# ### Sidebar

formulas:canonical_commutation_relations

$\color{black}{[}\color{royalblue}{\hat{p}_i},\color{firebrick}{\hat{x}_j}\color{black}{]} \color{black}{ = -i} \color{olive}{\hbar} \color{black}{\delta_{ij}}, \ \color{black}{[}\color{firebrick}{\hat{x}_i} , \color{firebrick}{\hat{x}_j}\color{black}{]} \color{black}{=} \color{darksalmon}{0}, \ \color{black}{[} \color{royalblue}{\hat{p}_i} , \color{royalblue}{\hat{p}_j} \color{black}{]} \color{black}{=} \color{darksalmon}{0}$

# Canonical Commutation Relations

## 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.

However, can measure the location on different axes - e.g. the location on the x-axis and the location on the y-axis - with arbitrary precision. Equally, we can measure the momentum in the direction of different axes with arbitrary precision.

The action, in the form h/2π, also occurs in the Heisenberg Uncertainty Relationship, and here it sets the limit of precision to which the conjugate coordinates (momentum and position; or energy and time) can be determined simultaneously for one particle. It also occurs in the Dirac commutator relations (discovered by the British theoretical physicist, Paul Dirac (1902–84)) which show that the order in which measurements of these conjugate coordinates are carried out does make a difference - a tiny difference of magnitude h/2π .

The Lazy Universe by Coopersmith

## Concrete

Quantum Mechanics

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

The mathematically rigorous theory of these commutation relations can be found in

• Jørgensen, P.E.T.; Moore, R.T.: Operator commutation relations. Dordrecht: Reidel, 1984
• Schmüdgen, K.: Unbounded operator algebras and representation theory. Birkhäuser Verlag, Basel, 1990

However, take note that the canonical commutation relations are mathematically problematic nevertheless.

One problem is that the operators involved are unbounded. If we want that they represent physical observables they have to be self-adjoint; but on their respective domains of self-adjointness the commutator on the left-hand side is undefined.

Another problem is that the canonical commutation relations rely on the possibility of choosing global coordinates on $\mathbb{R}^3$. Since this is, in general not possible on arbitrary configuration spaces it is unclear how they can then be generally valid.

The problem of finding an appropriate mathematical interpretation of the canonical commutation relations is the subject of quantization theory.

• A good discussion can be found here.

## 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. 