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# Canonical quantization and representation theory

When it comes to canonical quantization, the focus is quite often on the operators of the theory, and not on the Hilbert space itself.

## Some trivial examples

The simplest cases we can consider are finite-dimensional Hilbert spaces, which are simply finite-dimensional vector spaces. In the case of quantum mechanics, they are all isomorphic to $\mathbb{C}^n$ for some $n$.

## The Stone-von Neumann theorem

The most important theorem for what we want to do is the Stone-von Neumann theorem, which relates the Hilbert spaces we can apply to a

\begin{eqnarray} \left[ \hat{x}_i, \hat{p}_j \right] &=& -i\hbar \hat{I}\\ \left[ \hat{x}_i, \hat{x}_j \right] &=& 0\\ \left[ \hat{p}_i, \hat{p}_j \right] &=& 0 \end{eqnarray}

If $\mathcal{H}$ is finite-dimensional, there are no such operators, as can be checked by considering the trace, as $\text{Tr}(AB) = \text{Tr}(BA)$ :

\begin{eqnarray} \text{Tr}(\left[ \hat{x}_i, \hat{p}_j \right]) &=& \text{Tr}(\hat{x} \hat{p}) - \text{Tr}(\hat{p} \hat{x})\\ &=& \text{Tr}(\hat{x} \hat{p}) - \text{Tr}(\hat{x} \hat{p})\\ &=& 0 \text{Tr}(-i\hbar \hat{I}) &=& -i\hbar \text{Tr}(\hat{I})\\ &=& -i\hbar n \end{eqnarray}

Short of $n = 0$ (not the most interesting quantum theory) or $\hbar = 0$, this means that only infinite-dimensional Hilbert spaces can have such operators, as the trace operator isn't defined in general (and in particular not for $\hat{I}$).

We know that on $L^2(\mathbb{R}^n, d\mu)$, the operators defined by

\begin{eqnarray} \hat{x}_i \psi(\vec{x}) = x_i \psi(\vec{x})\\ \hat{p}_i \psi(\vec{x}) = -i\hbar \partial_i \psi(\vec{x})\\ \end{eqnarray}

on some appropriate dense subset of the Hilbert space do obey that condition. What we will see is that, in some sense, these are the only ones who do, at least for a finite number of such operators.

### The Heisenberg group

The canonical commutation relations are fulfilled by the Heisenberg algebra. For $n$ generators, they correspond to the $2n + 1$-dimensional algebra $H_{2n + 1}$, with generators $\hat{x}_1, \ldots, \hat{x}_n, \hat{p}_1 \ldots \hat{p}_n, -i\hbar \hat{I}$.

In the $n = 1$ case, the Heisenberg algebra is the group of real upper-triangular $3 \times 3$ matrices, ie for $A \in \mathfrak{h}$,

\begin{eqnarray} A = \begin{pmatrix} 0 & a & b\\ 0 & 0 & c\\ 0 & 0 & 0 \end{pmatrix} \end{eqnarray}

As a matrix Lie group, its Lie brackets are simply the commutator. We have

\begin{eqnarray} A_1 A_2 = \begin{pmatrix} 0 & a_1 & b_1\\ 0 & 0 & c_1\\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & a_2 & b_2\\ 0 & 0 & c_2\\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & a_1 c_2\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} \end{eqnarray}

and therefore

\begin{eqnarray} [A_1, A_2] = \begin{pmatrix} 0 & 0 & (a_1 c_2 - a_2 c_1)\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} \end{eqnarray}

and

\begin{eqnarray} [A_3, [A_1, A_2]] = 0 \end{eqnarray}

So that our algebra is indeed a Lie algebra. This algebra is fairly obviously $3$-dimensional, with a possible basis

\begin{eqnarray} A = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix},\ B = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix},\ C = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix} \end{eqnarray}

From what we've seen, this gives us $[A, B] = 0$, $[A, C] = B$ and $[B, C] = 0$. If we define $A = X$, $C = P$ and

\begin{eqnarray} B = \begin{pmatrix} 0 & 0 & -i\hbar\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} \end{eqnarray}

Then $(X, P, B)$ forms a basis of the Heisenberg group, with the proper commutation relations.

The corresponding group to this algebra is the Heisenberg group. If we simply use the exponential map, this is just the group given by

\begin{eqnarray} g = e^{A} \end{eqnarray}

### Representations of the Heisenberg group

Now that we have cast our canonical commutation relation in terms of a group, we ask ourselves : on what vector spaces does a representation of the Heisenberg group exists, and by extension, on what Hilbert space?

We already know that this space cannot be finite-dimensional, and for the purpose of quantum theory, we also require it to be separable. So that we can limit ourselves to infinite-dimensional Hilbert spaces with a countable basis. As $\hat{x}$ and $\hat{p}$ are observables, we would like them to be hermitian, and by extension, the Weyl operators to be unitary. We therefore are looking for unitary representations on infinite-dimensional separable Hilbert spaces, and since we can construct these from the irreducible such representations, this is what we will be focusing on.

As we've mentioned before, the standard operators of quantum mechanics are indeed such a representation, as we will now prove. Our Heisenberg group is $2n + 1$-dimensional, with parameters $(\vec{a}, \vec{b}, c)$. We will show that the family of operators $U_h(M(\vec{a}, \vec{b}, c))$ on $L^2(\mathbb{R}^n, d\mu)$, with the action

$$U_h(M(\vec{a}, \vec{b}, c)) \psi(\vec{x}) = e^{ih (\vec{b} \cdot \vec{x} + hc)} \psi(\vec{x} + h \vec{a})$$

is indeed such a representation.

First we show that $U$ is unitary, so that $\langle U \xi, U \psi \rangle = \langle \xi, \psi \rangle$ :

\begin{eqnarray} \langle U \xi, U \psi \rangle &=& \int (U_h(M(\vec{a}, \vec{b}, c)) \xi(\vec{x}))^* (U_h(M(\vec{a}, \vec{b}, c)) \psi(\vec{x})) d^nx\\ &=& \int (e^{ih (\vec{b} \cdot \vec{x} + hc)} \xi(\vec{x} + h \vec{a}))^* e^{ih (\vec{b} \cdot \vec{x} + hc)} \psi(\vec{x} + h \vec{a}) d^nx\\ &=& \int e^{-ih (\vec{b} \cdot \vec{x} + hc)} \xi^*(\vec{x} + h \vec{a}) e^{ih (\vec{b} \cdot \vec{x} + hc)} \psi(\vec{x} + h \vec{a}) d^nx\\ &=& \int \xi^*(\vec{x} + h \vec{a}) \psi(\vec{x} + h \vec{a}) d^nx\\ \end{eqnarray}

As the Lebesgue measure is translation-invariant, we indeed get

\begin{eqnarray} \langle U \xi, U \psi \rangle &=& \int \xi^*(\vec{x}) \psi(\vec{x}) d^nx\\ &=& \langle \xi, \psi \rangle \end{eqnarray}

Now to show that this representation is irreducible.

We would also like to show that the appropriate Lie algebra gives us the standard operators. Let's consider the three tangent vectors :

\begin{eqnarray} \frac{\partial}{\partial \vec{a}} \left[ U_h(M(\vec{a}, \vec{b}, c)) \psi(\vec{x}) \right] \bigg\rvert_{\vec{a}, \vec{b}, c = 0} &=& \left[ e^{ih (\vec{b} \cdot \vec{x} + hc)} \frac{\partial}{\partial \vec{a}} \psi(\vec{x} + h \vec{a})\right]\bigg\rvert_{\vec{a}, \vec{b}, c = 0}\\ &=& \left[\frac{\partial \vec{x} + h \vec{a}}{\partial \vec{a}} \psi'(\vec{x} + h \vec{a})\right]\bigg\rvert_{\vec{a}, \vec{b}, c = 0}\\ &=& h \frac{\partial}{\partial \vec{x}}\psi(\vec{x})\\ \frac{\partial}{\partial \vec{b}} \left[ U_h(M(\vec{a}, \vec{b}, c)) \psi(\vec{x}) \right] \bigg\rvert_{\vec{a}, \vec{b}, c = 0} &=& \left[ \frac{\partial}{\partial \vec{b}}(e^{ih (\vec{b} \cdot \vec{x} + hc)}) \psi(\vec{x} + h \vec{a})\right]\bigg\rvert_{\vec{a}, \vec{b}, c = 0}\\ &=& \left[ \vec{x} e^{ih (\vec{b} \cdot \vec{x} + hc)}) \psi(\vec{x} + h \vec{a}) \right]\bigg\rvert_{\vec{a}, \vec{b}, c = 0}\\ &=& \vec{x} e^{ih} \psi(\vec{x}) \end{eqnarray}

The hard part remaining is showing that this is true for all appropriate Hilbert spaces.

## Non-irreducible representations of the Heisenberg group

While having the irreducible representation is fine, that still leave other possible representations, and our quantization process may also involve other Poisson brackets.

### Products of Lie algebras

The specific example which made me write this article was the case of the relativistic point particle, which has, in addition to the classical Poisson bracket $[X^\mu, P_\nu] = \delta^\mu_\nu$, the BRST ghost-antighost pair $\left\{ c_\alpha, \bar{c}_\beta \right\} = \delta_{\alpha\beta}$. The resulting operator algebra is

\begin{eqnarray} \left[ \hat{X}^\mu, \hat{P}_\nu \right] &=& \delta^\mu_\nu \hat{I}\\ \left[ \hat{X}_\mu, \hat{X}_\nu \right] &=& 0\\ \left[ \hat{P}_\mu, \hat{P}_\nu \right] &=& 0 \end{eqnarray}
Last updated : 2020-01-09 12:27:47