# Malament's theorem and the edge-of-the-wedge theorem

There is a little known theorem of relativistic quantum mechanics related to the localizability of particles in a relativistic setting. Consider a quantum theory defined on Minkowski space, with a Hilbert space $\mathcal{H}$ on which acts a set of unitary symmetry transformations $\hat{U}(a, \Lambda)$, such that $\hat{U}$ is the associated unitary transformation of $\mathcal{H}$ associated with the PoincarĂ© transform $(a, \Lambda)$. Let's assume a subset $\Delta$ of Minkowski space such that, for a foliation of Minkowski space by spacelike hypersurfaces $\Sigma_t$, we have $\Delta_t = \Delta \cap \Sigma_t$ a compact set.

The idea of *localizability* is that we can say with certitude whether or not a particle is entirely localized within some region $\Delta$. In our case here, what we want is to associate an operator $\hat{E}_\Delta$ to our subset $\Delta$, with two possible eigenstates : either our particle is entirely localized within $\Delta$ with a probability of $1$ (with an eigenvalue of $1$), or that it is certain that it is not within $\delta$, also with probability $1$ (with an eigenvalue of $0$).

Malament's theorem relies on four assumptions :

- For any region $\Delta$ and any vector $a$, $$\hat{U}(a) \hat{E}_\Delta \hat{U}^\dagger(a) = \hat{E}_{\Delta + a}$$
- For any timelike translation $a$, the generator $H(a)$ of the one-parameter group $U(ta)$ has a spectrum bounded from below.
- If $\Delta$ and $\Delta'$ are two disjoint regions, then $$\hat{E}_\Delta \hat{E}_{\Delta'} = 0$$

Despite folk physics idea on the topic, there are most certainly relativistic quantum theories with particle-like objects, and in a similar manner, we also have the option of considering, for instance, a quantum theory for a scalar field for which we will pick as initial condition a wavefunction with compact support. So how does this relate to Malament's theorem?

## The Polyakov quantization

The simplest case to study is the quantization of a relativistic point particle. Despite its historical problems, this is entirely feasible, at least as far as free particles are concerned. To quickly recapitulate, we have the action

\begin{equation} S[X, e] = \int_{\tau_a}^{\tau_b} d\tau \frac{1}{2} (\frac{1}{e} \dot{X}^\mu(\tau) \dot{X}_\mu(\tau) - m^2 e) \end{equation}Its canonical momenta are

\begin{eqnarray} P_\mu &=& \frac{\partial L}{\partial \dot{X}^\mu} = \frac{1}{e} \dot{X}_\mu\\ p &=& \frac{\partial L}{\partial \dot{e}} = 0 \end{eqnarray}$e$ has no associated momentum. $P_\mu$ can be inverted simply as

\begin{eqnarray} \dot{X}_\mu(P) = e P_\mu \end{eqnarray}Our Hamiltonian is then

\begin{eqnarray} H &=& \dot{X}^\mu(P) P_\mu - \frac{1}{2} (\frac{1}{e} \dot{X}^\mu(P) \dot{X}_\mu(P) - m^2 e)\\ &=& \frac{e}{2} (P_\mu P^\mu + m^2) \end{eqnarray}This is roughly just the constraint of the momentum to be on the mass shell, as we can check by computing the equations of motion,

\begin{eqnarray} \dot{X}^\mu &=& \frac{\partial H}{\partial P_\mu} = e P^\mu\\ \dot{P}_\mu &=& - \frac{\partial H}{\partial X^\mu} = 0\\ \dot{e} &=& \frac{\partial H}{\partial p} = 0\\ \dot{p} &=& - \frac{\partial H}{\partial e} = \frac{1}{2} (P_\mu P^\mu + m^2) \end{eqnarray}...

The quantization of our theory simply from the Poisson brackets can be performed rather simply, using the Stone-von Neumann theorem, by picking the Hilbert space $\mathcal{H} = L^2(\mathbb{R}^4, d\mu)$, but we know all too well what happens using a naive quantization of our theory.

...

Now let's see how this relates to Malament's theorem. $\hat{X}$ is easy enough to use for the construction of a localization operator. Considering simply a region $\Delta_t$ at a time $t$, we'll say that our state is localized within $\Delta$ if $\langle \hat{X} \rangle_\psi$ has a spectrum entirely localized in $\Delta$.

Last updated :

*2019-10-21 15:57:28*