# Commutators in second quantization

## Main Question or Discussion Point

Hi. I've been trying to calculate a couple of commutators, namely $[\Psi(r),H]$ and $[\Psi^{\dagger}(r),H]$ where H is a free particle hamiltonian in second quantization. I have attached my attempts and I would greatly appreciate if anyone could tell me if I am right or if there is a better way to do it.

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A. Neumaier
2019 Award
Hi. I've been trying to calculate a couple of commutators, namely $[\Psi(r),H]$ and $[\Psi^{\dagger}(r),H]$ where H is a free particle hamiltonian in second quantization. I have attached my attempts and I would greatly appreciate if anyone could tell me if I am right or if there is a better way to do it.
There are mistakes already on line 5 and on line 6.
In substituting Psi in line 5, you forgot the part of the sum involving creation operators.
The part you have given vanishes since the commutator [a_k,a_k'] is zero, at least in the usual notation.

There are mistakes already on line 5 and on line 6.
In substituting Psi in line 5, you forgot the part of the sum involving creation operators.
The part you have given vanishes since the commutator [a_k,a_k'] is zero, at least in the usual notation.
It's not zero because it is a commutator of fermion operators. It would be zero if they were boson operators. For fermions it is the anti-commutator which is zero.

A. Neumaier
2019 Award
It's not zero because it is a commutator of fermion operators.
Nothing on your page had mentioned that. Still, things are not ok - in the equality on line 5, you forgot the creation part of Psi(r).

The best way to compute the commutators is to
(i) define a(f) := sum f(k) a_k,
(ii) compute for arbitrary functions f and g the anti-commutators
{a(f),a(g)}, {a(f),a^*(g)}, {a^*(f),a(g)}, and {a^*(f),a^*(g)},
(iii) write [A,BC]=ABC-BCA={A,B}C-B{A,C} to compute commutators of three Psi terms,
(iv) integrate over r.

Nothing on your page had mentioned that. Still, things are not ok - in the equality on line 5, you forgot the creation part of Psi(r).
Do we agree that Psi(r) ($\Psi(r)$) is the annihilation operator at position r? So what do you mean by the creation part of Psi(r) ? I have just written $\Psi(r)=\sum_k e^{ikr}a_k$

I tried doing it your way by calculating the anti-commutators, and it went a lot smoother. I still have a little problem with the second commutator though.

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A. Neumaier
2019 Award
Do we agree that Psi(r) ($\Psi(r)$) is the annihilation operator at position r? So what do you mean by the creation part of Psi(r) ? I have just written $\Psi(r)=\sum_k e^{ikr}a_k$
Well, since it is your problem, we can agree on anything you like, but nobody can guess what you mean if you use standard notation to designate nonstandard things.
You'd define all the terms you are using in a way differently from the traditional context (where iin the absence of other informations, bosons are the default, and where space-dependent fields are decomposed into creation and annihilation operators).

I actually thought I was using pretty standard notation(except for the boson vs. fermion default). I'm reading books on many particle theory such as G.D. Mahan's "Many Particle Physics".

$\Psi(r)$ is the annihilation operator at point $r$ and $\Psi^\dagger (r)$ is the corresponding creation operator. These operators obey the anti-commutation relation $\{\Psi(r),\Psi^\dagger (r')\}=\delta(r-r')$, so they are fermion operators. Using these operators the free particle Hamiltonian is written $H_0=\int dr\Psi^\dagger (r)\left(-\frac{1}{2m}\nabla^2 _r\right)\Psi(r)$. We have another set of fermion operators $a_k$ and $a^\dagger _k$ that annihilate/create states of momentum $k$. I can then express i.e. $\Psi(r)$ in terms of these as $\Psi(r)=\sum_k e^{ikr}a_k$. Actually I should divide by $\sqrt{V}$ here, where V is the volume, but I just leave that out.

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A. Neumaier
2019 Award
I actually thought I was using pretty standard notation(except for the boson vs. fermion default). I'm reading books on many particle theory such as G.D. Mahan's "Many Particle Physics".
This explains things. You work in a nonrelativistic setting. Then you indeed don't have the antiparticle part.

A. Neumaier