# Deriving commutation relations in second quantisation

I am trying to start from: \begin{align*} [\phi(x),\pi(x')] = i\hbar\delta(x-x') \\ [\phi(x),\phi(x')] = [\pi(x),\pi(x')]=0 \end{align*} to derive: \begin{align*} [a(k),a(k')^\dagger]=\delta_{kk'}\\ [a(k),a(k')]=[a(k)^\dagger,a(k')^\dagger]=0 \end{align*}

So starting with: \begin{align*} \phi(x) = \sum_k \left(\frac{\hbar c^2}{2\omega_k}\right)^\frac{1}{2}[a(k)u_k(x)+a(k)^\dagger u_k(x)^*] \end{align*} where $u_k(x) = \frac{1}{\sqrt{V}}e^{i(k \cdot x - \omega_k t)}$ and $\pi(x) = \frac{1}{c^2}\dot{\phi}(x)$ \begin{align*} &[\phi(x),\pi(x')] \\ &=-i\sum_{k,k'} \frac{\hbar}{2}\sqrt{\frac{\omega_k}{\omega_{k'}}}\left([a(k)^\dagger,a(k')]u_k(x)^*u_k(x')-[a(k),a(k')^\dagger]u_k(x)u_k(x')^*\right) \end{align*}

I'm not sure how to continue...

• Notice that, in your final identity, some $u_k^{(*)}$ should be $u_{k'}^{(*)}$ actually. Correct it and then make use of both your third identity and the fact that the $u_k$'s form an orthonormal complete set of functions... Commented Mar 5, 2014 at 7:56

I believe in the last line, the plane-wave functions $u_k(x)$ should carry different coordinates and momenta, e.g $$[a(k)^\dagger,a(k')]u_k(x)u_{k'}(x')$$ You may note that the commutator $[\phi(x),\pi(x')]=i\hbar\delta(x-x')$ holds if one choses $[a_k,a_{k'}^\dagger]=\delta_{kk'}$. However, this indirect reasoning is no proof that this choice is unique. It also won't tell you that $[a_k,a_{k'}]=0$
I recommend inverting the relation, that is expressing $a_k$ and $a_k^\dagger$ in terms of $\phi(x)$ and $\pi(x)$. Then you may check the CCR by direct calculation, because the commutation relations $$[\phi(x),\pi(x')]=i\hbar\delta(x-x')\\ [\pi(x),\pi(x')]=[\phi(x),\phi(x')]=0$$ are know.