Stationary States and time-independent states (aren't they the same?)

[MrMultiMedia]

I always thought they were the same, but now I am reading a question that says "which of he following time-independent functions describe stationary states of the corresponding quantum systems?"

Is there something I am missing? It's written like there is something to solve, but to me it seems like a trick question and all I really have to write is "if they are all time independent functions, then they all describe stationary states." Would this be a correct assumption?

 

[The_Duck]

Not all time-independent functions satisfy the time-independent Schrodinger equation. Probably the question wants you to say which of the functions are actually solutions to the time-independent Schrodinger equation.

 

[tom.stoer]

The time-dep. Schrödinger eq. reads

$$i\partial_t\,|\psi,t\rangle = \hat{H}\,|\psi,t\rangle$$

a stationary state is an eigenstate of the Hamiltonian, i.e.

$$\hat{H}\,|\psi,t\rangle = E_\psi\,|\psi,t\rangle$$

An example would be a state in an hydrogen atom labelled by nlm

$$|\psi,t\rangle = |nlm,t\rangle$$

Note that this state is not time-independent.

But the time dependency is "trivial" as can be seen by the usual ansatz

$$|\psi,t\rangle = e^{-iEt}|\psi,0\rangle$$

which solves the time-dep. Schrödinger eq. provided that we use an (time-indep.) eigenstate

$$|\psi,0\rangle$$

of the Hamiltonian. This state is time-independent, but it is not a solution of the time-dependent Schrödinger eq. but of the time-independent Schrödinger eq.

In which sense is

$$|\psi,t\rangle = e^{-iEt}|\psi,0\rangle$$

time-dependent but "stationary"?

The state defines a one-dim. subspace of the Hilbert space, and the time-evolution does not leave this subspace; the time-evolution is described by a "trivial phase factor" e^-iEt therefore the time-dependency does not change the "direction" of the state vector. In that sense the direction is stationary.

 

[Alesak]

So, if I understand it correctly, if we have time-indep potential we can use time-indep Schrödinger equation(=eigenvalue problem for Hamiltonian) to derive set of solutions for t=0, from which any other state can be formed(=they form orthonormal basis) and which can be developed into any later time by expression

$$|\psi,t\rangle = e^{-iEt}|\psi,0\rangle$$

Correct?


What is time-indep state anyway? I guess it could be state which does not change with time, but it would be probably rare case.

I've found "www.mit.edu/~tokmakof/.../1._Introduction_3-15-10.pdf" useful link on the topic.

In which sense is

$$|\psi,t\rangle = e^{-iEt}|\psi,0\rangle$$

time-dependent but "stationary"?

The state defines a one-dim. subspace of the Hilbert space, and the time-evolution does not leave this subspace; the time-evolution is described by a "trivial phase factor" e^-iEt therefore the time-dependency does not change the "direction" of the state vector. In that sense the direction is stationary.

In other words, probability amplitude is constant. It should also be stressed that linear superposition of two stationary states is not stationary.

 

[tom.stoer]

What is time-indep state anyway? I guess it could be state which does not change with time

If you mean "time-indep. w.r.t. the time-dep. SE" this means that you have to have an eigenvalue E=0.

It should also be stressed that linear superposition of two stationary states is not stationary.

This is true only if the two states are not degenerate. Think about the hydrogen atom and the quantum numbers nlm. A state like a|nlm>+a'|n'lm> for n≠n' is not an energy eigenstate; but a state like a|nlm>+a'|nlm'> is. Therefore the latter state is stationary.