3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is $f$ here a function? Is it $\mathbb{R} \to \mathbb{R}$? Your expression for creation operator with integral is unfamiliar to me. $\endgroup$ – Minethlos Jul 26 '15 at 17:11
  • $\begingroup$ @Minethlos $f$ is an element of the one-particle Hilbert space. In most cases, it is thus a square integrable function $f\in L^2(\mathbb{R}^d,\mathbb{C})$ $\endgroup$ – yuggib Jul 26 '15 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.