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Why isn’t time-dependent Schrodinger equation separable?

  1. Feb 27, 2012 #1
    Greetings everyone,

    I haven’t done any quantum in a while, and was reviewing my textbook, Griffiths Ed. 1. The form of the Schrodinger equation I’m using is:

    i[itex]\hbar[/itex][itex]\partial[/itex][itex]\Psi[/itex]/[itex]\partial[/itex]t = -[itex]\hbar[/itex]2/2m * [itex]\partial[/itex]2[itex]\Psi[/itex]/[itex]\partial[/itex]x2 + V[itex]\Psi[/itex]

    The book says if V is a function of x only, then the wave function is separable. This makes sense to me, but what I wanted to know is, why is it not generally separable otherwise? My understanding is that any linear homogeneous equation is separable (could someone please verify that?), so that if L([itex]\Psi[/itex]) = 0, then [itex]\Psi[/itex] should be separable, where L is a linear operator. Referring to the Schrodinger equation, L would be:

    i[itex]\hbar[/itex][itex]\partial[/itex]/[itex]\partial[/itex]t + [itex]\hbar[/itex]2/2m * [itex]\partial[/itex]2/[itex]\partial[/itex]x2 - V.

    I’m not going to worry about the first two terms right now, since it seems obvious enough that they're linear. By definition of linearity:

    L(c1u1 + c2u2) = c1L(u1) + c2L(u2)

    It seems like V(x, t) should be a linear operator, since as far as I can tell it won't behave so much as an operator, but instead as a function that just gets multiplied by given solutions of the wave function:

    V(x, t)(c1[itex]\Psi[/itex]1 + c2[itex]\Psi[/itex]2) = c1V(x, t)[itex]\Psi[/itex]1 + c2V(x, t)[itex]\Psi[/itex]2

    This seems like it should be true by basic algebra to me – what’s my mistake here? It seems like I'm either applying the definition of linearity wrong, or the potential function can, in fact, be something that will actually "affect" the wavefunction it operates on, like something ending with partials - but this doesn't make sense to me, since the potential function should just give numbers for the potential energy.

    A simple example of a potential function for which the above doesn’t work and/or an explanation of why I'm using the definition of linearity wrong would be very much appreciated.

    Thanks very much for any help you can give.

    -HJ Farnsworth
     
  2. jcsd
  3. Feb 27, 2012 #2
  4. Feb 28, 2012 #3
    Thank you for the references - but unfortunately I still do not understand what I did wrong.

    If an equation is linear and homogeneous, am I right in thinking it is separable as well? (I just got this from a textbook, but I'm not sure if it was referring to a specific example or the general case).

    If so, what was my mistake? I want to try to understand this in the scope of what I did above, so I can see what I did wrong, in addition to how to do it right.

    Thanks for the help.

    -HJ Farnsworth
     
  5. Feb 28, 2012 #4

    kith

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    Science Advisor

    That's not right, because of the dispersion relation: see here.

    I don't have much time, so just a few thoughts:
    -separable means that ψ(x,t) can be written as a product χ(t)φ(x)
    -try this separation ansatz for the potential in question; if you can't produce an equation of the structure L1(x)φ(x)=L2(t)χ(t), the equation is not separable
    -regardless of V, ψ(x,t) can only be written as χ(t)φ(x) if it is an eigenfunction of the Hamiltonian; so this quote is misleading:

     
  6. Feb 28, 2012 #5
    These are just random references that use seperability in completely irrelevant ways. The first one is about seperability of a metric space, the second one is about spatial seperability in relation to quantum entanglement. Here we are talking about differential equations that have seperable solutions. Unfortunately, in math and science we often use the same word to refer to totally different things.
     
  7. Feb 28, 2012 #6
    Thanks for the responses.

    So, my mistake was just in thinking that linearity and homogeneity of an equation implied that it was separable. This is in fact not the case, so the stuff I wrote afterwards was correct, but irrelevant. Is this right?

    -HJ Farnsworth
     
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