# Deriving the bosonic action of annihilation and creation operators in QFT

The definition of bosonic annihilation and creation operators are given as:

$$a^{\dagger}_{\vec{k}}\lvert n_{\vec{k}}\rangle =\sqrt{n+1}\lvert (n+1)_{\vec{k}}\rangle$$ $$a_{\vec{k}}\lvert n_{\vec{k}}\rangle =\sqrt{n}\lvert (n-1)_{\vec{k}}\rangle$$

How do I derive them in the context of QFT besides just accepting it as a definition?

• This is no definition, you can derive these relations by assuming some constant $\alpha$ such that $a^\dagger |n\rangle = \alpha |n+1\rangle$, just as in usual quantum mechanics. May 24, 2021 at 18:01
• If you have had a course on quantum mechanics then you must have studied the harmonic oscillator. If not the take a step back and don’t try to run before you can walk. May 24, 2021 at 18:07

I'll drop the $$k$$s for brevity.
An alternative to @Youran's approach with less up-front combinatorics uses$$[a,\,a^\dagger]=1,\,[A,\,BC]=[A,\,B]C+B[A,\,C]\implies[a,\,(a^\dagger)^{m+1}]=(a^\dagger)^m+a^\dagger[a,\,(a^\dagger)^m]$$to prove by induction $$[a,\,(a^\dagger)^n]=n(a^\dagger)^{n-1}$$ for $$n\ge1$$, so$$a(a^\dagger)^n|0\rangle=[a,\,(a^\dagger)^n]|0\rangle=n(a^\dagger)^{n-1}|0\rangle.$$Now normalizing $$|n\rangle\propto(a^\dagger)^n|0\rangle$$ and $$|n-1\rangle\propto(a^\dagger)^{n-1}|0\rangle$$, $$|n\rangle=\tfrac{1}{\sqrt{n}}a^\dagger|n-1\rangle$$. You can do the other direction similarly, or just note it follows from a corollary of the above, $$|n\rangle=\tfrac{1}{\sqrt{n!}}(a^\dagger)^n|0\rangle$$.
First, we need to define $$|n_k\rangle$$. It is reasonable that $$|n_k\rangle \propto (a^\dagger_k)^n|0\rangle$$ with a proportional factor to be fixed s.t. $$\langle n_k|n_k\rangle=1$$. Similarly, the proportional factor of $$|(n+1)_k\rangle$$ should be fixed s.t. $$\langle (n+1)_k|(n+1)_k\rangle=1$$. So what you want to prove, in the language of QFT, is $$\frac{\langle n_k|a_ka_k^\dagger|n_k\rangle}{\langle n_k|n_k\rangle}=n_k+1.$$ This is easy to see since the enumerator has $$(n+1)!$$ ways to contract while the denominator has only $$n!$$.