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Position variables in the wave function

  1. Feb 3, 2014 #1
    I would like some clarification as to the mathematical and physical definitions of the position variables in the wave function. I often see that it is treated as a variable independent of time; this is utilized in the separation of variables technique. However, the Schrodinger equation implies that this cannot be so. My argument (for the one-dimensional case) is as follows:

    We can rearrange the Schrodinger equation to show that [itex](\hat{H}- i\hbar \frac{\partial{}}{\partial{t}})\psi = 0[/itex]. Since [itex]\psi[/itex] is not identically zero, it follows that we can say that [itex]\hat{H} = i\hbar \frac{\partial{}}{\partial{t}}[/itex] for all solutions to the Schrodinger equation. One consequence of this is that commutation relations between these two operators and a third must produce equivalent results; we will examine commutation with [itex]\hat{x}[/itex]. On one hand, we have:
    [itex][\hat{x}, i\hbar \frac{\partial{}}{\partial{t}}]\psi = x i\hbar \frac{\partial{}}{\partial{t}}\psi - i\hbar \frac{\partial{}}{\partial{t}}(x\psi)[/itex]
    [itex]= x i\hbar \frac{\partial{}}{\partial{t}}\psi - x i\hbar \frac{\partial{}}{\partial{t}}\psi - \psi i\hbar \frac{\partial{}}{\partial{t}}x[/itex]
    [itex]= -\frac{\hbar}{i}\psi \frac{\partial{x}}{\partial{t}}[/itex].
    On the other hand:
    [itex][\hat{x}, \hat{H}]\psi = x(-\frac{\hbar^2}{2m} \frac{\partial^2{}}{\partial{x}^2} + V)\psi - (-\frac{\hbar^2}{2m} \frac{\partial^2{}}{\partial{x}^2} + V)(x\psi)[/itex]
    [itex]= -x\frac{\hbar^2}{2m} \frac{\partial^2{\psi}}{\partial{x}^2} + Vx\psi + \frac{\hbar^2}{2m} \frac{\partial^2{(x\psi})}{\partial{x}^2} - Vx\psi = -x\frac{\hbar^2}{2m} \frac{\partial^2{\psi}}{\partial{x}^2} + \frac{\hbar^2}{2m} \frac{\partial^2{(x\psi})}{\partial{x}^2}[/itex]
    [itex]= -x\frac{\hbar^2}{2m} \frac{\partial^2{\psi}}{\partial{x}^2} + x\frac{\hbar^2}{2m} \frac{\partial^2{\psi}}{\partial{x}^2} + \frac{\hbar^2}{2m} \frac{\partial{\psi}}{\partial{x}} + \frac{\hbar^2}{2m} \frac{\partial{\psi}}{\partial{x}} = \frac{\hbar^2}{m} \frac{\partial{\psi}}{\partial{x}} = \frac{i\hbar}{m}\hat{p}\psi[/itex]
    From the equality of the operators, we must have:
    [itex]\frac{i\hbar}{m}\hat{p}\psi = -\frac{\hbar}{i}\psi \frac{\partial{x}}{\partial{t}} \rightarrow \hat{p}\psi\ = m\psi \frac{\partial{x}}{\partial{t}}[/itex].

    From this, it follows that if the position has no explicit time dependence, then the application of the momentum operator to any wave function must yield zero in all cases. This is clearly inconsistent with reality. This brings me to my questions: Where did I go wrong in the argument? If I did not make a mistake, how is the position variable defined and how are its use with the separation of variables technique and its explicit time dependence reconcileable?
     
    Last edited: Feb 3, 2014
  2. jcsd
  3. Feb 3, 2014 #2

    Avodyne

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    Science Advisor

    No: this is true only if acting with the third operator on a solution of the Schrodinger equation yields another solution of the Schrodinger equation. If it does not, then acting with the third operator takes you out of the space of functions on which [itex]\hat{H} = i\hbar \frac{\partial{}}{\partial{t}}[/itex] is true.

    In particular, if ##\psi(x,t)## is a solution of the Schrodinger equation, then in general ##x\,\psi(x,t)## is not.
     
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