ, t 0 ; t ep iĥt ψ, t 0 ρ E de c E ep iĥt E ρ E de c E e i Et E where ρ E characterizes the distritubion of energy eigenstates and the orthonormality relation is ρ E E E δ E E Consider the correlation between the initial state, ψ, t 0, and the state at time t, C t ψ, t 0 ψ, t 0 ; t ψ, t 0 ep iĥt ψ, t 0 ρ E de c E E ρ E de c E e i Et E 7

8 ρ E ρ E dede c E c E e i Et E E de ρ E de c E e i Et ρ E de c E e i Et Suppose the initial distribution is peaked around some energy E 0. Write C t as C t ρ E de c E e i Et e i E0t ρ E de c E e i E E0t For energies far from E 0, the eponential e i E E0t oscillates rapidly and the value of the integral is small as long as ρ E and c E vary slowly, and E E 0 t. On the other hand, there will be coherent contributions for any time interval t such that E E 0 t Therefore, times t and energies E E 0 + E contribute significantly, where E t This is a very different statement than the necessary relationship between the uncertainties in conjugate observables. In particular, recall that there is no Hermitian operator corresponding to a measurement of time, so the general result for noncommuting operators does not hold. 7 The Heisenberg picture Heisenberg developed a different formulation of quantum mechanics, originally based on matrices, in which the dynamical equations of Hamiltonian mechanics take a very similar quantum form. Since the timedependent dynamical variables of mechanics are replaced by operators in the quantum realm, those operators are time-dependent in the Heisenberg picture. We take a different approach from the historical development. In keeping with the parallel between active and passive transformations, we begin by placing the time dependence in the basis kets. Start with the time-dependent wave function and let the basis carry the time dependence. ψ, t ψ t, t ψ For this to agree with the Schrödinger picture result we set ψ, t 0 ψ and require Since ψ is arbitrary, t ψ Û t, t 0 ψ Û t, t 0 ψ and for the kets, Û t, t 0, t, t Û t, t 0, t 0 Notice that while in the active Schrödinger picture states rotated with the action of Û t, t 0 e i Ĥt t0, the basis transforms in the reverse sense with the adjoint, Û t, t 0 e i Ĥt t0. 8

9 7. Time dependent operators and the Heisenberg equation of motion Net, consider how this affects the definition of the position operator through its eigenvalue equation. At the initial time, we have X Multiplying on the left by Û t, t 0 and inserting the identity, Û t, t 0 Û t, t 0 this becomes Û XÛÛ Û Û XÛ, t, t We see that the time-dependent basis kets are eigenkets of the time-dependent operator, X t Û t, t 0 X t 0 Û t, t 0 A similar argument holds for any Schrödinger picture observable,  S, and we let all operators change their time-dependence in this same way,  H t Û t, t 0 ÂSÛ t, t 0 Notice that ÂS may or may not already have some additional eplicit time dependence. From the Schrödinger equation for the time translation operator, eq., we immediately have dâh t Û Â S Û + Û ÂS Û + Û Û Â S i ÂS Ĥ t ÂSÛ + Û Û i  S Ĥ t Û Û Û i ] Ĥ t, Û ÂS Û + Û ÂS Û Since Û commutes with the Hamiltonian, i ] i ] Ĥ t, Û ÂS Û Ĥ t,  H Writing the last term as Û Â Â Û we have the Heisenberg equation of motion, S H dâh t i Ĥ t,  H t] +  t H This has eactly the form of the classical Hamiltonian equation of motion with the Poisson bracket {H, A} replaced by i times the commutator, i Ĥ, Â]. If a operator ] has no time dependence in the active picture ÂSt 0, and commutes with the Hamiltonian Ĥ,  H 0, then it is conserved dâh t 0 again resembling the classical conservation result. 7. Ehrenfest s Theorem In the Heisenberg picture, let the Hamiltonian be given by Ĥ m + V X 9

10 Then the rate of change of the momentum operator is given by d and we show in the net section that, V X], Ĥ] i, i m + V, V X] i X ] i V X Therefore, we reproduce an operator form of the classical equation of motion, d V X This holds only in the Heisenberg picture, but if we take epectation values, these are the same in both the Schrödinger and Heisenberg pictures, and we have the Ehrenfest theorem, d V X 3 In the special case of a free particle, V 0 and we may integrate, d 0 t 0 The position operator follows from so integrating d X ] X, i m m t m 0 Notice that and there is an uncertainty relation X t X 0 + m 0 t Xi 0, X j t] Xi 0, m j 0] t it m δ ij X t X t 0 4m Any initial spread in the epectation of the position therefore grows in time. 0

11 8 Some useful commutation relations First, consider the commutator of the position operator with a power of the momentum operator: ]  k X, k Using the identity Â, BĈ]  BĈ BĈ  BĈ BÂĈ + BÂĈ BĈ ] Â, B Ĉ + B Â, Ĉ] and writing k as the product k this becomes ]  k X, k ] X, k + Âk k + i ] k X, This gives a recursion relation Âk Âk k + i for any Âk in terms of Âk. iterating, we have  k Âk + i Âk + i Âk + i Âk 3 + i k k k k 3 Âk 3 3 k + 3i and we suspect that continuing in this way will lead to + i + i  k ik ] To prove this conjecture, notice that it works for k since  X, i. Now suppose Ân i n n. Then the recursion relation gives k  n Ân n + i i n in n n k n + i completing the proof by induction. Now, consider the commutator of the position operator with a function of the momentum operator: X, f ] k where we define a function of an operator to be the Taylor series for the function, with the variable replaced by the operator. First restricting our attention to the -direction, we only need the Taylor series in the form f p, p y, p z k0 k f 0, p y, p z p k k! p

12 since X commutes with y and k z. Setting p f f k, and replacing p, Therefore, f k0 k! f k 0, y, z k X, f ] k0 k0 k0 k! f k 0, y, ] z X, k k! f k 0, y, k z ik k! f k 0, y, k z i Substituting back into our original commutation relation, X, f ] k0 k0 i k! f k 0, y, ] z X, k k! f k 0, y, k z ik k0 k! f k 0, y, z k and we recognize the Taylor series for f p. Therefore, X, f ] f i p The corresponding result holds for each component, so the derivative becomes a gradient, X, f ] i f Eercise: Prove that, f X] i Xf