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Commuting Hamiltonian with the projection of position

  1. Nov 24, 2015 #1
    Hi all,

    This is the problem I want to share with you.
    We have the hamiltonian H=aP+bm, which we are commuting with the position x and take:

    [x,H]=ia, (ħ=1)

    Ok. Now if we take, instead of x, the operator

    X=Π+ x Π+--
    where Π± projects on states of positive or negative energy
    the result will be:

    Can somebody explain me how this result comes out.
    Thanks in advance

    Attached Files:

  2. jcsd
  3. Nov 24, 2015 #2


    Staff: Mentor

    Why exactly do you think this is a position operator?

    As far as I can see you have defined an operator you call X, then find that it doesn't behave the same as a position operator, and for some reason it concerns you. Why has me beat. And exactly what do you mean by projecting on states with positive and negative energy? I don't know of any such operator.

    Bottom line here is you need to post a LOT more detail. Without that detail I don't think anyone can really help you - at least I can't. Someone else may see what you are on about, in which case I am all ears, but for me it doesn't really make any sense.

  4. Nov 25, 2015 #3


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    Staff: Mentor

    I really don't see how you can get such a commutator. Can you show the intermediate steps?
  5. Nov 25, 2015 #4
    You seem to be claiming that this ##X## operator is identical to ##x## and therefore the results should match. Well, that's wrong. You can get away with not including energy=0, but you forgot the cross terms too. If you split up a space into two complementary subspaces with the corresponding projectors, then you get the identity
    $$x=(P_1+P_2)x(P_1+P_2) = P_1 x P_1 + P_2 x P_2 + P_1 x P_2 + P_2 x P_1$$
    Your expansion is missing the cross terms.


  6. Nov 25, 2015 #5
    First I want to thank you all for devoting some time to answer me.

    I had to be more comprehensive, so let me provide some more details.

    We take the Dirac Hamiltonian H=aP+mb in the non-manifestly-covariant formulation.
    According to this Hamiltonian the velocity of a free Dirac particle is given by:
    so the commutator [H,u] will not be zero. (a is actually a matrix).

    Now, I read in a paper that it is possible to get the commutator between the Hamiltonian H and the velocity "V" equal to zero if we start with the position operator
    Probably this operator is different from x, but is still associated to position. (At least this is claimed by the author)
    The projectors Π± project on states of positive (+) or negative (-) energy states and can be defined as Π±=1/2(1±Λ) with Π+-=1 and Λ=f(H) a normalized function of Hamiltonian.

    Thus in the paper author says that the velocity V of the new position operator X is given by:
    dX/dt=V=-i[X,H]=P/H, hence the commutator [H,V] will be equal to zero:

    The last part is the easy one. My query is how comes up the commutation V=-i[X,H]=P/H.

    I hope to become more clear now.

  7. Nov 25, 2015 #6


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    Staff: Mentor

    Could you please give the reference? It will greatly help us help you.
  8. Nov 25, 2015 #7


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    Science Advisor

    If by "energy" you mean "eigenvalue of H", and if ##a## commutes with ##H##, then we can show that ##[X,H]=ia##.

    ##[X,H]=\Pi_+[x,H]\Pi_++\Pi_-[x,H]\Pi_- = ia(\Pi^2_++\Pi^2_-)=ia(\Pi_++\Pi_-)=ia##.
  9. Nov 26, 2015 #8
    You can find the paper at "arXiv:1002.0066 [quant-ph]" (subsection 3.1)

    Thank you Avodyne but that exactly is my query, the result that is appearing in the paper is the inverse of this.
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