Commuting Hamiltonian with the projection of position

[Quantum child]

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₊ +Π_-xΠ_-
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

 

[bhobba]

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

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

 

[DrClaude]

X=Π₊ x Π₊ +Π_-xΠ_-
where Π_± projects on states of positive or negative energy
the result will be:
[X,H]=iP/H

I really don't see how you can get such a commutator. Can you show the intermediate steps?

 

[Jazzdude]

X=Π₊ x Π₊ +Π_-xΠ_-

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

 

[Quantum child]

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₊xΠ₊+Π_-xΠ_-
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₊+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

 

[DrClaude]

I read in a paper

Could you please give the reference? It will greatly help us help you.

 

[Avodyne]

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

$[X,H]=\Pi_+[x,H]\Pi_++\Pi_-[x,H]\Pi_- = ia(\Pi^2_++\Pi^2_-)=ia(\Pi_++\Pi_-)=ia$.

 

[Quantum child]

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.