# Associating a Unitary operator to proper Lorentz transformations?

If one reads eg page 32 of Srednicki where he says:

In quantum theory, symmetries are represented by unitary (or antiunitary) operators. This means that we associate a unitary operator U(Λ) to each proper, orthochronous Lorentz transformation Λ. These operators must obey the composition rule...

Where does all this come from? Is it Wigners theorem? Probability conservation?

For instance equation 2.15 in the same link:

$$U(\Lambda)^{-1}P^{\mu}U(\Lambda) = \Lambda^{\mu}_{\nu}P^{\nu}$$

I understand the right hand side no problem: The Lorentz matrix acts on the four vector and "rotates it" end of story. But why do we need the left hand side? What does the left hand side even mean, what is $U$, except being a unitary operator?

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Answering this question properly would require your understanding of the mathematics and principles behind Lie Algebras, group theory, and Field Theory. However, we are in the awkward predicament that knowing these things would automatically answer your question. Thus, you will have to forgive me in advance because I am not very good at summarizing in a few words an entire, math-heavy chapter of a textbook. But I will try.

This statement has foundations in Lie algebra. If you were to work through it (and there is a compendium of literature available on the subject), you might eventually come upon the concept that every observable in nature corresponds to a symmetry and that every symmetry can be described as actions of a unitary operator. So the LHS means the same as the RHS but using unitary operators.

That's the easy part. Why we need it is because in field theory, unitary operators are much easier to manipulate than our non-unitary representations. In field theory, we promote states to operators and then "sandwich" the resulting combination in field states. For example:

$$U(\Lambda)^{\dagger}P^{\mu}U(\Lambda)|\alpha>~=~\Lambda_\nu^\mu P^\nu |\alpha>$$

It may not be apparent from this why we use the unitary operator notation, but when I "sandwich" it between two states it becomes apparent:

$$<\alpha|U(\Lambda)^{\dagger}P^{\mu}U(\Lambda)|\alpha>~=~<\alpha|\Lambda_\nu^\mu P^\nu |\alpha>$$

The RHS is difficult to interpret (is the operator changing? will I have 2 different states at the end?), however we can look at the LHS and see that $U(\Lambda)$ acts on $|\alpha>$ and $U^\dagger (\Lambda)$ acts similarly on $<\alpha|$. Assuming this corresponds to a rotation, we can see that the total system is rotated so that now we are finding the expectation value of the momentum four-vector for the rotated state, which is exactly what we should interpret it as.

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No problem, I've seen all this stuff but it's always nice to see some other dude explain in his own words. Thanks for the reply. – Faraday Apr 23 '13 at 16:27

The fact that symmetries are represented by unitary or anti-unitary operators is, indeed, due to Wigner's theorem.

When we write $U(\Lambda)$ what we really mean is that $U$ is mapping from the Lorentz group to the set of unitary operators on the Hilbert space $\mathcal H$ of the quantum field theory; $$U: \mathrm{SO}(3,1)^+\to \mathrm U(\mathcal H)$$ Here I have restricted to the proper, orthochronous Lorentz group (the identity component). The equation you wrote down does not just fall out of the sky, it can be proven. Weinberg vol.1 section 2.4 has a lot of detail on this (and so do books on Lie groups and Lie algebras). The basic idea is to write $U(\Lambda)$ as an operator exponential of generators (operators) of Lorentz transformations and to use the following lemma about operator exponentials: $$e^X Y e^{-X} = e^{\mathrm{ad}_X}Y, \qquad \mathrm{ad}_XY = [X,Y],$$ and the desired result follows.

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Nice answer. I have loved it. – omehoque Jul 2 '13 at 23:36
@Ome Well thanks. – joshphysics Jul 3 '13 at 1:14
I am learning QFT from Srednicki's book. In the book eqn (33.1) - (33.4) are closely related with this topic. Can you please tell me where I can find a clear treatment of the unitary representation of Lorentz group? I wanted to mail you, but I don't have it. :P – omehoque Jul 3 '13 at 20:23
I would say that the best resource is Weinberg's QFT, but I don't think his treatment is that great either to be honest. I think the book Group Theory in Physics by Wu-Ki Tung has a decent discussion of this topic as well. Other than that, you'll have to piece lots of different resources together to really understand this I'd say. – joshphysics Jul 3 '13 at 22:23
Thanks a lot :) – omehoque Jul 3 '13 at 22:25

This is not as such related to field theory. The same story happens in regular quantum mechanics, but then just rotations or gallilei transformations. I do not think it is necessary to resort to Lie groups in order to understand what is going on.

I would interpret $U(\Lambda)$ as the operator which translates a quantum state into the frame of reference identified by $\Lambda$. So you could define new states by $|\tilde\alpha\rangle = U(\Lambda) |\tilde\alpha\rangle$ and use the old momentum operators $P^\mu$. This is you left-hand-side. Since you are shifting to a new frame of reference you could also simply transform the operator according to you right-hand-side and leave the states unchanged.

For rotations this corresponds to either rotating the quantum states (LHS) or rotating the operators you use to calculate expectation values (RHS). It is similar to the relationsship betwen the Schrödinger picture and the Heisenberg picture of time-evolution. Transform either the states or the operators, but only one of them. In the case of rotations (which is what I am familiar with), there are explicit formulas for $U$ given a 3d rotation matrix. These are called the Wigner D-matrices.

The formal way of constructing the unitary operators can probably be understood most deeply by going into Lie groups as noted above, and there are certain relevant mathematical structures which makes the relationship between $U$ and $\Lambda$ more clear ("the map $\Lambda\mapsto U$ is a unitary representation of the Lorentz group").

In this case the momentum operator transforms using $\Lambda$ directly, but more general transformations can occur. A good introduction, which is not too heavy on group theory stuff is Chapter 3 in Sakurai, Modern Quantum Mechanics where you can also find much about Wigner D matrices and how they look.

I am not sure if you think this was a restatement of the above, but I hope it helped a bit.

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+1 this is an excellent addendum to my interpretation – Jim Apr 24 '13 at 0:11