Unitarity angular momentum operators

[Yoran91]

Hi,

I'm confused by a sentence in a set of lecture notes I have on quantum mechanics. In it, it is assumed there is some representation $\pi$ of $SO(3)$ on a Hilbert space. This representation is assumed to be irreducible and unitary.

It is then said that the operators $J_i$, which are said to be the infinitesimal generators of the rotation group satisfying $[J_i,J_j]=i \epsilon_{ijk}$, are Hermitian as a consequence of the unitarity of this representation.

This confuses me. Shouldn't they say that the operators $\pi (J_i)$ are Hermitian? Are they writing $J_i$ for both the infinitesimal generators of the group and the operators they are mapped to?

 

[DrDu]

This confuses me. Shouldn't they say that the operators $\pi (J_i)$ are Hermitian? Are they writing $J_i$ for both the infinitesimal generators of the group and the operators they are mapped to?

Yes, they should do so. But it is quite common to write J for some representation.

 

[Yoran91]

Thanks for your quick answer.

Does that, then, imply that the Casimir operator is actually $\sum_i \pi(J_i)^2$?
Do the commutation relations satisfied by the Lie group generators carry over to the operators under the representation map? I can see that

$[\pi(J_i),\pi(J_j)] = \pi(J_i)\pi(J_j) - \pi(J_j)\pi(J_i) = \pi(J_i J_j) -\pi(J_j J_i)$,

but I don't see why I could conclude that the last line equals $\pi([J_i,J_j])$

EDIT: The only way I could see that happen if the representation is linear, but I haven't seen that assumed in its definition. Should that be included?

 

[dextercioby]

The representation is linear, if the representation space is linear (and a separable Hilbert space is).

 

[Yoran91]

How so? I mean : how do you know the representation is linear whenever the representation space is?

 

[dextercioby]

In QM it's the result of the theorem of Wigner. The representation operators can only be linear or antilinear (i.e. conjugate linear).