Klein-Gordon Canonical Commutation Relation (CCR)

In the complex Klein-Gordon field we regard as dynamical variables the field $\phi$, the complex conjugate of the field $\phi^*$, and the momenta $\pi$, $\pi^*$. I can't see how should arise the (equal time) canonical commutation relations, in particular:

$$\left[\phi\left(t, {\bf x}\right),\ \pi\left(t,{\bf y} \right)\right]=i\hbar\delta^3\left({\bf x}-{\bf y}\right)$$

because the momenta are time derivatives of the field, and hence a function of $(ct,{\bf x})$, so the commutator above should be zero because both are functions of the space time coordinates, $\pi$ is not a differential operator.

Where am I wrong?

The field and its conjugate momentum are operators. They act on the Hilbert space of the theory, which is not the space of square-integrable functions of position, as it is in single particle quantum mechanics. Rather, the configuration space of the theory is the set of all field configurations and the wavefunction is actually a functional, which ascribes an amplitude $\Psi[\phi(x^\mu)]$ for the field to be in a configuration $\phi(x^\mu)$.

QFT is just quantum mechanics applied to an infinite collection of degrees of freedom labeled by positions $x$. Once you get used to the switch it comes to seem pretty simple, but the set of all field configurations makes a huge space so naturally it is very difficult to make things mathematically rigorous. For the most part physicists can live without rigourously defining the configuration space.

The dictionary from ordinary quantum mechanics to quantum field theory is

$$\begin{array}{lcl} \mathrm{QM} && \mathrm{QFT}\\ t,i &\to& \left(t,x,y,z\right)\equiv\left(t,{\bf x}\right)\equiv x^{\mu}\\ q_{i}\left(t\right) &\to& \phi\left(x^{\mu}\right)\\ p_{i}\left(t\right) &\to& \pi\left(x^{\mu}\right)\\ \psi\left(t,q_{1},q_{2},\cdots,q_{N}\right) &\to& \Psi\left[\phi\left(x^{\mu}\right)\right]\\ \left[q_{i}\left(t\right),\ p_{j}\left(t\right)\right]~=~i\hbar\delta_{ij} &\to& \left[\phi\left(t, {\bf x}\right),\ \pi\left(t,{\bf y} \right)\right]=i\hbar\delta^3\left({\bf x}-{\bf y}\right)\\ \hat{H}\left(t\right) &\to& \hat{H}\left(t\right)=\int\mathrm{d}^{3}x\ \hat{\mathcal{H}}\left(x^{\mu}\right) \end{array}$$

You can represent the field momenta explicitly by functional derivatives:

$$\pi(x^\mu) ~=~ - i\hbar \frac{\delta}{\delta \phi(x^\mu)},$$

and check that this satisfies the commutation relationship.

• So, just as in Non-relativistic quantum mechanics commutation relations are part of the postulates of the theory, the same thing is true for quantum field theory isn't it? – Jorge Lavín Mar 11 '13 at 15:30
• @Nivalth Yes, the commutation relations are part of the definition of the theory. – Michael Brown Mar 11 '13 at 15:37
• I've always felt that the notation $\Psi[\phi(x^\mu)]$ is a bit misleading (if not strictly speaking meaningless even though it's common in physics) because $\Psi$ eats a field configuration $\phi$ and outputs a number, it does not eat the value $\phi(x^\mu)$ of the field configuration at the point $x^\mu$. In particular, I would imagine that this might confuse someone who is attempting to take literally the translation between $\psi$ and $\Psi$ in your list. How do you feel about this Michael? – joshphysics Jul 10 '13 at 9:24
• @joshphysics Of course you're right. I wrote $\phi(x^\mu)$ to indicate that $\phi$ is a function and the coordinates $x^\mu$ are relegated to the equivalent of indices. The $[\ ]$ on $\Psi$ are meant to indicate that it is a functional. One should feel free tweak the notation in whatever way makes sense to them. – Michael Brown Jul 10 '13 at 10:42