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Solutions of of the time-independent Schrodinger eq.

  1. Oct 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Supposed that [tex]\psi[/tex]1 and [tex]\psi[/tex]2 are two different solutions of the TISE with the same energy E.

    a) show that [tex]\psi[/tex]1 + [tex]\psi[/tex]2 is also a solution with energy E.
    b) show that c*[tex]\psi[/tex]1 is also a solution with energy E.

    2. Relevant equations

    TISE: (-[tex]\hbar[/tex]/2m)*[tex]\nabla[/tex]^2*[tex]\psi[/tex](r) +V(r)*[tex]\psi[/tex](r) = E*[tex]\psi[/tex](r)

    3. The attempt at a solution

    My first thought was to solve by substitution but since it doesn't give us the 2 solutions as a function I can't do that. Should I just find 2 different solutions and plug them in to the TISE to see if they satisfy? I feel I am missing a rule or a postulate to solve this with.

    P.S. I am very new to QM, be gentle
     
  2. jcsd
  3. Oct 2, 2010 #2

    fzero

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    If [tex]\psi_1[/tex] is a solution of the TISE with energy E, what must

    [tex]\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_1(r) [/tex]

    equal? You don't need to know the functional form of [tex]\psi_1(r)[/tex] to answer this.
     
  4. Oct 2, 2010 #3
    [tex]\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_1(r)[/tex] = E*[tex]\psi[/tex]1

    I don't see what this proves though, isn't the equation just re-arranged?
     
  5. Oct 2, 2010 #4

    fzero

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    Well it's reiterating what information you're given. The problem wants you to compute

    [tex]\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) (\psi_1(r)+\psi_2(r))[/tex]

    and

    [tex]\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) ( c \psi_1(r)).[/tex]
     
  6. Oct 2, 2010 #5
    [tex]\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_1(r) = E\psi_1(r)[/tex]
    [tex]\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_2(r) = E\psi_2(r)[/tex]

    therefore [tex]\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) (\psi_1(r)+\psi_2(r)) = E(\psi_1(r)+\psi_2(r)) ?[/tex]

    It just doesn't feel like I'm proving anything.

    Edit*Perhaps I just have to show that E[f(x)] is a linear operator? (where f(x) would be [tex]\psi[/tex])
     
    Last edited: Oct 2, 2010
  7. Oct 2, 2010 #6

    fzero

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    It's not a really deep problem. You're just verifying that the wave equation is linear.

    E is a scalar, I don't think you'd be expected to prove that multiplication by a scalar is a linear operation.
     
  8. Oct 2, 2010 #7

    vela

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    You're supposed to show how what's after "therefore" follows from the first two lines.
     
  9. Oct 3, 2010 #8
    [tex]
    \left[\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_1(r) = E\psi_1(r)\right]+
    [/tex] [tex]
    \left[\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_2(r) = E\psi_2(r)\right]
    [/tex]
    [tex]
    \left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \left(\psi_1(r)+\psi_2(r)\right) = E\psi_1(r) + E\psi_2(r)
    [/tex]
    Since they both have the same E, it can be factored out:
    [tex]
    =\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \left(\psi_1(r)+\psi_2(r)\right) = E\left(\psi_1(r) + \psi_2(r)\right)
    [/tex]

    ??
     
    Last edited: Oct 3, 2010
  10. Oct 3, 2010 #9

    vela

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    It's clear you understand conceptually what's going on, but you want to write it down in a clear, organized way. A lot of students have trouble writing down a proof because it seems so obvious.

    To show [itex]\psi_1(\vec{r})+\psi_2(\vec{r})[/itex] is a solution means to show that

    [tex]\left[-\frac{\hbar}{2m}\nabla^2 + V(\vec{r}) \right] \left(\psi_1(\vec{r})+\psi_2(\vec{r})\right) = E(\psi_1(\vec{r})+\psi_2(\vec{r}))[/tex]

    For problems like this, it's good to start with one side of the equation, apply concepts one at a time, and show what you get after each step until you end up at the other side of the equation.

    [tex]\begin{align*}
    \left[-\frac{\hbar}{2m}\nabla^2 + V(r) \right] \left(\psi_1(r)+\psi_2(r)\right) &= \cdots \\
    & = \cdots \\
    && \vdots \\
    &= E(\psi_1(\vec{r})+\psi_2(\vec{r}))
    \end{align*}[/tex]

    In the first step, you'd use the linearity of

    [tex]\left[-\frac{\hbar}{2m}\nabla^2 + V(r) \right][/tex]

    In the next step, you'd use the fact that [itex]\psi_1[/itex] and [itex]\psi_2[/itex] are solutions with energy E, and so on.
     
  11. Oct 3, 2010 #10
    I understand now, thanks for your help!
     
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