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I Schrodinger equation and stationary states

  1. Mar 17, 2016 #1
    Is the solution of the time-independent schrodinger equation always a stationary state?
    Can it be non-stationary?
     
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  3. Mar 17, 2016 #2

    DrClaude

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    Eigenstates of an Hamiltonian are always stationary with respect to time evolution with that same Hamiltonian. It should take you three lines (at most) to prove it.
     
  4. Mar 17, 2016 #3

    dextercioby

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    With regards to the 1st question (whose answer automatically implies the 2nd), I don't approve of the phrase "time-independent schrodinger equation" (not because Schrödinger is misspelled), and I invite you to read the very definition of a (pure quantum) stationary state.
     
  5. Mar 18, 2016 #4

    PeroK

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    Instead of Schrödinger's cat, you could have Schrödinger's Umlaut. You don't know whether it's there or not until you look!
     
  6. Mar 19, 2016 #5
    Can we name this equation in general [itex]i\hbar\frac{\partial}{\partial t}|\psi \rangle=\hat H|\psi \rangle[/itex] a "time-independent Schroedinger equation"?
    Noting that I choose here the potential V(r) independent of time.
    I think in general we can't.
     
  7. Mar 19, 2016 #6

    PeroK

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    The partial derivative with respect to time is not consistent with "time independence"!
     
  8. Mar 19, 2016 #7
    If you choose [itex]|\psi \rangle[/itex] here to be the eigenstate of [itex]\hat H[/itex] then the solution of the equation will be a stationary state, and the probability density of a stationary state is independent of time.If you choose [itex]|\psi \rangle[/itex] to be the eigenstate of [itex]\hat H[/itex], the Schroedinger equation will reduce to the form [itex]\hat H|E\rangle=E|E\rangle[/itex]. Which is the time independent Schroedinger equation.
    So we can say also that [itex]i\hbar\frac{\partial}{\partial t}|\psi\rangle=\hat H|\psi\rangle[/itex] is not "time dependent" in general.
     
    Last edited: Mar 19, 2016
  9. Mar 19, 2016 #8

    vanhees71

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    Well, you have to assume that the Hamiltonian is not explicitly time dependent. Then it's a one-liner, indeed.
     
  10. Mar 19, 2016 #9

    PeroK

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    You have a differential equation that you solve by separation of variables. That leads to two differential equations: one for the function of time (that is position-independent) and one for the function of position (that is time independent).

    Now, if the potential is time-independent, then the time equation is standard and you get the same time function for all (time-independent) potentials. Namely: ##e^{-i Et/ \hbar}##, where ##E## is a constant and turns out to be the energy level associated with a particular solution.

    And, this means that the time-independent equation for the function of position is where all the action is, because that equation depends on the potential. Moreover, it turns out to be an equation for eigenfunctions of the Hamiltonian:
    ##H \psi(x) = E \psi(x)##.
    And that equation can't have any time derivatives in it. That would make no sense whatsoever.
     
  11. Mar 19, 2016 #10

    PeroK

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    PS It's just my opinion, but this is the problem with using shorthand notation before you have mastered what you are doing. It's easy to write:

    ##i\hbar\frac{\partial}{\partial t}|\psi\rangle=\hat H|\psi\rangle##

    And see nothing wrong. But if you write:

    ##i\hbar\frac{\partial}{\partial t}\psi (x)=\hat H\psi(x)##

    Then, you can see at a glance that something is wrong.
     
  12. Mar 19, 2016 #11

    vanhees71

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    What should be wrong? It should only be made very clear that you have two different formalisms for the same thing.

    The first equation is the equation of motion of the vector representing a pure state in the Schrödinger picture of time evolution, and in the 2nd case you have written the Schrödinger equation for the wave function in the position representation. The connection between the equations is that
    $$\psi(x)=\langle x|\psi \rangle.$$
    The symbol ##\hat{H}##, the Hamilton operator, has different meanings in the two equations. In the first equation it's an abstract self-adjoint operator on the abstract Hilbert space, and in the second equation it's the Hamilton operator in the position representation.

    The most simple approach for general calculations is the abstract formulation. An energy eigenvector is defined by the equation
    $$\hat{H} |u_{E} \rangle=E |u_E \rangle.$$
    Assuming that ##\hat{H}## is not time dependent. This implies that if you prepare the system at time ##t=0## in such an eigen vector of the Hamiltonian, it stays in this state. Indeed, in this case the time-evolution equation in the Schrödinger picture is solved by
    $$|\psi(t) \rangle=\exp \left (-\frac{\mathrm{i}}{\hbar} t E \right ) \rangle |u_E \rangle.$$
    Indeed we have
    $$\mathrm{i} \hbar \partial_t |\psi(t) \rangle = E \exp \left (-\frac{\mathrm{i}}{\hbar} t E \right) |u_E \rangle = \hat{H} \exp \left (-\frac{\mathrm{i}}{\hbar} E t \right ) |u_E \rangle = \hat{H} |\psi(t) \rangle.$$
    Now, the ##\exp(-\mathrm{i} t/\hbar)## is only a phase factor and thus does not change the state, because a (pure) state is not defined by the state ket itself but any state ket that's just multiplied by a phase factor represents the same state, i.e., the state is represented by a whole ray in Hilbert space. In this sense the time evolution doesn't change the state, if it is initially an eigenstate of the Hamiltonian (provided it is time independent).
     
  13. Mar 19, 2016 #12
    [itex]\psi(r)[/itex] can depend on time explicitly as it is can be like this[itex]\psi(r)=\langle r|E\rangle e^{-iEt/\hbar}[/itex]where[itex]|\psi\rangle=|E(t)\rangle[/itex]
    @PeroK @vanhees71
    Here now what is really confusing me:
    We know that the solution of [itex]i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle=\hat H|\psi(t)\rangle[/itex] may not be a stationary state wave function.
    But there is a proof that as [itex]\hat H[/itex] is hermitian, the probability density of the wave function (solution of this Schroedinger equation) [itex]|\psi(t)\rangle[/itex] will be conserved.Which means that any[itex]|\psi(t)\rangle[/itex](the solution of this equation) is stationary !!
    what goes wrong here?
     
  14. Mar 19, 2016 #13

    vanhees71

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    It's not the probability density that is conserved but the norm of the vector, i.e., the total probability adds up to 1 at any time as it must. That's very easy to verify in the abstract formalism:
    $$\frac{\mathrm{d}}{\mathrm{d} t} \langle \psi(t)|\psi(t) \rangle=(\frac{\partial}{\partial t} \langle \psi(t))|\psi(t) \rangle + \langle \psi(t) |\partial_t \psi(t) \rangle.$$

    Now you have
    $$\mathrm{i} \hbar \partial_t |\psi(t) \rangle=\hat{H} |\psi(t) \rangle \; \Rightarrow \; -\mathrm{i} \hbar \partial_t \langle \psi(t)| = \langle \psi(t) |\hat{H}^{\dagger} = \langle \psi(t) |\hat{H}.$$
    So you have
    $$\frac{\mathrm{d}}{\mathrm{d} t} \langle \psi(t)|\psi(t) \rangle=\frac{i}{\hbar} (\langle \psi(t)|\hat{H}|\psi(t)-\langle \psi(t)|\hat{H} \psi(t) \rangle)=0.$$
    Of course, the wave function is a function of ##t## and ##\vec{x}##. The (generalized) position eigenstate is time-independent (in the here used Schrödinger picture). So you have
    $$\psi(t,\vec{x})=\langle \vec{x}|\psi(t) \rangle.$$
    As I showed above, in the case of a time-independent Hamiltonian, if the state ket is an eigenvector of the Hamiltonian (i.e., an energy eigenstate) at ##t=0##, the time dependence of the state is only multiplying this state by the factor ##\exp(-\mathrm{i} E t/\hbar)##, but multiplying a vector, representing the state of the system, by a factor describes the same state of the system.

    The meaning of the state ket is that multiplying it with a (generalized) eigenvector of an operator representing an observable (e.g., the position) gives the corresponding wave function, e.g., ##\psi(t,\vec{x})##. Now the physical meaning of this wave function is that its modulus squared gives the probability distribution to find the particle's position to be at ##\vec{x}##, i.e., the only physical meaning is in
    $$P(t,\vec{x})=|\psi(t,\vec{x})|^2.$$
    Now in the case that ##\psi(t=0,\vec{x})=u_E(\vec{x})## I've shown in my previous posting that ##\psi(t,\vec{x})=\exp(-\mathrm{i} t E/\hbar) u_E(\vec{x})##, but then
    $$P_(t,\vec{x})=|u_E(\vec{x})|^2,$$
    and this is constant in time, i.e., in this case you have a time-independent (or stationary) state. This is why the eigenvectors of the Hamiltonian and only those vectors represent the stationary pure states of the system.
     
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