- #1

thoughtgaze

- 74

- 0

This is discussed in Weinberg's Quantum Theory of Fields, in the chapter on Relativistic Quantum Mechanics.

The point I am somewhat confused about occurs on page 63 - 64, if you have the book.

He operates on a single particle state with the unitary homogeneous lorentz transformation operator.

He labels the single particle states with their momentum, and a σ to denote any other degrees of freedom. In the case we consider, σ is purely discrete.

Okay, when he operates on such a state with the lorentz transformation, we get the state with the transformed momentum, but then he says in general it should be written as a linear combination of the σ states with the corresponding momentum. In mathematical language, this point is written as

U([itex]\Lambda[/itex])[itex]\Psi[/itex]

Where [itex]\Lambda[/itex] is the Lorentz transformation and the einstein summatioin convention is used.

This part makes sense.

The part that confuses me is he later "defines" [itex]\Psi[/itex]

[itex]\Psi[/itex]

Where L(p) is the homogeneous lorentz transformation that takes k to p.

To me, this definition is the crucial step in the method of induced representations, but I don't necessarily see how it's justified. It seems to conflict with the notion that operating on the state with a homogeneous lorentz transformation should in general give us some sum over the σ states.

The whole point, as far as I can tell, is that we are trying to find how general U([itex]\Lambda[/itex]) operates on the single particle states (or at least reduce it to the problem of finding the little group representations as Weinberg points out)

FYI, "k" is an arbitrary "standard" momentum. Chosen for each class of p

The point I am somewhat confused about occurs on page 63 - 64, if you have the book.

He operates on a single particle state with the unitary homogeneous lorentz transformation operator.

He labels the single particle states with their momentum, and a σ to denote any other degrees of freedom. In the case we consider, σ is purely discrete.

Okay, when he operates on such a state with the lorentz transformation, we get the state with the transformed momentum, but then he says in general it should be written as a linear combination of the σ states with the corresponding momentum. In mathematical language, this point is written as

U([itex]\Lambda[/itex])[itex]\Psi[/itex]

_{p,[itex]\sigma[/itex]}= C_{σ',σ}([itex]\Lambda[/itex],p)[itex]\Psi[/itex]_{[itex]\Lambda[/itex]p,σ'}Where [itex]\Lambda[/itex] is the Lorentz transformation and the einstein summatioin convention is used.

This part makes sense.

The part that confuses me is he later "defines" [itex]\Psi[/itex]

_{p,[itex]\sigma[/itex]}to be[itex]\Psi[/itex]

_{p,[itex]\sigma[/itex]}[itex]\equiv[/itex] N(p)U(L(p))[itex]\Psi[/itex]_{k,[itex]\sigma[/itex]}Where L(p) is the homogeneous lorentz transformation that takes k to p.

**Why are we allowed to "define" this transformation in such a way without a sum over the [itex]\sigma[/itex] states?**To me, this definition is the crucial step in the method of induced representations, but I don't necessarily see how it's justified. It seems to conflict with the notion that operating on the state with a homogeneous lorentz transformation should in general give us some sum over the σ states.

The whole point, as far as I can tell, is that we are trying to find how general U([itex]\Lambda[/itex]) operates on the single particle states (or at least reduce it to the problem of finding the little group representations as Weinberg points out)

FYI, "k" is an arbitrary "standard" momentum. Chosen for each class of p

^{2}and fixed sign of p_{0}
Last edited: