Commutator of density operator with kinetic energy operator

[element_zero]

Hey guys,

maybe you can help me with the following problem. I have to calculate the commutator relations in position representation:
a) [V,ρ]
b) [p,ρ]
c) [p^2,ρ]

Note that <q'|ρ|q>=ρ(q',q) is a matrix element of the density operator

I already solved the first one. You just have to apply the potential operator on a matix element of the density operator.
[V,ρ]=<q'|Vρ|q>-<q'|ρV|q>=...=(V(q')-V(q))*ρ(q',q)

The rest however is more tricky as the momentum operator is not diagonal in this domain.
[p,ρ]=<q'|pρ|q>-<q'|ρp|q>=...?
I got the hint that I should try an integration over an auxiliary variable which should lead to something like <q'|ρ|q''>~δ(q'-q'') (Delta functions)

A Fourier transformation is NOT necessary as far as I know.

The result of c should be something like:
-(d^2/dq'^2-d^2/dq^2)*ρ(q',q)

Thanks a lot for your help!

 

[member 553083]

</code>For b), you can use integration by parts to get the desired result. Specifically, let <q'|pρ|q> = ∫dq'' ρ(q',q'') p(-q'')Then, integrating by parts, and using the fact that the boundary terms vanish since ρ(q',q'') is assumed to be zero at q'=±∞ and q''=±∞, we obtain<q'|pρ|q> = -∫dq'' (∂/∂q'')ρ(q',q'') (-q'') = -(∂/∂q'') ∫dq'' ρ(q',q'') (-q'')= -(∂/∂q'')δ(q'-q'') = (d/dq'') δ(q'-q'')Similarly,<q'|ρp|q> = (d/dq) δ(q'-q)Therefore,[p,ρ] = <q'|pρ|q>-<q'|ρp|q> = (d/dq'')δ(q'-q'') - (d/dq)δ(q'-q)= (d/dq'' - d/dq)δ(q'-q)For c), note that[p^2, ρ] = <q'|p^2ρ|q> - <q'|ρp^2|q>= -(d/dq'')^2 δ(q'-q'') - (-d/dq)^2 δ(q'-q)= -(d^2/dq'^2 - d^2/dq^2)δ(q'-q)