# Irreducible Representations Of Lorentz Group

+ 6 like - 0 dislike
1424 views

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.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user hans

+ 6 like - 0 dislike

For Poincare algebra there are (as far as I know) two different approaches to find its representations. In the first approach one begins from a finite dimensional representation of (complexified) Lorentz algebra, and using it one constructs a representation on the space of some fields on Minkowski 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 maximal subgroup of (the identity component) of the 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 representation of $G_k$ (or double cover 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 (the identity component of) the 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 the 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 the 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 group starting from finite dimensional irreducible representations of the Little group corresponding to a fixed four momenta.

$^{**}$ If $V_k$ is not a proper representation of $G_k$ but is a representation of the double cover $\mathcal{G}_k$ of $G_k$ then we'll also need to specify a section $G_k\to \mathcal{G}_k$ of the covering map so that we know how $G_k$ acts on $V_k$.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user user10001
answered May 7, 2013 by (635 points)
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.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user hans
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.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user user10001
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.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user hans
As I said $V_p$ has nothing in common with $V_k$ except for the dimension. And since its dimension is same as that of $V_k$ so I can choose any arbitrary basis of $V_p$ and label it with $\sigma$; But this doesn't mean that $|p,\sigma>$ is eigenvalue of (say) $J_3$. In fact $G_k$ was defined such that its elements fix momentum $k$; so in particular there is no guarantee that elements of $G_k$ will also fix some $p$ different from $k$. In general they will not. So elements of $G_k$ (in general) may not have any eigenvectors in $V_p$ for $p$ different from $k$.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user user10001
I understand that $G_k$ may not have any eigenvectors in $V_p$. I don't understand at what stage your Hilbert space corresponds to physical reality. When Hilbert space corresponds to reality, you should be able to find a complete set commuting observables such that $P^{\mu} |p,\sigma\rangle = p^{\mu} |p,\sigma\rangle$ and $J |p,\sigma\rangle = \sigma |p,\sigma\rangle$.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user hans
Ok, thank you very much.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user hans
+1: I had parallel doubts and this answer helped me a lot. (also thanks to the OP for coincidentally coming up with similar doubts!)

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user 1989189198
+ 5 like - 0 dislike

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.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user Frederic Brünner
answered May 7, 2013 by (1,120 points)
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 $|p,\sigma\rangle = N(p)U(L(p))|k,\sigma\rangle$.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user hans
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.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user user1504
@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 $p \neq 0$.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user hans
That doesn't contradict anything I said. It doesn't have to be the eigenvalue of $J_z$.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user user1504
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 $J |p \neq k, \sigma\rangle \neq \sigma |p \neq k, \sigma\rangle$.

This post imported from StackExchange Physics at 2015-03-30 13:59 (UTC), posted by SE-user hans

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar\varnothing$sicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.