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In Weinberg's The Theory of Quantum Fields Volume 1, he considers classification one-particle states under inhomogeneous Lorentz group. My question only considers pages 62-64.

He define states as $P^{\mu} |p,\sigma\rangle = p^{\mu} |p,\sigma\rangle $, where $\sigma$ is any other label. Then he shows that, for a Lorentz Transformation : $$P^{\mu}U(\Lambda)|p,\sigma\rangle = \Lambda^{\mu}_{\rho} p^{\rho}U(\Lambda)|p,\sigma\rangle $$ Therefore: $$U(\Lambda)|p,\sigma\rangle = \sum_{\sigma'} C_{\sigma' \sigma}(\Lambda,p)|\Lambda p,\sigma'\rangle.$$ Then he wants to find $C$ in irreducible representations of the inhomogeneous Lorentz group. For any $m$ he chooses a $k$ such that $k^{\mu}k_{\mu} = - m^2$. Then defines express $p$'s with mass m, according to $p^{\mu} = L^{\mu}_{v}(p)k^v$.

Then he defines $$|p,\sigma\rangle = N(p)U(L(p))|k,\sigma\rangle$$ (where $N(p)$ are normalization constants). I didn't understand this last statement. Is $\sigma$ an eigenvalue of the corresponding operator, or just a label? I mean, if $J |k,\sigma \rangle = \sigma |k,\sigma\rangle $ then is it true, $J |p,\sigma\rangle = \sigma |p,\sigma\rangle$. If so how can we say that if $$U(\Lambda)|k,\sigma\rangle = \sum_{\sigma'} C_{\sigma' \sigma}(\Lambda,k)|\Lambda k=p,\sigma'\rangle$$

Thanks for any help. First pages of these notes on General Relativity from Lorentz Invariance are very similar to Weinberg's book.

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2 Answers 2

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For Poincare algebra there are (as far as I know) two different approaches to find its representations. In first approach one begins from a finite dimensional representation of (complexified) Lorentz algebra, and using it one constructs a representation on space of some fields on Minkoski space. Representation so obtained is usually not irreducible and an irreducible representation is obtained from it through some differential equation. E.g. space of massive Dirac fields satisfying Dirac equation form an irreducible representation of Poincare group (added later : last statement is not quite correct).

Another approach is to find (irreducible, unitary) Hilbert space representation of identity component of Poincare algebra by so called "Little group method". This is what Weinberg is doing in pages 62-64 in volume 1 of his QFT book. Idea of this approach is following --

In momentum space fix a hyperboloid $S_m=\{p|p^2=m^2,p_0 \geq 0\}$ corresponding to a given (nonnegative) mass $m$. (note : here I am using signature $(1,-1,-1,-1)$)

Choose a 4-momentum $k$ on $S_m$. Let $G_k$ be the subgroup of (identity component) of Lorentz group such that $G_k$ fixes $k$. i.e. for each Lorentz transformation $\Lambda\in G_k$ we have $\Lambda k=k$. $G_k$ is called little group corresponding to 4-momentum $k$.

Let $V_k$ be a fixed finite dimensional irreducible Unitary representation of $G_k$. Fix a basis of this vector space $|k,1>,|k,2>,...,|k,n>$ where $n$ is (complex) dimension of $V_k$ {note that $k$ is a fixed vector, and not a variable}

Now for every other $p\in S_m$ introduce a vector space $V_p$ which is spanned by the basis $|p,1>,|p,2>,...,|p,n>$.

Hilbert space representation of (identity component of) Poincare group is now constructed by gluing these vector spaces $V_p$'s together. This is done as follows :-

i) Define $H$ to be direct sum of $V_p$'s.

ii) For every $p\in S_m$ fix a Lorentz transformation $L_p$ that takes you from $k$ to $p$, i.e. $L_p(k)=p$. Also fix a number $N(p)$ (this is used for fixing suitable normalization for basis states). In particular, take $L_k=I$.

iii) Define operator $U(L_p)$ corresponding to $L_p$ on $V_k$ as :-

$U(L_p)|k,\sigma>=N(p)^{-1}|p,\sigma>,\:\sigma=1,...,n\tag1$

This only defines action of $L_p$'s on subspace $V_k$ of $H$. But in fact this definition uniquely extends to the action of whole of (identity component of) Poincare group on the whole of $H$ as follows --

Suppose $\Lambda$ be ANY Lorentz transformation in the identity component of Lorentz group, and $|p,\sigma>$ be any basis state. Then (all the following steps are from Weinberg's book):

$U(\Lambda)|p,\sigma> = N(p) U(\Lambda) U(L_p)|k,\sigma>$ {using def. (1)}

$= N(p) U(\Lambda.L_p)|k,\sigma>$ (from requiring $U(\Lambda) U(L_p)=U(\Lambda.L_p)$)

$= N(p) U(L_{\Lambda p}.L_{\Lambda p}^{-1}.\Lambda.L_p)|k,\sigma> $

$= N(p) U(L_{\Lambda p})U(L_{\Lambda p}^{-1}.\Lambda.L_p)|k,\sigma>$

Now note that $L_{\Lambda p}^{-1}.\Lambda.L_p $ is an element of $G_k$ {check it} and $V_k$ is irreducible representation of $G_k$. So $U(L_{\Lambda p}^{-1}.\Lambda.L_p)|k,\sigma>$ is again in $V_k$; and from (1) we know how $U(L_{\Lambda p})$ acts on $V_k$; thus we know what is $U(\Lambda)|p,\sigma>$.

Summarizing, the idea of little group method is to construct irreducible Hilbert space representations of the identity component of Poincare starting from finite dimensional irreducible representations of Little group corresponding to a fixed four momenta.

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Thanks for answer. But I think your definition of $U(L_p)$ and $|p,\sigma\rangle$ are different from book. Correct me if I am wrong, but I think you can't define, at least it is not trivial, an operator $U$ which transforms $|k,\sigma\rangle$ to $|p,\sigma\rangle$ and corresponds to Lorentz transformation $L(p)$. In the book he defines $|p,\sigma\rangle$ as the result of $U(L(p))$ acting on $|k,\sigma\rangle$. I believe you can't choose $U(L(p))$ as you want, because once we choose $L(p)$, we also define a new observer for who $|k,\sigma> \rightarrow U(L(p)) |k\sigma\rangle$, which is fixed. –  hans May 7 '13 at 21:34
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This answer is actually reverse of Weinberg's approach. Weinberg begins with a Hilbert space H that is assumed to be irreducible representations of Poincare algebra; and then finds a basis of it using little group method. Here we begin from a finite dimensional representation of little group and construct a Hilbert space representation of Poincare group using it. In particular, while in Weinberg (1) is a definition of basis element $|p,\sigma>$ in our case it defines action of $U(L_p)$. These two approaches are completely equivalent. –  user10001 May 7 '13 at 21:57
    
But in your case for some operator $J$, $J |p,\sigma\rangle = \sigma |p,\sigma\rangle$, in Weinberg's definition this is not true in general for $p \neq k$. –  hans May 7 '13 at 22:20
    
No. In basis $|k,\sigma>$, label $\sigma $ can be taken as eigenvalue of (say) $J_3$; but for $p$ different from $k$, $\sigma$ appearing in basis $|p,\sigma>$ is not in general spin eigenvalue or anything like that. Initially action of $G_k$ is only defined on $V_k$. $V_p$'s are completely different vector spaces which have nothing in common with $V_k$ except for the dimension. To know how $G_k$ acts on $V_p$ (and hence what are eigenvectors of (say) spin in $V_p$) one should follow the definition of action of Lorentz operators that we have given later. –  user10001 May 7 '13 at 22:39
    
I can't see why you can't define $|p,\sigma\rangle$ as eigenvectors of $J_3$ at the beginning in your approach.(I think you shouldn't be able to as I said in my first comment). If you can't define then (as I understand) your approach is equivalent to Weinberg's approach. –  hans May 7 '13 at 22:55

With respect to the discussion of momentum-eigenstates and the following derivation in Weinberg's book, $\sigma$ is just a label that denotes any degree of freedom that is not momentum. Even though it can be identified with spin, its nature is not relevant for the discussion at hand.

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Thanks for your comment, I know that he uses $\sigma$ for anything other than momentum but my question don't have anything to do with spin, I used $J |p,\sigma\rangle = \sigma |p,\sigma\rangle$ for any observable. I ask if this relation is true after definition of $|p,\sigma\rangle = N(p)U(L(p))|k,\sigma\rangle $. –  hans May 7 '13 at 14:16
    
Yes, it's true. The $\sigma$ are eigenvalues of some operators that commute with the $P$ operators. It wouldn't make any sense to use them to label eigenkets otherwise. –  user1504 May 7 '13 at 20:48
    
@user1504 But in this notes link, he says that (at page 2 between (7) and (8) ) $\sigma$ is not an eigenvalue of $J_z$ for $p \neq 0$. –  hans May 7 '13 at 21:29
    
That doesn't contradict anything I said. It doesn't have to be the eigenvalue of $J_z$. –  user1504 May 7 '13 at 21:55
    
But it is an eigenvalue of $J_z$ for $|k,\sigma>$. I mean the operator does not important here, in the link it says that for some operator $J$ $J |k,\sigma\rangle = \sigma |k,\sigma\rangle$ but $J |p \neq k, \sigma\rangle \neq \sigma |p \neq k, \sigma\rangle$. –  hans May 7 '13 at 22:14

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