# Representations of Lorentz group on Hilbert spaces

Question 1:

Representation theory and group theory are quite new subjects for me and I'm having troubles in understanding why a representation of a group on Hilbert space act on operators as (here I take the example of the Lorentz group acting on a scalar field operator):

$$\hat{\phi}'(x) = e^{\frac{i}{2} \omega_{\mu\nu}J_r^{\mu\nu}} \hat{\phi}(x) e^{-\frac{i}{2} \omega_{\mu\nu}J_r^{\mu\nu}}$$

While for classical fields I have:

$$\phi'(x) = e^{-\frac{i}{2} \omega_{\mu\nu}J_{r'}^{\mu\nu}} \phi(x)$$

where $r$ and $r'$ are the correct representations on the respective spaces.

My attempt to answer this question is the following. In the case of operators the vector space where the representation is realised is the Hilbert space, let's say $\mathbb{H} = \text{span}\{\vert k \rangle \}_k$. Therefore, the representation of the Lorentz group act as:

$$e^{-\frac{i}{2} \omega_{\mu\nu}J_{r}^{\mu\nu}} \vert k \rangle$$

Then as for the Heisenberg and Shrödinger picture and time translations, I can choose to transform either the operators or the states. If I want to transform the operators I need to have:

$$e^{\frac{i}{2} \omega_{\mu\nu}J_r^{\mu\nu}} \hat{\phi}(x) e^{-\frac{i}{2} \omega_{\mu\nu}J_r^{\mu\nu}}$$

in order to have in both cases the same matrix elements under transformation:

$$\langle k \vert e^{\frac{i}{2} \omega_{\mu\nu}J_r^{\mu\nu}} \hat{\phi}(x) e^{-\frac{i}{2} \omega_{\mu\nu}J_r^{\mu\nu}} \vert k' \rangle$$

Question 2:

Is the representation $r'$ the same as $r$? I know that $J_{r'}^{\mu\nu} = i(x^\mu\partial^\nu - x^\nu\partial^\mu)$. Can I obtain $J_{r}^{\mu\nu}$ just by replacing $\hat p^{\mu} = i\partial^\mu$?

For question 1, you ask why an operator $A$ is transformed under a unitary transformation $U$ as $A' = U A U^\dagger$? If yes, the answer is simply that the 2 states $|a>$ and $|b>$ being transformed in $|a'> = U |a>$ and $|b'>= U |b>$, you want

$$<a|A|b> = <a'|A'|b'>$$

Since $<a'|A'|b'> = <a | U^\dagger A' U |b>$, the equality above being true for any states $|a>$ and $|b>$, you deduce: $A = U^\dagger A' U$. The unitary of $U$ tells you that $U^{-1} = U^\dagger$ and thus $A' = U A U^\dagger$.

• Why do you require the mean value of A to be the same after transformation? – Worldsheep Jan 2 '15 at 14:23
• because the $U$ transformation is supposed to be a good symmetry of your theory. – Paganini Jan 2 '15 at 14:44
• And if it is not the case? Why for time translations we transform either the states (Shrodinger picture) or the operators (Heisenberg picture)? – Worldsheep Jan 2 '15 at 14:47
• the way you treat the time evolution of your theory (in states for Schrodinger, in operators for Heisenberg and a mix of the 2 for the interaction picture) shouldn't have any consequence on physics. Consequently, if the prime quantities are the ones in the Heisenberg picture and the no-prime the ones in the Schrodinger, you still want: $<a|A|b> = <a'|A'|b'>$. – Paganini Jan 2 '15 at 14:54
• I'm getting confused by the fact that in your example you transform generators together with states, while in the time evolution picture you transform one or the other... – Worldsheep Jan 2 '15 at 16:18

The answer to your first question can be given by appealing to the notion of symmetry in quantum mechanics. I don't think any discussion of the different time evolution pictures is really necessary.

Let a transformation $T$ be a quantum symmetry, then given a transition probability built out of states and operators, if we replace states $\psi$ that appear in this probability with $T\psi$, and if we replace operators $O$ that appear with $TOT^{-1}$, then the transition probability will not change.

This fact makes defining operator transformation as $O\mapsto TOT^{-1}$ a useful concept. Let's see why the above fact is true.

Let a quantum system with Hilbert space $\mathcal H$ be given. The transition probability from state $\phi\in\mathcal H$ to state $\psi\in\mathcal H$ is defined as \begin{align} P(\psi,\phi) = \left|\frac{(\psi,\phi)}{\|\psi\|\|\phi\|}\right|^2. \end{align} Recall that states must have nonzero norm, so we don't have to worry about dividing by zero on the right hand side. Transition probabilities are what we compute in quantum mechanics to make predictions of probabilities of outcomes of measurements.

A symmetry of this system is any invertible transformation $T:\mathcal H \to\mathcal H$ which preserves transition probablities. Explicitly, $T$ is a symmetry provided it is a bijection and \begin{align} P(T\psi, T\phi) = P(\psi,\phi) \end{align} for all nonzero $\psi,\phi\in\mathcal H$. Now, consider a nonzero state $\phi$ such that $\phi = O\xi$ for some $\xi\in\mathcal H$. The condition for symmetry guarantees that \begin{align} P(T\psi, TO\xi) = P(\phi, O\xi) \end{align} Since $T$ is invertible, we can insert $T^{-1}T = I$ in between the $O$ and the $\xi$ without changing the left hand side, so the condition guarantees that \begin{align} P(T\psi, TOT^{-1}T\xi) = P(\phi, O\xi). \end{align} A similar remark could be said if $O$ were a string $O_1O_2\cdots O_n$ of composed operators instead. In that case, we would have \begin{align} P(T\psi, TO_1T^{-1}TO_2T^{-1}\cdots TO_nT^{-1}T\xi) = P(\phi, O_1O_2\cdots O_n\xi). \end{align} This is a mathematical way of writing the fact we wanted to demonstrate.

The representations $r$ and $r'$ are not the same. As you've written them, $r$ is a representation of the symmetry group $G$ acting on $\mathcal H$, while $r'$ is a representation of $G$ acting on the target space $V$ of the theory. The target space is simply the space in which the values of the classical field lives. For a real scalar field, we would have $V=\mathbb R$. For a Lorentz vector field, we would have $V =\mathbb R^{3,1}$, etc.
The replacement $\hat p^\mu = i\partial^\mu$ is not the correct method of determining $r$. Notice, in particular, that $\hat p^\mu$ is an operator on $\mathcal H$, but $i\partial^\mu$ is (for each $\mu$) an operator on the set of admissible classical field configurations.