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

  1. Oct 28, 2012 #1
    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?
  2. jcsd
  3. Oct 28, 2012 #2
    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.
  4. Oct 29, 2012 #3


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    The time-dep. Schrödinger eq. reads

    [tex]i\partial_t\,|\psi,t\rangle = \hat{H}\,|\psi,t\rangle[/tex]

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

    [tex]\hat{H}\,|\psi,t\rangle = E_\psi\,|\psi,t\rangle[/tex]

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

    [tex]|\psi,t\rangle = |nlm,t\rangle[/tex]

    Note that this state is not time-independent.

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

    [tex]|\psi,t\rangle = e^{-iEt}|\psi,0\rangle[/tex]

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


    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

    [tex]|\psi,t\rangle = e^{-iEt}|\psi,0\rangle[/tex]

    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.
  5. Oct 29, 2012 #4
    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

    [tex]|\psi,t\rangle = e^{-iEt}|\psi,0\rangle[/tex]


    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" [Broken] useful link on the topic.

    In other words, probability amplitude is constant. It should also be stressed that linear superposition of two stationary states is not stationary.
    Last edited by a moderator: May 6, 2017
  6. Oct 29, 2012 #5


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

    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.
    Last edited: Oct 29, 2012
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