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Homework Help: Commutator problem with momentum operators

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the commutator [tex]

    2. Relevant equations
    [tex] \hat{p_{x}}=\frac{\hbar}{i}\frac{\partial}{\partial x} [/tex]

    [tex] \hat{p_{y}}=\frac{\hbar}{i}\frac{\partial}{\partial y} [/tex]

    3. The attempt at a solution
    [tex][\hat{p}_{x}, \hat{p}_{y}]=\hat{p}_{x}\hat{p}_{y}-\hat{p}_{y}\hat{p}_{x}[/tex]

    [tex]=\frac{\hbar}{i}\frac{\partial}{\partial x}\frac{\hbar}{i}\frac{\partial}{\partial y}-\frac{\hbar}{i}\frac{\partial}{\partial y}\frac{\hbar}{i}\frac{\partial}{\partial x}[/tex]

    [tex]=(\frac{\hbar}{i})^2\frac{\partial^2}{\partial x\partial y}-(\frac{\hbar}{i})^2\frac{\partial^2}{\partial y\partial x}[/tex]

    Is this the correct commutator? I don't know what else I can do to further complete this. Any help is appreciated.
  2. jcsd
  3. Jan 24, 2010 #2


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    It's fine. But partial derivatives commute in the physics sense. Your two partial derivatives are equal.
  4. Jan 24, 2010 #3
    Okay thanks for the response, i have a similar problem to that of the first that I put up, but I am very confused.

    Show that [tex] [\hat{L}_{x},\hat{P}_{y}]=i\hbar\hat{P}_{z}.
    (Note: \hat{L}_{x}=y\hat{P}_{z}-z\hat{P}_{y})[/tex]
    Televant equations
    [tex] \hat{p_{x}}=\frac{\hbar}{i}\frac{\partial}{\partial x} [/tex]

    [tex] \hat{p_{y}}=\frac{\hbar}{i}\frac{\partial}{\partial y} [/tex]

    [tex] \hat{p_{z}}=\frac{\hbar}{i}\frac{\partial}{\partial z} [/tex]

    So i started out by just writing out the commutator in the general form:

    (1) [tex] \hat{L}_{x}\hat{P}_{y}-\hat{P}_{y}\hat{L}_{x}[/tex]

    Then I slowly substituted some equations
    (2) [tex](y\hat{P}_{z}-z\hat{P}_{y})\frac{\hbar}{i}\frac{\partial}{\partial y}\psi-\frac{\hbar}{i}\frac{\partial}{\partial y}(y\hat{P}_{z}\psi-z\hat{P}_{y}\psi)[/tex]

    (3)[tex](y\frac{\hbar}{i}\frac{\partial}{\partial z}-z\frac{\hbar}{i}\frac{\partial}{\partial y})\frac{\hbar}{i}\frac{\partial}{\partial y}\psi-\frac{\hbar}{i}\frac{\partial}{\partial y}(y\frac{\hbar}{i}\frac{\partial}{\partial z}\psi-z\frac{\hbar}{i}\frac{\partial}{\partial y}\psi)[/tex]

    I decided to factor out \frac{\hbar}{i}
    [tex](4)(\frac{\hbar}{i})^2(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})\frac{\partial \psi}{\partial y}-(\frac{\hbar}{i})^2\frac{\partial}{\partial y}(y\frac{\partial \psi}{\partial z}-z\frac{\partial \psi}{\partial y}) [/tex]

    Now i'm not sure what to do..
  5. Jan 24, 2010 #4


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    Now start expanding the derivatives in the second term using the product rule. Look for cancellations with symmetric partial derivatives like dy*dz vs dz*dy.
    Last edited: Jan 25, 2010
  6. Jan 25, 2010 #5
    okay so then i got this

    (5) [tex](\frac{\hbar}{i})^2(y\frac{\partial}{\partial z}\frac{\partial\psi}{\partial y}-z\frac{\partial}{\partial y}\frac{\partial\psi}{\partial y})-(\frac{\hbar}{i})^2(\frac{\partial\psi}{\partial z}+y\frac{\partial}{\partial y}\frac{\partial\psi}{\partial z}-\frac{\partial z}{\partial y}\frac{\partial\psi}{\partial y}+z\frac{\partial}{\partial y}\frac{\partial\psi}{\partial y})[/tex]

    So then would [tex]y\frac{\partial}{\partial z}\frac{\partial\psi}{\partial y} y\frac{\partial}{\partial y}\frac{\partial\psi}{\partial z}[/tex] cancel out? I'm not sure what you mean by cancel out haha, i have a really bad math foundation if you could please explain that would help a lot. Also did i do that last line correctly?

    [tex]z\frac{\partial}{\partial y}\frac{\partial\psi}{\partial y}[/tex] This term would also go away right?

    Thanks for all your help!
  7. Jan 25, 2010 #6
    1- The last term has a wrong sign.
    2- By 'canceling out' one means finding a term with exactly the same form of a given term in the equation but with reversed sign so both would disappear. As in x+y-z-x=0, x and -x cancel out each other and the equation reduces to y-z=0.
    2- Remember that derivatives have no order in the sense that [tex]\frac{\partial}{\partial x}\frac{\partial f(x,y)}{\partial y} = \frac{\partial}{\partial y}\frac{\partial f(x,y)}{\partial x}[/tex].
    4- Now start canceling out terms and I'd recall that the derivative of coordinates wrt each other vanishes.

  8. Jan 26, 2010 #7
    what about the (partial z/ partial y) (partial psi/ partial y)?
    What does that become?
  9. Jan 26, 2010 #8
    Didn't you notice my fourth remark?!

  10. Jan 28, 2010 #9
    ahh I see what you mean. Thanks for all the help!
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