# What is Merzbacher's justification in this step? (Second-Quantization)?

$$\newcommand{\d}{^{\dagger}}$$

[Short Version] How do you prove the creation operator relations? ($+$ fermions, $-$ bosons)

$$a_i\d a_j\d \pm a_j\d a_i \d=0$$

[Long Version]

In chapter 20 of Merzbacher's "Quantum Mechanics", the algebra of the annihilation and creation operators is derived. Specifically,

$$a_i\d a_j\d \pm a_j\d a_i \d=0$$

where, of course, the plus sign corresponds to operators on fermions and the opposite for bosons. However, there is one particular step in the derivation that I can't understand. I've quoted it below, shortening some parts for brevity.

... Hence, for any state many-particle state $|\psi\rangle$, $$a_i\d a_j\d|\psi\rangle=\lambda a_j\d a_i\d|\psi\rangle$$ To show that $\lambda$ doesn't depend on $|\psi\rangle$ or the indices $i,j$, we switch to another basis through a unitary transformation. $$(a_i\d a_j\d-\lambda a_j\d a_i\d)|\psi\rangle=\sum_{k,l} c_{ki}c_{lj}(b_k\d b_l\d-\lambda b_l\d b_k\d)|\psi\rangle=0 \tag{1}$$ where $c_{ij}$ are the matrix elements of the unitary transformation, constrained by the unitarity-condition, $$\sum_k c_{ik}^*c_{kj}=\delta_{ij}$$ These matrix elements are just complex numbers. If the theory is to have the same form in any representation (unitary symmetry), equation (1) above holds for every $|\psi\rangle$ only if for each pair $k,l$, $$b_i\d b_j\d-\lambda b_j\d b_i\d=0$$

The bold text is what I don't understand. I obviously know that if a state is equal to zero, you can expand that state in terms of a linearly independent basis and equate each component to zero (by the definition of linear independence). I also know that any creation operator merely "bumps up" the state it acts on, apart from a constant. But a priori we don't know whether this normalization depends on the specific representation we use. That's what we're trying to deduce! This is how I think the derivation above should have actually gone.

\begin{align} (a_i\d a_j\d-\lambda(\psi,a_i,a_j) a_j\d a_i\d)|\psi\rangle&=\sum_{k,l} c_{ki}c_{lj}(b_k\d b_l\d-\lambda(\psi,a_i,a_j) b_l\d b_k\d)|\psi\rangle\\ &=\sum_{k,l,m} c_{ki}c_{lj}(b_k\d b_l\d-\lambda(\psi,a_i,a_j) b_l\d b_k\d)\underbrace{\langle [n_b]_m|\psi\rangle}_{D_m} |[n_b]_m\rangle\\ &=\sum_{k,l,m} c_{ki}c_{lj}D_m(\lambda([n_b]_m,b_l,b_k)-\lambda(\psi,a_i,a_j))b_l\d b_k\d|[n_b]_m\rangle\\ &=0 \end{align}

and now we can conclude that each $m^{\textrm{th}}$ component is 0.

$$\sum_{k,l} c_{ki}c_{lj}(\lambda([n_b]_m,b_k,b_l)-\lambda(\psi,a_i,a_j))=0$$

I have used the fact that

$$b_i\d |n_1,\ldots,n_i,\ldots\rangle\propto |n_1,\ldots , n_i+1,\ldots\rangle$$