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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:
    [X,H]=iP/H

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

    Attached Files:

  2. jcsd
  3. Nov 24, 2015 #2

    bhobba

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    Gold Member

    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.

    Thanks
    Bill
     
  4. Nov 25, 2015 #3

    DrClaude

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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.

    Cheers,

    Jazz
     
  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:
    dx/dt=u=-i[x,H]=a,
    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
    X=Π++--
    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:
    [H,V]=P-P=0.

    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.

    QC
     
  7. Nov 25, 2015 #6

    DrClaude

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

    Avodyne

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    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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