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Schrodinger's wave equation

  1. Mar 23, 2010 #1
    In what conditions do we use time dependent and time independent Schrodinger's wave equations?
     
  2. jcsd
  3. Mar 23, 2010 #2
    The time independent Schrodinger's wave equation is used only for a stationary state.
     
  4. Mar 23, 2010 #3
    The time-dependent version is always the right choice.

    However if your potential does not depend on time, but position only [itex]V(x)[/itex] (unlike [itex]V(x,t)[/itex]), then you can use the trial solution [itex]\psi(x,t)=\phi(x)e^{-i E t/\hbar}[/itex] and derive an equation for the special part [itex]\phi[/itex] of the wavefunction. You effectively get the time-independent version of the Schrödinger equation. From this time-indepedent special case you can first find [itex]\phi[/itex] and E and finally put it back into the full wavefunction [itex]\psi(x,t)[/itex]. Please try that above exercise with the trial solution.
     
  5. Mar 23, 2010 #4

    SpectraCat

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    I don't agree. It makes sense to use the time-dependent Schrodinger equation (TDSE) when solving a problem for a time-dependent potential. However quite often what one is most interested in is the energy spectrum and/or eigenstates of a time-independent potential. In such a case, it is preferable to solve the time-independent Schrodinger equation (TISE), either analytically (this is only rarely possible for meaningful real-world problems), or numerically (i.e. using the variational method and diagonalizing a large matrix).

    Once the eigenstates are known, one then gets the time-dependent wavefunctions for free, since the time-dependent part of the wavefunction for the jth eigenstate is just:
    [tex]f_j(t)=e^{-i\frac{E_j}{\hbar}t}[/tex]
    where Ej is just the jth eigenvalue. In fact, any time dependent wavefunction can be written as a linear combination of these solutions, since they form a complete basis.

    I guess it is largely a matter of taste, but I find the latter approach simpler to understand and use.
     
  6. Mar 23, 2010 #5
    It's not a matter of taste. What you want to use merely depends on whether your potential is [itex]V(x,t)[/itex] or [itex]V(x)[/tex]. What time-independent eigenstates do you want to find if you haven't even given a time-independent potential to deal with?

    Maybe you are refering to some perturbation theory approximations?
     
  7. Mar 23, 2010 #6

    SpectraCat

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    Read my first sentence again .. you said it is *always* best to use the TDSE. I said it is best to use the TISE when solving a time independent potential.

    To answer your question, yes, I would also tend to use the TISE first in cases where it is appropriate to treat the time-variant part of the potential as a (weak) perturbation.

    In cases where there is truly a strongly-coupled time-dependent potential to deal with, I would probably not use either form of the Schrodinger equation, but instead I would formulate solutions in the interaction picture, where quantum propagators are used to describe the time-evolution of some initial state, which I would represent as a linear combination of eigenstates. Of course that choice is largely due to my own education and training, and the sorts of problems I run into in my research. Honestly, outside of fairly straightforward derivations and pedagogic illustrations, I have found little use for the TDSE.

    I understand that others will have a different take .. I was just providing a counter-example to your statement that, "The time-dependent version is always the right choice."
     
  8. Mar 23, 2010 #7
    I wanted to emphasize that there is only one equation that describes the physics. Everything else is a special case. This is to prevent misconceptions about "dualities" and "special cases".
    I had the impression the author of the initial question was thinking there are two different equations and you have to make a lucky guess which one to use. I prefer to teach that there are no dualities, i.e. conflicting definitions in physics. Absolutely everything in physics can be transformed consistently into each other.

    You are right. I didn't mean to say that you should actually chose the most general equation to handle all problems.
     
  9. Mar 23, 2010 #8
    If you are interested in a) the possible results of an energy measurement and b) the probability distribution of those results, then you must solve the energy eigenvalue equation where the energy operator is the Hamiltonian [tex]\hat H = - {{\hbar ^2 } \over {2m}}{{\partial ^2 } \over {\partial x^2 }} + V(x)[/tex] in one dimension for a time independent potential. This energy eigenvalue equation is what we call the time independent Schrodinger equation.

    The solution gives us the energy eigenvalues [tex]E_n[/tex], which are the only possible results of an energy measurement. The state function, [tex]\psi (x)[/tex], determined by the state preparation procedure, must then be written in the energy representation where the basis functions [tex]\left| {E_n } \right\rangle[/tex] are the eigenfunctions of the Hamiltonian. Now [tex]\left| {\left\langle {{E_n }}
    \mathrel{\left | {\vphantom {{E_n } \psi }}
    \right. \kern-\nulldelimiterspace}
    {\psi } \right\rangle } \right|^2[/tex] gives us the probability of getting the energy value [tex]E_n[/tex] when we know the system is in state [tex]\psi (x)[/tex]. Note that time is not a relevant parameter here.

    However, all state functions will vary in time according to the time dependent Schrodinger equation [tex]i\hbar {{\partial \psi (x,t)} \over {\partial t}} = \hat H\psi (x,t)[/tex]. The probabilities might change with time, depending on the situation.
     
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