# Canonical commutation relation field theory

This is a question on computation. If we set the quantum field canonical commutation relations as

$$[\hat{\pi}(x),\hat{\phi}(x')] = -i\hbar \delta(x-x'),$$

the goal is to find the Fourier transform of this commutation relation:

$$[\hat{\pi}_k,\hat{\phi}_{k'}] = -i\hbar \delta_{k,k'}.$$

My Fourier transforms are defined as

$$\hat{\phi}_k = \frac{1}{L^{1/2}}\int_0^L dx e^{-ikx} \hat{\phi}(x),$$

$$\hat{\pi}_k = \frac{1}{L^{1/2}}\int_0^L dx e^{ikx} \hat{\pi}(x).$$

Direct substitution yields

$$[\hat{\pi}_k,\hat{\phi}_k'] = \frac{1}{L}\int_0^Ldx\int_0^L dx' e^{ikx-ik'x'}[\hat{\pi}(x),\hat{\phi}(x')] = \frac{-i\hbar}{L}\int_0^L dx e^{ix(k-k')},$$

where this last integral is a delta function. My question is, is it true that

$$\int_0^L\! dx ~e^{ix(k-k')} ~=~ L~\delta_{k,k'}?$$

(I get always confused with the normalization constants.)

• The commutator you start with are usually called the "equal time commutation relations", because most field theorists use the Heisenberg or interaction pictures. – Sean E. Lake Dec 6 '17 at 0:03

Yes, $\int_0^L dxe^{ix(k-k')} = L\delta_{k,k'}$ which is the Kroenecker delta for discrete Fourier series. For true Fourier transform (where $L\rightarrow\infty$), the relation holds true but the Kroenecker $\delta_{k,k'}$ delta gets replaced by a Dirac delta $\delta(k-k')$.

Proof for discrete Fourier series (since you defined your transform with definite integrals):

$$\int_0^L dxe^{i(k-k')x} = \frac{e^{i(k-k')L} -1}{i(k-k')}$$ Now, for Fourier series, this transform is defined only for $k=\frac{2\pi n}{L}$ for any $n\in\mathbb{N}$. Thus, if $k=k'$, the integral just gives $L=L\delta_{k,k'=k}$ ($\delta_{k,k'}=1$ if $k=k'$ and $0$ if not). For $k\neq k'$, we get:

$$\frac{e^{i(k-k')L} -1}{i(k-k')} = L\frac{e^{2\pi i(n-n')} -1}{2\pi i(n-n')}=0=L\delta_{k,k'\neq k}$$

since $e^{2\pi in}=1$ for any $n\in\mathbb{N}$. Thus you have your result:

$$\int_0^L dxe^{i(k-k')x} = L\delta_{k,k'}$$ for any $k$.

For the continuous case the identiy holds but you replace the Kroenecker delta by a Dirac delta (proof not necessarily obvious). Since it is beyond the scope of this question I won't draw it here :).