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Homework Help: Another quick QM question

  1. Sep 28, 2010 #1
    For a particle in a potential V(x), calculate [H,p]

    [tex]H = \frac{-\hbar^2}{2m}\frac{\delta^{2}}{\delta x^{2}} + V(x)[/tex]
    [tex]p = -i\hbar \frac{\delta}{\delta x}[/tex]

    [tex][H,p] =[/tex]
    [tex] Hp - pH =[/tex]
    [tex]\frac{-\hbar^2}{2m}\frac{\delta^{2}}{\delta x^{2}} * -i\hbar \frac{\delta}{\delta x} + V(x)-i\hbar \frac{\delta}{\delta x} - -i\hbar \frac{\delta}{\delta x}\frac{-\hbar^2}{2m}\frac{\delta^{2}}{\delta x^{2}} - -i\hbar \frac{\delta}{\delta x}V(x) = [/tex]
    [tex]i\frac{\hbar^{3}}{2m}\frac{\delta^{3}}{\delta x^{3}} - V(x)ih\frac{\delta}{\delta x} - i\frac{\hbar^{3}}{2m}\frac{\delta^{3}}{\delta x^{3}} + i\hbar \frac{\delta}{\delta x}V(x) = [/tex]
    [tex]i\hbar \frac{\delta}{\delta x}V(x) - V(x)ih\frac{\delta}{\delta x}[/tex]

    Given [tex]\Delta A \Delta B \geq |<C>|/2[/tex] with [tex]|<C>| = [A,B][/tex], find
    [tex]\Delta A \Delta B[/tex]

    [tex]<C> = \int \Psi^{*}(i\hbar \frac{\delta}{\delta x}V(x) - V(x)ih\frac{\delta}{\delta x})\Psi dx[/tex]

    Can I go from there? Is any of this correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 29, 2010 #2

    fzero

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    It looks fine so far, but you need to remember this is an operator expression, so the derivative in the first term acts on everything to the RHS. Specifically, this means that

    [tex] [H,p]\psi(x) = i\hbar \frac{\delta}{\delta x}(V(x) \psi(x)) - V(x)ih\frac{\delta \psi(x)}{\delta x}[/tex]
    [tex] = i\hbar \left(\frac{\delta V(x)}{\delta x}\right) \psi(x) + i\hbar V(x) \frac{\delta \psi(x) }{\delta x} - V(x)ih\frac{\delta \psi(x)}{\delta x}[/tex]
    [tex] = i\hbar \left(\frac{\delta V(x)}{\delta x}\right) \psi(x) .[/tex]

    This is important for computing [tex]\langle [H,p] \rangle[/tex] in the 2nd part.
     
  4. Sep 29, 2010 #3
    alright, so the second part I would get:

    [tex]<C> = \int \Psi^{*} \Psi i\hbar \frac{\delta}{\delta x}V(x)dx = [/tex]
    [tex] i\hbar \int \Psi{*} \Psi \frac{\delta}{\delta x}V(x)dx[/tex]

    Anyway to simplify this?
     
  5. Sep 29, 2010 #4

    fzero

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    If you don't know the potential and wavefunction, I don't see how you could do more than just call this [tex]i\hbar\langle \delta V(x)/\delta x \rangle[/tex].
     
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