Intuitive

The canonical commutation relations tell us that we can't measure the momentum and the location of a particle at the same time with arbitrary precision.

Concrete

Quantum Mechanics

\begin{equation} \label{eq:commquantummech} [\hat{p}_i,\hat{x}_j] = -i \delta_{ij} .\end{equation}

This equation tells us that, for example, a measurement of momentum in the x-direction changes what we can expect for a measurement of the location on the x-axis. Take note that only for measurements along the same axis is the commutator non-zero. A measurement of momentum in the y-direction has no influence on what we can expect for the position on the x-axis. In other words, this means that we can't know momentum and position in the same direction at the same time with arbitrary precision.

Every time we measure the location of a particle the momentum becomes uncertain and vice versa. This is known as Heisenberg's uncertainty principle. Analogous observations can be made for angular momentum along different axes, because the commutator for the corresponding operator is non-zero, too. In general, we can check for any two physical quantities if they commute with each other. If they don't, we know that they can't be measured at the same time with arbitrary precision.

Derivation

In quantum mechanics, we identify the conserved quantities like energy and momentum with the generators of the corresponding symmetry. This is motivated by Noether's theorem which connect each conserved quantity with a symmetry.

For example, invariance under translation yields conservation of momentum and therefore

\[ \text{momentum } \hat{p}_i \rightarrow \text{ generator of spatial-translations } -i \partial_i \]

Analogous, the invariance under the action of the generator of time-translations leads us to the conservation of energy. Consequently \[ \text{energy } \hat{E} \rightarrow \text{ generator of time-translations } i \partial_o. \]

With these operators at hand, we can derive the canonical commutation relation \[[\hat{p}_i,\hat{x}_j] \Psi = (\hat{p}_i\hat{x}_j -\hat{x}_j \hat{p}_i ) \Psi = (-\partial_i\hat{x}_j + \hat{x}_j \partial_i )i \Psi \] \begin{equation}\underbrace{=}_{\text{product rule}} -(\partial_i\hat{x}_j)i \Psi - \ \hat{x}_j ( \partial_i \Psi ) + \hat{x}_j \partial_i \Psi \underbrace{=}_{\text{because } \partial_i\hat{x}_j = \frac{\partial x_j}{\partial x_i} } - i \delta_{ij} \Psi .\end{equation}

This equation holds for arbitrary $\Psi$, because we made no assumptions about $\Psi$ and therefore we can write the equation without it

Quantum Field Theory

The canonical commutation relation in quantum field theory reads

\begin{equation} \label{qftcomm} [\Phi(x), \pi(y)]=\Phi(x) \pi(y) - \pi(y) \Phi(x) = i \delta(x-y) \end{equation} where $\delta(x-y)$ is the Dirac delta distribution and $\pi(y) = \frac{\partial \mathscr{L}}{\partial(\partial_0\Phi)}$ is the conjugate momentum.

It tells us that the fields in quantum field theory $\Phi(x)$ can‘t be simply a function, but must be operators. In contrast, ordinary functions and numbers commute:

For example $f(x)=3x$ and $g(y)= 7y^2 +3$ clearly commute $$ [f(x) , g(x)]= f(x)g(x) - g(x) f(x) = 3x (7y^2 +3) -(7y^2 +3) 3x =0. $$

Therefore, we have to conclude that quantum fields are operators that act, like every operator in quantum theory, on abstract states. Upon closer inspection, we see that the action of the fields can be interpreted as the creation and destruction of particles.

The conjugate momentum is the conserved quantity that follows via Noether‘s theorem from invariance under translations of $\Phi(x)$ itself: $ \Phi(x) \rightarrow \Phi(x) + \epsilon$.

For example, for spin $0$ fields the conjugate momentum is$$\pi(x) = \frac{\partial \mathscr{L}}{\partial(\partial_0\Phi(x))} = \frac{\partial }{\partial(\partial_0\Phi(x))} \frac{1}{2}( \partial _{\mu} \Phi(x) \partial ^{\mu} \Phi(x) -m \Phi^2(x)) $$ $$ = \partial_0\Phi(x). $$

Take note that, for brevity, it‘s conventional to write $\Phi(x)$ instead of $\Phi(x,t)$ , which means we include $t$ in $x$: $x_0=t$, $x_1=x$, $x_2=y$ and $x_3=z$.

Derivation

Analogous to what we do in quantum mechanics, we identify the conserved quantities like energy and momentum with the generators of the corresponding symmetry.

The invariance under displacements of the field itself $\Phi \rightarrow \Phi - i \epsilon$ yields a new conserved quantity, called conjugate momentum $\Pi$. So completely analogous to what we discussed above for quantum mechanics, we now identify the conjugate momentum density with the corresponding generator

\[ \mathrm{conj. \ mom. \ density \ } \pi(x) \rightarrow {\mathrm{gen. \ of \ displ. \ of \ the \ field \ itself: \ }} -i \frac{\partial}{\partial \Phi(x)} \]

We can then calculate

\[[\Phi(x), \pi(y)] \Psi = \left[\Phi(x), -i \frac{\partial}{\partial \Phi(y)}\right] \Psi \] \begin{equation} \label{eq:field-conjmom-commutator} \underbrace{=}_{\text{product rule}} -i \Phi(x) \frac{\partial \Psi}{\partial \Phi(y)} + i \Phi(x) \left(\frac{\partial\Psi}{\partial \Phi(y)} \right) + i \left( \frac{\partial \Phi(x)}{\partial \Phi(y)} \right) \Psi = i \delta(x-y) \Psi \end{equation} Again, the equations hold for arbitrary $\Psi$ and we can therefore write

\begin{equation} [\Phi(x), \pi(y)] = i \delta(x-y) \end{equation}

Abstract

Why is it interesting?

The canonical commutation relations encapsulate what quantum theory is all about.

In quantum mechanics, it tells us that we need to work with measurement operators.

In quantum field theory it tells us that our fields are operators.

History

“I went back to Cambridge at the beginning of October 1925, and resumed my previous style of life, intense thinking about these problems during the week and relaxing on Sunday, going for a long walk in the country alone. The main purpose of these long walks was to have a rest so that I would start refreshed on the following Monday. It was during one of the Sunday walks in October 1925, when I was thinking about this (uv- vu), in spite of my intention to relax, that I thought about Poisson brackets. I remembered something which I had read up previously, and from what I could remember, there seemed to be a close similarity between a Poisson bracket of two quantities and the commutator. The idea came in a flash, I suppose, and provided of course some excitement, and then came the reaction “No, this is probably wrong”. I did not remember very well the precise formula for a Poisson bracket, and only had some vague recollections. But there were exciting possibilities there, and I thought that I might be getting to some big idea. It was really a very disturbing situation, and it became imperative for me to brush up on my knowledge of Poisson brackets. Of course, I could not do that when I was right out in the countryside. I just had to hurry home and see what I could find about Poisson brackets. I looked through my lecture notes, the notes that I had taken at various lectures, and there was no reference there anywhere to Poisson brackets. The textbooks which I had at home were all too elementary to mention them. There was nothing I could do, because it was Sunday evening then and the libraries were all closed. I just had to wait impatiently through that night without knowing whether this idea was really any good or not, but I still think that my confidence gradually grew during the course of the night. The next morning I hurried along to one of the libraries as soon as it was open, and then I looked up Poisson brackets in Whitackers Analytical Dynamics, and I found that they were just what I needed.” Dirac