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I'm slightly confused about something in volume 1 of Weinberg.

He says $U(\Lambda)\Psi_{p,\sigma}=\sum_{\sigma'}C_{\sigma'\sigma}(\Lambda,p)\Psi_{\Lambda p,\sigma'}$. Then,

"In general, it may be possible by using suitable linear combinations of the $\Psi_{p,\sigma}$ to choose the $\sigma$ labels in such a way that the matrix $C_{\sigma'\sigma}$ is block-diagonal; in other words, so that the $\Psi_{p,\sigma}$ with $\sigma$ within any one block by themselves furnish a representation of the inhomogeneous Lorentz group. It is natural to identify the states of a specific particle type with the components of a representation of the inhomogeneous Lorentz group which is irreducible, in the sense that it cannot be further decomposed in this way."

I'm not sure if it's just the way he has it worded but I don't understand what he's trying to say here.

Next, he says "Note that the only functions of $p^\mu$. that are left invariant by all proper orthochronous Lorentz transformations $\Lambda^\mu_\nu$ are the invariant square $p^2 =\eta_{\mu\nu}p^\mu p^\nu$, and for $p^2 < 0$, also the sign of $p^0$. Hence, for each value of $p^2$, and (for $p^2 < 0$) each sign of $p^0$, we can choose a 'standard' four-momentum, say $k^\mu$, and express any $k^\mu$ of this class as $p^\mu=L^\mu_\nu k^\nu$ where $L^\mu_\nu$ is some standard Lorentz transformation that depends on $p^\mu$, and also implicitly on our choice of the standard $k^\mu$. We can then define the states $\Psi_{p,\sigma}$ of momentum $p$ by

$\Psi_{p,\sigma} \equiv N(p)U(L(p))\Psi_{k,\sigma}$

My questions here are - how does this follow from the talk about the invariance of $p^2$?

How does this imply that the transformation in question doesn't change the $\sigma$ index?

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    $\begingroup$ I think more relevant than these quantities being invariant is that starting from a certain sign of $k^0$ and a value of $k^2$ you can get any other momentum vector $p$ with the same square and sign with a Lorentz transform. I believe that here, after a choice of standard momentum $k$ for each such pair, and a choice of labelling for the degeneracies of this momentum state ($\sigma$) the idea is that the labelling for all other momentum states is induced. $\endgroup$ – s.harp Aug 25 '15 at 17:41
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In the first paragraph you quoted he is saying that in general it may be the case that the $\Psi_{p,\sigma}$ fields are reducible. Say $\sigma$ is an index going from 1 to 3, maybe after you do a suitable transformation, the first component is the state of a spinless particle and the next two are the states of a spinor. Or maybe all three components refer to spinless particles but with different masses or charges. If this is the case no matter what Lorentz transformation you do a spin 0 particle won't transform to a spin 1/2 (etc.) so the $C$ matrix will break up into blocks.

Weinberg is saying if the representation is reducible like this we should really consider it to be states of distinct particles, so from now on he will consider only an irreducible representation, and you can reach any index $\sigma$ from any other index $\sigma^\prime$ by a suitable Lorentz transformation.

Then in your next quoted paragraph, he is saying only invariants are the mass $p^2$ and the sign of the time component. If they have the same mass and sign of time component, we can reach any momentum $p$ from any other $k$ by some suitable Lorentz transformation. So if we have the irreducible set of states $\Psi_{k,\sigma}$ for some standard $k$, we can reach any other state of the particle by a Lorentz transformation.

Now the problem is how do we relate the $\sigma$ indices of states with different momentum? Let's pick some standard Lorentz transformation to get to $p$ from $k$, $L(p)$. This let's us define what the spin components are for a boosted momentum. That is why the $\sigma$ index doesn't change, we are defining it so it doesn't change. $$\Psi_{p,\sigma} \equiv N(p)U(L(p))\Psi_{k,\sigma}$$ is a definition of why say the third component of $\Psi_{p,\sigma}$ is called the third component and not the second, or for that matter why it is not a linear combination of the third and second (and so forth).

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  • $\begingroup$ Okay I'm more clear than I was before but I'm not fully sure yet. You said the $\Psi_{p,\sigma}$ are reducible. These are state vectors in the physical Hilbert space, so what does it mean for them to be reducible? I'm only familiar with the concept of matricies being reducible. $\endgroup$ – Okazaki Aug 30 '15 at 14:44
  • $\begingroup$ Secondly, you said we are defining the $\sigma$ index so it doesn't change. This seems strange to me. Why can we make this definition? It seems more clear to me if we were defining $L(p)$ to not change the $\sigma$ index, as the way things are currently formulate makes it seems like there's nothing special about $L(p)$. For example, we could make a general lorentz transformation on $\Psi_{p,\sigma}$, which is a linear combination of the states with all possible $\sigma$s, and by the composition rule this would just be another general lorentz transformation on $\Psi_{k,\sigma}$, a contradiction $\endgroup$ – Okazaki Aug 30 '15 at 14:48
  • $\begingroup$ Maybe I abused the language a little. The state vectors $\Psi_{p,\sigma}$ form a vector space, and the Lorentz transformations are linear transformations on this vector space. We can represent these linear transformations as matrices and use the concept of reducibility you are familiar with. In fact the $C$ matrix is just the part of this matrix that ignores the $p$ index (which Weinberg already discussed). $\endgroup$ – octonion Aug 30 '15 at 15:08
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    $\begingroup$ Defining the $\sigma$ index is the same thing as picking a basis for the vector space of all states with momentum $p$, certainly we are free to do that. Your argument is not a contradiction because we are not saying a general Lorentz transformation doesn't change the $\sigma$ index only the transformations $L(p)$. We can pick these $L(p)$ however we want as long as they take $k$ to $p$. $\endgroup$ – octonion Aug 30 '15 at 15:13
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To answer the first part of your question, in the transformation rule $U(\Lambda)\Psi_{p,\sigma}=\sum_{\sigma'}C_{\sigma\sigma'}(\Lambda,p)\Psi_{\Lambda p,\sigma'}$ the matrices $C_{\sigma\sigma'}(\Lambda, p)$ are invertible, and lie within some continuous group of operators (looking ahead, it's OK to think of this group as a finite dimensional Lie group). When Weinberg talks about representations, he means a representation of this group (or more accurately, the 'simplest' group that can be represented by the $C(\Lambda,p)$ matrices). Note that for consistency, the $C(\Lambda,p)$ matrices must satisfy a group-like composition rule: $$ C(\Lambda_2,\Lambda_1 p)_{\sigma\sigma'}C(\Lambda_1,p)_{\sigma'\sigma''}=C(\Lambda_2\Lambda_1,p)_{\sigma\sigma''}. $$ Physically, the ability to find a basis where all $C$ matrices are block diagonal means that there is a way to keep track of various types of particles while performing Lorentz transformations (e.g. apart from mass, particles can be separated by their couplings to weak force fields and relative spin).

To answer your second question, the invariance of $p^2$ alone is unimportant. It is important that the only functions of $p$ that are invariant under proper, orthochronous Lorentz transformations are $\text{sign}(p^0)$ and $p^2$. This means that for $p^2=-m^2<0$, the manifold $p^2+m^2=0$ has only two connected components, positive and negative energy hyperbolas. This means that if $k^\mu$ is a momentum 4-vector with positive energy, then all other positive energy momenta $p^\mu$ satisfy $p^\mu=\Lambda(p) k$ for some proper orthochronous $\Lambda(p)$. As for whether the $\sigma$-components are unchanged, the equation $\Psi_{p,\sigma}\equiv N(p)U(L(p))\Psi_{k,\sigma}$ should be read as the definition of a basis vector for the Hilbert space of 1-particle states. The transformation of these states* is then extended by linearity to define the transformation rule for an arbitrary state. The spin index $\sigma$ is mostly for bookkeeping purposes, since it's just used to define a consistent transformation rule. Physically, you can think of it as the spin you would measure if you `trapped' the particle (slowed it down to your own reference frame) in some prescribed and reproducible way. The arbitrariness in the spin associated with the choice of $L(p)$ corresponds to the arbitrariness in how the particle is decelerated--some 'torque' could be applied in the process. The 'actual' interpretation isn't quite as simple as the heuristic idea outlined above, but it gives the basic gist. (The additional subtlety is from the requirement that the little group transformation associated with $\lambda$ agree with $\Lambda$ when $\Lambda$ is a pure rotation; this can be done, but it is a little awkward to describe).

*recall that $$U(\Lambda)\Psi_{p,\sigma}= U(L(\Lambda p))\cdot U(L(\Lambda p)^{-1}\Lambda L(p))\Psi_{k,\sigma}=D(R(\Lambda,p))_{\sigma,\sigma'} U(L(\Lambda p))\Psi_{k,\sigma'}=D(R(\Lambda,p))_{\sigma,\sigma'}\Psi_{\Lambda p,\sigma'}$$

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What I understand is that Weinberg, via eq. 2.5.5 defines the states $\Psi_{p,\sigma}$. To my understanding, the operator $U(L(p))$ that performs such a transformation should belong to the factor group $SO(3,1)^{\uparrow}/SO(3)$, for it does not affect the spin index. Hence, it should be a pure Lorentz boost (not including rotations).

However, in the time-like case, $k^\mu=(0,0,0,M)$, Weinberg gives a convenient choice for $L(p)$, in equation 2.5.24. This is, for sure, not a Lorentz boost for it has $i,k$ components.

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