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Do spin and momentum commute?

  1. Apr 17, 2010 #1
    I did some maths and I found that angular momentum operator does not commute with normal mometum:
    [tex][ J_{\alpha \beta}, P_{\gamma}] = \eta_{\alpha \gamma} P_{\beta} - \eta_{\beta \gamma} P_{\alpha}[/tex]

    Now, the "third" component of angular momentum:
    [tex]J_{z} := J_{x y}[/tex]
    [tex][J_{z}, P_{x}] = -P_{y}[/tex]
    [tex][J_{z}, P_{y}] = P_{x}[/tex]
    [tex][J_{z}, P_{z}] = 0[/tex]
    It does not commute with two components od momentum!

    So how it is possible to measure both momentum and spin? Why is spin the "third" component, in the direction of momentum, when the act of determining momentum should make measurement of spin impossible?

    Am I wrong somewhere?
    Last edited: Apr 17, 2010
  2. jcsd
  3. Apr 17, 2010 #2
    Angular momentum and linear momentum don't commute because the angular momentum operator contains the position operator in its definition. The spin operator isn't defined in terms of r x p or anything like that. In other words, the value of a particle's spin does not depend at all on the spatial distribution of its wavefunction.
  4. Apr 17, 2010 #3
    Spin is not the angular momentum of a particle. The total angular momentum of a particle is spin plus orbital angular momentum. The orbital component doesn't commute with momentum, but the spin component does.
  5. Apr 17, 2010 #4
    Thanks, I didn't know. Could you provide some maths? What is spin then?
  6. Apr 17, 2010 #5
  7. Apr 17, 2010 #6
    The spin operators act on a different Hilbert space (let us call it [itex]\mathcal{H}_{2}[/itex]) than the momentum and position operators (let us call it [itex]\mathcal{H}_{1}[/itex]). That is why they commute. The total angular momentum [itex]\mathbf{J}[/itex] is indeed the sum of the orbital angular momentum ([itex]\mathbf{L}[/itex]) and the spin angular momentum ([itex]\mathbf{S}[/itex]). However, we must note that any operator A acting on the direct product of the two Hilbert spaces [itex]\mathcal{H} = \mathcal{H}_{1} \otimes \mathcal{H}_{2}[/itex] is of the form [itex]A \equiv A_{1} \otimes A_{2}[/itex]. If some operator does not act on one or the other space, it's operator part for that space is simply the unit operator in that space. For example:

    \mathbf{L} = \mathbf{L}_{1} \otimes 1_{2}

    \mathbf{S} = 1_{1} \otimes \mathbf{S}_{2}

    \mathbf{J} = \mathbf{L} + \mathbf{S} = \mathbf{L}_{1} \otimes 1_{2} + 1_{1} \otimes \mathbf{S}_{2}

    Next, the commutator of two operators A and B acting on this direct product space is defined as:

    [A, B] & = & A B - B A = (A_{1} \otimes A_{2})(B_{1} \otimes B_{2}) - (B_{1} \otimes B_{2})(A_{1} \otimes A_{2}) \\

    & = & (A_{1} B_{1}) \otimes (A_{2} B_{2}) - (B_{1} A_{1}) \otimes (B_{2} A_{2}) \\

    & = & (A_{1} B_{1} - B_{1} A_{1}) \otimes (A_{2} B_{2} - B_{2} A_{2}) \\

    & = & [A_{1}, B_{1}] \otimes [A_{2}, B_{2}]

    So, the momentum operator [tex]\mathbf{P} = \mathbf{P}_{1} \otimes 1_{2}[/tex] and the spin operator [tex]\mathbf{S} = 1_{1} \otimes \mathbf{S}_{2}[/tex] have a commutator:

    [S_{i}, P_{j}] = [1_{1}, (P_{j})_{1}] \otimes [(S_{i})_{2}, 1_{2}] = 0_{1} \otimes 0_{2} = 0

    i.e. they commute.
    Last edited by a moderator: Mar 3, 2012
  8. Apr 18, 2010 #7
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