All states are stationary, all observables are constant.


by hearnden
Tags: constant, observables, states, stationary
hearnden
hearnden is offline
#1
Jan21-12, 07:36 PM
P: 2
The following lecture from University of Oxford contains an explanation of the constancy of probability distributions for all observables when a system is in a stationary state: http://www.youtube.com/watch?v=0yvX4jhzblY#t=15m35s. However, the derivation of the vanishing amplitude does not actually use the fact that the state is stationary, leading to the conclusion that all observables have fixed probability distributions for all states, stationary or otherwise. Clearly this is false, but where is the error in the maths?

For a system in state [itex]\psi[/itex], the probability amplitude [itex]A[/itex] of measuring the value of an observable [itex]Q[/itex] as [itex]q_n[/itex] (the [itex]n[/itex]th eigenvalue of [itex]Q[/itex]) is [itex]\langle q_n|\psi\rangle[/itex]. The time derivative of this amplitude goes as:
\begin{align}
i\hbar\frac{\partial A}{\partial t}
= i\hbar\frac{\partial \langle q_n|}{\partial t}|\psi\rangle
+ i\hbar\langle q_n|\frac{\partial |\psi\rangle}{\partial t}
\end{align}
From the Schrödinger equation applied to [itex]|\psi\rangle[/itex], [itex]i\hbar\frac{\partial |\psi\rangle}{\partial t} = H|\psi\rangle[/itex], and from its conjugate form applied to [itex]\langle q_n|[/itex], [itex]-i\hbar\frac{\partial \langle q_n|}{\partial t} = \langle q_n|H[/itex]. So
\begin{align*}
i\hbar\frac{\partial A}{\partial t}
&= -\langle q_n|H|\psi\rangle + \langle q_n|H|\psi\rangle \\
&= 0
\end{align*}
This has concluded that for any state [itex]\psi[/itex], the probability amplitude for an observable [itex]Q[/itex] is constant in time. Clearly this is false, but which step in the derivation was dodgy?
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The_Duck
The_Duck is offline
#2
Jan21-12, 11:44 PM
P: 790
I think you're right that the calculation is nonsensical.

I believe for this calculation you should take |qn> to be a time-independent state vector. If you let|qn> evolve with time it is in general going to evolve into something that is no longer an eigenstate of Q, in which case computing <qn|psi> is not relevant to the results of measuring the observable Q.

Taking |qn> to be time independent the first term drops out and we can solve the resulting differential equation to get (in the case of a stationary state of energy E)

A(t) = exp(-i E t/hbar) A(t=0)

i.e., the amplitude only changes by a phase, so the probability of obtaining any given value of Q doesn't change with time, because the probability is given by the square-modulus of the amplitude.
Ken G
Ken G is offline
#3
Jan22-12, 08:26 AM
PF Gold
P: 3,072
I'd say the problem is that using the "conjugate form" is only true if qn is an energy eigenfunction, but I don't see just why! Perhaps they should really have used H* in that expression, so it only works if H is acting like a real operator, i.e., acting on an energy eigenstate.

kith
kith is offline
#4
Jan22-12, 09:43 AM
P: 684

All states are stationary, all observables are constant.


I agree with The_Duck's second paragraph.

Maybe it is insightful to compare the Schrödinger picture with the Heisenberg picture for this. At t=0, we prepare a system in state |ψ(0)>. We want to do a measurement of observable Q(0) with eigenvalues |qn(0)>. Before we do this, we let the system evolve from time 0 to time t. The time-dependent probability amplitudes <qn|ψ>t can now be interpreted in two ways:

1) <qn(0)|ψ(t)>
This is the approach in the Schrödinger picture. Our state evolves from initial state |ψ(0)> to final state |ψ(t)>. Then we ask "what is the probability to find the system in the state |qn(0)>?". The time derivative of this specific state is zero, because it doesn't depend on time.

2) <qn(t)|ψ(0)>
This is the approach in the Heisenberg picture. Our state remains constant, but our observable evolves from Q(0) to Q(t). Then we ask "what is the probability to find the system in the state |qn(t)>?". This state depends on the time, we do the measurement.

So in this sense, one can say that the base kets |qn> are constant in the Schrödinger picture and time-dependent in the Heisenberg picture. This gets obvious, if one looks at the eigenvalue equations Q(0)|qn(0)>=qn(0)|qn(0)>, which doesn't change in time, and Q(t)|qn(t)>=qn(t)|qn(t)>, which does.
Ken G
Ken G is offline
#5
Jan22-12, 01:26 PM
PF Gold
P: 3,072
Ah, I see, The_Duck and kith have made it clear. What that proof actually shows is that <ψ1(t)|ψ2(t)> does not vary in time, but that does not imply that the outcomes of any measurements (that don't correspond to stationary eigenstates) will not vary in time, because ψ1(t) will not in general remain an eigenstate of the observable, as was said above. Indeed, more generally we can say that <ψ1(t)|A|ψ2(t)> does not vary in time, for any operator A that has no explicit t dependence and commutes with H, by direct application of the argument in that proof. If A does not commute with H, the argument fails. Indeed, the argument is a direct demonstration of the well-known result
d/dt(<ψ1(t)|A|ψ2(t)> ) = i/hbar*(<ψ1(t)|HA-AH|ψ2(t)> ), and that certainly also works if we choose ψ1(t)=ψ2(t)=ψ(t), showing that expected results of measurement A will indeed vary if A does not commute with H.
hearnden
hearnden is offline
#6
Jan22-12, 06:59 PM
P: 2
Thanks everyone, that's explained the error quite well.

I think the intent of the "proof" in the context of that lecture was to demonstrate that stationary states are not physically realizable, because in that state nothing ever changes (all observables have constant expectation values). It probably should have gone in the direction The_Duck took it, but perhaps the lecturer got lost and had to handwave his way out.

The lesson that I took out of that exercise is that the property of remaining-an-eigenstate and the property of evolving-according-to-Schrödinger's-equation are, in the general case, not necessarily compatible. Only observables that commute with H have eigenstates that both remain eigenstates throughout time and also evolve according to Schrödinger's-equation.

Further, as illustrated by kith, it's very easy for time information to be lost in conventional ket notation, so you just have to be careful and keep track of which of several possible type signatures applies to |q>. It could be just a spatial function q(x), a space/time function q(x, t), or a snapshot of a space/time function at a particular time q(x, 0). It's left to semantics rather than syntax to disambiguate: the notation does not help do it for you.
Ken G
Ken G is offline
#7
Jan22-12, 08:10 PM
PF Gold
P: 3,072
Quote Quote by hearnden View Post
I think the intent of the "proof" in the context of that lecture was to demonstrate that stationary states are not physically realizable, because in that state nothing ever changes (all observables have constant expectation values).
I hope not! It is very important to quantum mechanics that stationary states are physically realizable, this is used all the time. The proof should have been used to show that the expectation value of any observable that commutes with the Hamiltonian (in particular, the energy), doesn't change under the evolution of the Schroedinger equation. It can also be used to show that observables that don't commute with Hamiltonian do have changes in their expectation values, even during unitary evolution. Of course measurement is usually considered to be a break from unitary evolution, and the evolution during measurement is not governed by the Schroedinger equation, so subsequent measurements can be affected.

Only observables that commute with H have eigenstates that both remain eigenstates throughout time and also evolve according to Schrödinger's-equation.
Yes, that's the key lesson.


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