# Weinberg discussion on induced representations in his QFT Vol.1 Ch. 2

#### thoughtgaze

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($\Lambda$)$\Psi$p,$\sigma$ = Cσ',σ($\Lambda$,p)$\Psi$$\Lambda$p,σ'

Where $\Lambda$ is the Lorentz transformation and the einstein summatioin convention is used.

This part makes sense.

The part that confuses me is he later "defines" $\Psi$p,$\sigma$ to be

$\Psi$p,$\sigma$ $\equiv$ N(p)U(L(p))$\Psi$k,$\sigma$

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 $\sigma$ 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($\Lambda$) 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 p2 and fixed sign of p0

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#### thoughtgaze

oh thanks, it appears to be a much more concise version of my question. thanks i'll take a look and see what i find

#### Avodyne

It's just a definition. For a massive particle, the spin label σ on the particle state is usually defined to be the eigenvalue of Sz in the rest frame. In other words, we choose to define the spin labels so that the matrix C is the identity matrix. It's not necessary, but it is convenient.

#### thoughtgaze

It's like constructing Euler angles - for each value of p he wants to choose a particular basis for the σ label.

Using Lorentz transformations, the values of p you can generate all have the same value of p2 = m2. Out of these he picks one, a "standard" four-momentum k. This is like choosing a particular rest frame, k = (m,0,0,0). Next, for every other p value, he picks a "standard" way of getting there from k, which he calls L(p). L(p) could be, for example, a simple boost in the k-p plane.

Now L(p) not only maps k into p, it also uniquely maps the basis for σ states at k into a set of σ states for p. That is, Eq(2.5.5). So if he uses this to define ψ he doesn't need the matrix Cσ'σ any more - he's diagonalized it.
I think the thing in bold is one of the crucial points. Also what Avodyne said about the fact that the label represents what would be measured in a particular frame defined by the choice of k.

This makes sense.

But, what prevents this unique mapping from being true for general four-momentum and general homogeneous lorentz transformations? Is there an intuitive way to see this?

In other words, why couldn't I use the same argument for U(Λ)Ψp,σ = Cσ',σ(Λ,p)ΨΛp,σ'

and simply diagonalize Cσ',σ(Λ,p) with the same argument?

I am guessing it is because it may not satisfy the unique mapping that Bill_K has referred to, but if this is the case? How do I see this?

Also, Avodyne said "It's not necessary, but it is convenient."

But it does seem necessary in order to reduce the problem to finding the little group representations WEinberg speaks about.

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#### dextercioby

Homework Helper
The U's are different in the 2 equations. The U in the first is a representation operator for a general Lorentz transformation that will <rotate> the spin space, while the U in the second will not <rotate> the spin space, it only boosts from a particular (conveniently chosen) 4-momentum to a general one.

The <rotate> means mixing the unit vectors in the spin space.

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#### thoughtgaze

thanks dextercioby, i think it is clear now

this clears up for me what bill_k said about ""standard" way of getting there from k, which he calls L(p)."

the way of getting to p from k is restricted to simple boosts. that's an interesting distinction that i wish weinberg made more explicit but perhaps it's necessarily implied anyway from his definition.

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Why are we allowed to "define" this transformation in such a way without a sum over the $\sigma$ states?
I think the key is in the fact that $k$ is fixed here. So, the order of reasoning is this:

1) We fix $U$, so we know what $U(L)$ are.
2) We fix $k$ ("standard four-momentum")
3) We define $\Psi_{p,\sigma}$, for all other momenta using what is on the RHS.
4) Now we know, from this definition, what are $\Psi_{p,\sigma}$ for any $p$

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