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Schrodinger equation

  1. Feb 5, 2005 #1
    can you explain this statement "if psi is a solution of a schrodinger equation, then so is kpsi, where k is any constant".

    why is that multplying psi by a constant does not its value?
     
  2. jcsd
  3. Feb 5, 2005 #2

    dextercioby

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    Try to see whether that's true or not by simply plugging the new state vector into Schroedinger's equation.It shouldn't be too hard...

    Daniel.
     
  4. Feb 6, 2005 #3

    jtbell

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    Just to remind you, Schrödinger's equation in one spatial dimension looks like this:

    [tex]- \frac {\hbar^2} {2 m} \frac {\partial^2 \Psi} {\partial x^2} + V \Psi = i \hbar \frac {\partial \Psi} {\partial t} [/tex]

    All the terms contain either psi itself, or one of its derivatives, therefore we call this a homogeneous differential equation. Also, no term contains powers of psi or of one of its derivatives, or combinations of psi and its derivatives, therefore we call this a linear differential equation.

    The fact that Schrödinger's equation is linear and homogeneous guarantees that if if a particular psi is a solution, then k*psi is also a solution.
     
  5. Feb 6, 2005 #4

    dextercioby

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    How about
    [tex] \frac{d|\Psi\rangle}{dt}=\frac{1}{i\hbar}\hat{H}|\Psi\rangle [/tex]

    ,where [itex] \hat{H} [/itex] is a densly-defined,self adjoint LINEAR operator (in agreement with the second principle)... :wink:

    Daniel.
     
    Last edited: Feb 6, 2005
  6. Feb 6, 2005 #5
    If you look at the equation dextercioby has put up (that is the most general form of Schrodinger's equation), you will see that both operators that act upon [tex]|\psi\rangle[/tex] are linear. We have the result that, for linear operators

    [tex]\hat{L}(\lambda\mathbf{a} + \mu\mathbf{b}) = \lambda\hat{L}\mathbf{a} + \mu\hat{L}\mathbf{b}[/tex]

    What this means is that in the Schrodinger equation, if you replace [tex]|\psi\rangle[/tex] by [tex]k|\psi\rangle[/tex] you can take all the [tex]k[/tex]s out and so cancel them.
     
  7. Feb 6, 2005 #6

    dextercioby

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    I emphasized the word "linear" and left to the OP to see what implications that fact would have.After all,it's better for him to figure out things by himself,as i think that would give him a feeling of satisfaction,too... :wink:


    Daniel.
     
  8. Feb 6, 2005 #7

    selfAdjoint

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    Daniel, I certainly agree with your reasoning here, but I humbly suggest that in that case the thread be moved to Homework. I sort of like to see different posters working things out one by one down a thread on this QM thread. Just my prejudice though.
     
  9. Feb 6, 2005 #8

    dextercioby

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    SA,if u want u can move it to HM section,no problem on my behalf. :smile:

    But,in fact,they are.Maybe my posts are too numerous,but i can't help it... :wink: :tongue2:

    Daniel.
     
  10. Feb 6, 2005 #9
    Yes I do understand the need of the person to work it out for himself. But since we received no acknowledgement of his/her understanding the problem, I thought I should explain it in detail.
     
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