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Basic qm derivation

  1. Feb 6, 2005 #1
    Suppose that [tex]\psi (x)[/tex] is some solution to the time-independent Schrödinger equation;

    [tex]-\frac{h^2}{2m}\frac{\partial^2\psi(x)}{\partial x} + V(x)\psi(x) = E\psi(x)[/tex].​

    I want to show that if the potential V(x) is an even function, then [tex]\psi(-x)[/tex] is also a solution to same equation (same E and V).

    I know I'm supposed to combine the facts that [tex]\psi(x)[/tex] is a solution and that V(x) = V(-x), but I can't see how. I've noted that

    [tex]\frac{\partial^2\psi(-x)}{\partial x} = \frac{\partial^2\psi(x)}{\partial x}[/tex],​

    but that's pretty much it :confused:
     
  2. jcsd
  3. Feb 6, 2005 #2

    dextercioby

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    Your last formula contains 2 typo's.You may wanna repair it,because it's pretty important to the proof itself...

    Daniel.
     
  4. Feb 6, 2005 #3
    Try simply making the substitution [tex]x\to-x[/tex] in the SWE, then using the fact that [tex]V(-x)=V(x)[/tex]. The new form should then show directly that [tex]\psi(-x)[/tex] is a solution as well, since it satisfies the wave equation.
     
  5. Feb 6, 2005 #4

    dextercioby

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    Yes,the way it's written and the condition imposed upon the potential energy,then the total Hamiltonian is parity invariant and of course the parity operator and the Hamiltonian commute,ergo they admit a complete set of eigenvectors...End proof... :wink:

    Daniel.
     
  6. Feb 6, 2005 #5

    Galileo

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    The proof also needs to show that the energies are nondegenerate.
     
  7. Feb 6, 2005 #6

    dextercioby

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    What???Please,explain...I may be tired and i may not see it...

    Daniel.
     
  8. Feb 6, 2005 #7

    Galileo

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    Nevermind. I didn't read the actual question. I thought it said 'every solution to the SE is either even or odd'.
     
  9. Feb 6, 2005 #8
    I've tried the substitution-thing - that was my first approach, I couldn't make it work. I'm too tired now, maybe I'll work it out tomorrow - thanks for your replies.
     
  10. Feb 6, 2005 #9

    dextercioby

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    Weird,the way i see it,it's immediate... :rolleyes: Anyway,i see that u didn't noticed.
    [tex] \frac{\partial^{2} \psi}{\partial x} [/tex]

    is not correct.An essential "2" is missing...

    Daniel.
     
  11. Feb 6, 2005 #10
    Oh, yea, of course. My problem is that I don't know exactly what to end up with actually. Should I prove this:

    [tex]\frac{\partial^2\psi(-x)}{\partial x^2} + V(x)\psi(-x) = E\psi(-x)[/tex]​

    or this:

    [tex]\frac{\partial^2\psi(-x)}{\partial x^2} + V(-x)\psi(-x) = E\psi(-x)[/tex]​

    The essential difference being the minus in the potential function. Maybe it's because I'm tired (I am!), but I can't quite figure out?
     
  12. Feb 6, 2005 #11

    dextercioby

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    They're equal,because the potential is parity invarint,viz.
    [tex] V(x)=V(-x) [/tex]

    Daniel.
     
  13. Feb 6, 2005 #12
    Oh, yea, I'm going to lie down now :smile:
     
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