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Questions about Commutators

  1. Jun 28, 2014 #1
    Hi all,
    My motivation is understanding some derivations in Quantum Mechanics, but I think my questions are purely algebraic. I have a general question and then a specific one:

    General Question - when writing the commutator of commuting vector and a scalar operators (for instance angular momentum and some Hamiltonian) - [itex][\vec A,H]=0[/itex] - what is meant by this *exactly*? I see two possible answers:

    1. [itex][A_i,H]=0[/itex] for [itex]i=1,2,3[/itex]
    2. [itex][A_1+A_2+A_3,H]=0[/itex] in which case we could have [itex][A_i,H]\ne0[/itex] for some [itex]i[/itex] .

    It seems to me that in the QM context almost always what is meant is the first option but I'm not certain...

    Specific Question - if [itex]\vec A[/itex] and [itex]\vec B[/itex] commute with [itex]H[/itex], does [itex]\vec A \cdot \vec B[/itex] also necessarily commute? If the answer to the question above is #1, then obviously it does. If the answer is #2 then I guess not?

    Would greatly appreciate the clarifications. Thanks!
    Last edited: Jun 28, 2014
  2. jcsd
  3. Jun 28, 2014 #2


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    General question:
    Definitively option 1.
    Never in physics you will see such a thing as A1+A2+A3.
    Such a quantity has no general meaning.
    It is not independent of the frame of reference and physics is independent of the choice of the frame of reference.

    Specific question:
    The answer is no, not necessarily.
    Do you think that A and B being vectors would play a role in the answer to your question?
    At least, for scalar operators, the answer is clearly no.
  4. Jun 28, 2014 #3


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    H is a scalar operator, i.e:

    [tex][\vec{A}, H ] = \vec{A}H - H\vec{A} = (A_1H,A_2 H , A_3 H) - (HA_1,HA_2,HA_3)[/tex]

    So indeed it's the first case.
  5. Jun 28, 2014 #4
    Can you please elaborate on the second part. What do you mean by " for scalar operators, the answer is clearly no."?
    If [itex]A[/itex] and [itex]B[/itex] commute with [itex]H[/itex] then: [tex][AB,H]=A[B,H]+[A,H]B=0[/tex]
    If the answer to my first question was #1, then for vector operators [itex]\vec A[/itex] and [itex]\vec B[/itex] that commute with [itex]H[/itex] we would have: [tex][\vec A \cdot \vec B,H]=[A_1B_1,H]+[A_2B_2,H]+[A_3B_3,H][/tex] and equals zero by the previous case.
  6. Jun 28, 2014 #5


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    yes, because
    [tex]\left[\sum_iA_iB_i,H\right] = \sum_i\left(A_i[B_i,H]+[A_i,H]B_i\right) = 0[/tex]
    as you said.

    (except for quantization anomalies in quantum field theories)


    Where's the problem? Can you please provide a counterexample?
    Last edited: Jun 28, 2014
  7. Jun 28, 2014 #6


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    Bad reading of the initial post!
  8. Jun 28, 2014 #7


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    proof style!
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