# Commutation relations in second quantization

I know that for operators $a(\chi_1), a(\chi_2)$ of the same type (fermionic or bosonic)

$$[a(\chi_1), a(\chi_2)]_{-\xi} = [a^\dagger (\chi_1), a^\dagger (\chi_2)]_{-\xi} = 0 \tag{1}$$

where

$$\xi = \begin{cases} +1 &\text{for bosons} \\ -1 &\text{for fermions} \end{cases} \tag{2}$$

and $[.]_{-1}$ is commutator and $[.]_{+1}$ is anticommutator. I also know how those operators act on arbitrary Fock states:

$$a^\dagger (\chi) | \phi_1, \dots, \phi_N \rangle = | \chi, \phi_1, \dots, \phi_N \rangle \tag{3}$$

$$a (\chi) | \phi_1, \dots, \phi_N \rangle = \sum_j \xi^{j-1} \langle \chi | \phi_j \rangle | \phi_1, \dots, \hat{\phi}_j, \dots, \phi_N \rangle \tag{4}$$

where $\hat{\psi_k}$ denotes absence of a particular wavefunction.

How do I derive relation

$$[a (\chi_1), a^\dagger (\chi_2)]_{-\xi} = a (\chi_1) a^\dagger (\chi_2) - \xi\, a^\dagger (\chi_2) a (\chi_1) = \langle \chi_1 | \chi_2 \rangle \tag{5} ?$$

P.S. I'm following these notes (section 1.5) and I can't understand what's meant in this phys.SE post.

Edit (28.07): Say $|\Psi \rangle = | \phi_1, \dots, \phi_N \rangle$. I tried

$$a (\chi_1) a^\dagger (\chi_2) |\Psi \rangle = \sum_k \xi^{k} \langle \chi_2 | \phi_j \rangle | \chi_1, \phi_1, \dots, \hat{\phi}_j, \dots, \phi_N \rangle + \langle \chi_1 | \chi_2 \rangle |\Psi \rangle$$

$$- \xi\, a^\dagger (\chi_2) a (\chi_1) |\Psi \rangle = - \sum_k \xi^{k} \langle \chi_1 | \phi_j \rangle | \chi_2, \phi_1, \dots, \hat{\phi}_j, \dots, \phi_N \rangle$$

Adding the two above lines I should get desired result. It seems that the sums should cancel, but I can't figure out why.

For bosons: We put ourselves in a suitable common domain, i.e. the finite particle vectors. Then $$\bigl(a^*(f)\Psi\bigr)_n(X_n)=\frac{1}{\sqrt{n}}\sum_{j=1}^n f(x_j)\Psi_{n-1}(X_n\setminus{x_j})\\ \bigl(a(f)\Psi\bigr)_n(X_n)=\sqrt{n+1}\int \bar{f}(x)\Psi_{n+1}(x,X_n)dx$$ Hence by definition $$\bigl(a(g)a^*(f)\Psi\bigr)_n(X_n)=\sum_{j=1}^{n+1}\int \bar{g}(x_1)f(x_j)\Psi_n(X_{n+1}\setminus x_j)\\\bigl(a^*(f)a(g)\Psi\bigr)_n(X_n)=\sum_{j=1}^{n}\int \bar{g}(x)f(x_j)\Psi_n(x,X_{n}\setminus x_j)\; .$$ The result immediately follows subtracting. For fermions is similar.
• Is $f$ here a function? Is it $\mathbb{R} \to \mathbb{R}$? Your expression for creation operator with integral is unfamiliar to me. – Minethlos Jul 26 '15 at 17:11
• @Minethlos $f$ is an element of the one-particle Hilbert space. In most cases, it is thus a square integrable function $f\in L^2(\mathbb{R}^d,\mathbb{C})$ – yuggib Jul 26 '15 at 17:36