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Calculating Commutator of Differential Angular Momentum

  1. May 9, 2014 #1
    Hi there!

    I have tried for hours to calculate the commutator of angular momentum in the differential form, but I cannot get the correct answer. This is my first experience with actually checking if two operators commutes, so there may be some beginner's misunderstandings that causes the problem.

    I know from my textbook the correct answer:

    [\hat{L}_x,\hat{L}_y] = \frac{\hbar}{i}\hat{L}_z = \frac{\hbar}{i} \left( x \frac{\partial \psi}{\partial y} - y \frac{\partial \psi}{\partial x} \right)

    The angular momenta are defined as

    \hat{L}_x = y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} \\
    \hat{L}_y = z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \\
    \hat{L}_z = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}

    Sets up the commutator expression, and writes it out:

    [\hat{L}_x,\hat{L}_y] = \left[ y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} , z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \right]

    From what I assume is a well-known algebraic rule for handling commutators, I get the following:

    \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x} \right] - \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, x \frac{\hbar}{i} \frac{\partial}{\partial z} \right]
    - \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, z \frac{\hbar}{i} \frac{\partial}{\partial x} \right]
    + \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z} \right]

    It was straight forward to find that the two middle terms commute, i.e. they equal zero. Hence, we are left with:

    \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x} \right]
    + \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z} \right] \phantom{................} [1]

    Upon evaluating these terms, I let them act on a function [itex] \psi [/itex]. Looking at the first term:
    First I factored the complex fraction out. Then I wrote out the commutator and let each differential fraction act on [itex] \psi [/itex]:

    \frac{\hbar}{i} \left(y \frac{\partial \psi z}{\partial z} \frac{\partial \psi}{\partial x} - yz \frac{\partial \psi }{\partial x} \frac{\partial \psi}{\partial z} \right)

    The y in the second term above can be factored out from the differentials, because the differential is not done with respect to y. In the first term, however, the z must be differentiated along with psi by using the product rule. Doing this, and cancelling equal terms of opposite sign, yields:

    \frac{\hbar}{i} y \psi \frac{\partial \psi}{\partial x}

    Repeating this calculation for the second term in [1] yields

    -\frac{\hbar}{i} x \psi \frac{\partial \psi}{\partial y}

    Substituting into [1] yields

    [\hat{L}_x,\hat{L}_y] \psi = \frac{\hbar}{i} \left( y \frac{\partial \psi}{\partial x} - x \frac{\partial \psi}{\partial y} \right) \psi

    Now, comparing with the correct answer:

    i) My signs are opposite of what they should be
    ii) More importantly, I have a [itex] \psi [/itex] in the differentials which should not be there. My question is, "why, and how do I rid my answer of it?".

    I hope I have sufficiently showed what I have done. I have checked my notes several times, and I cannot get rid of those extra psi's in my answer.

    I would appreciate any help!

  2. jcsd
  3. May 9, 2014 #2


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    Science Advisor

    [A,B] ψ = A(Bψ) - B(Aψ)

    Looks like you forgot the parentheses. Also there should be two factors of ħ/i in front, not one. (And just one ψ!)
  4. May 9, 2014 #3


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    Gold Member

    For such cases, because it is a bit difficult not to make a mistake during your calculations, it's better to write some things in a more compact, yet known form....
    For example, you could avoid so many [itex]\frac{h}{i} \frac{d}{dx_{i}}[/itex] by substituting them with [itex]p_{i}[/itex], for which you know its commutation relations with the coordinates [itex]x_{i}[/itex]

    So for example the first parenthesis is:
    [itex] [y p_{z},z p_{x}]= y [p_{z},zp_{x}]= y [p_{z},z] p_{x} = -ih y p_{x}[/itex]
    The 2nd is:
    [itex] [z p_{y},x p_{z}]= [z, xp_{z}] p_{y}= x [z,p_{z}] p_{y}= ih x p_{y}[/itex]

    Also I don't understand why you have h/i twice in the commutators...where did they come from?
    [itex] a [\hat{A},\hat{B}] \ne [a \hat{A},a \hat{B}][/itex]
    Last edited: May 9, 2014
  5. May 9, 2014 #4
    Your main problem is that you are operating too soon on the wavefunction. Leave it right until the end.
  6. May 9, 2014 #5


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    Gold Member

    You need to show
    [itex]L_{z}= \frac{i}{h} [L_{x},L_{y}]=\frac{i}{h}[ y \frac{\partial}{\partial z}-z\frac{\partial}{\partial y}, z \frac{\partial}{\partial x}-x\frac{\partial}{\partial z}][/itex]
    Now you should replace [itex]\frac{\partial}{\partial x_{j}} = \frac{i}{h} p_{j}[/itex]

    [itex]\frac{i}{h} [L_{x},L_{y}]=\frac{i}{h}[ y \frac{i}{h} p_{z}-z\frac{i}{h} p_{y}, z \frac{i}{h} p_{x}-x\frac{i}{h} p_{z}][/itex]

    [itex]\frac{i}{h} [L_{x},L_{y}]=(\frac{i}{h})^{3}[ y p_{z}-z p_{y}, z p_{x}-x p_{z}][/itex]

    [itex]\frac{i}{h} [L_{x},L_{y}]=(\frac{i}{h})^{3}([ y p_{z}, z p_{x}]+ [ z p_{y}, x p_{z}])[/itex]

    [itex]\frac{i}{h} [L_{x},L_{y}]=(\frac{i}{h})^{3}(-ih y p_{x}+ ih x p_{y})[/itex]

    [itex]\frac{i}{h} [L_{x},L_{y}]=(\frac{i}{h})^{3}(-ih y \frac{h}{i} \frac{\partial}{\partial x}+ ih x \frac{h}{i} \frac{\partial}{\partial y})[/itex]

    [itex]\frac{i}{h} [L_{x},L_{y}]=-\frac{i}{h}(- y \frac{\partial}{\partial x}+ x \frac{\partial}{\partial y})=\frac{i}{h} L_{z}[/itex]

    But obviously your angular momenta are not well defined.... For example, you know that angular momenta have dimensions of [itex]h[/itex], also you show partial derivatives in their definition (reminding me of momenta) which is not correct....
    Another way to see that is by remembering that angular momenta are:
    [itex]\vec{L}=\vec{r} \times \vec{p} [/itex]
    which after the quantization of the momenta will get:
    [itex]\vec{L}=\frac{h}{i} \vec{r} \times \vec{∇} [/itex]
    with the cross giving the things you give as: xd_y - y d_x etc....
    In fact the definitions you have given correspond to:
    [itex] \frac{i L_{i}}{h}[/itex] in order to make them quantum operators...
    with that definition:
    [itex]\frac{i}{h} [\frac{ i L_{x}}{h}, \frac{i L_{y}}{h}]= \frac{i}{h} \frac{i L_{z}}{h}[/itex]
    which will give you the
    [itex] [L_{x},L_{y}]= \frac{h}{i}L_{z}[/itex]
    Last edited: May 9, 2014
  7. May 11, 2014 #6
    I see where I went wrong. I have managed to get the correct answer. Thank you for your help!
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