# Operator in second quantization

1. Jan 26, 2010

### Alexios

Hello,

I'm struggling with the second quantization formalism. I'd like to derive the hamiltonian of a system with non-interacting particles
$$\hat{H}=\int dx\,a(x)^\dagger \left[\frac{\hat{P}}{2m}+V(x)\right]a(x),$$
where $$a(x) = \hat{\Psi}(x)$$.

I know the second quantized representation of a single-particle operator $$\hat{O}$$ which is diagonal in the basis $$\{|\alpha\rangle\}$$:

$$\hat{O}=\sum_i o_{\alpha_i} a_{\alpha_i}^\dagger a_{\alpha_i}$$

My idea was, as a first step, to derive the expression of the linear momentum operator:

$$\hat{P}=\sum_i p_{p_i} a_{p_i}^\dagger a_{p_i} =\sum_i \int dx\, \langle x|p_i\rangle a^\dagger(x) \int dx\, \langle p_i|x\rangle a(x) ??$$

The above is probably wrong (and I don't know how to proceed in order to derive an expression which is of similar form as $$\hat{H}$$)

Any help is much appreciated.

2. Jan 26, 2010

### peteratcam

It is wrong, but not very wrong. I guess you are trying to use the change of basis:
$$a_p = \int dx\, \langle p_i|x\rangle a(x)$$
$$a_p^\dagger = \int dx\, \langle x|p_i\rangle a^\dagger(x)$$

to expand the diagonal representation
$$\hat{P}=\sum_p p a_{p}^\dagger a_{p}$$

Your biggest mistake is failing to use different labels for dummy variables of integration. You should instead have written:
$$\hat{P}=\sum_p p \int dx_2\, \langle x_2|p_i\rangle a^\dagger(x_2) \int dx_1\, \langle p_i|x_1\rangle a(x_1)$$
$$\hat{P}=\iint dx_2dx_1a^\dagger(x_2) a(x_1)\sum_p p \langle x_2|p_i\rangle \langle p_i|x_1\rangle$$

This takes you closer. Can you manipulate it further to derive what you want?

3. Jan 26, 2010

### Alexios

Thanks, that makes sense to me. As far as I know, the momentum operator in 3 dimensions should be
$$\hat{P}=-i\hbar \int d^3 x a^\dagger (x) \nabla a(x)$$

$$\int \int dx_1\, dx_2\, a^\dagger (x_2)\langle x_2|\left(\sum \hat{P}_i |p_i\rangle \langle p_i|\right)|x_1\rangle a(x_1)$$

The inner term simplifies to $$\langle x_2|\hat{P}|x_1\rangle = -i\hbar \nabla \langle x_2|x_1\rangle$$?! Because $$\langle x_2|x_1\rangle = \delta(x_1-x_2)$$ the above double integration indeed gives the desired result.

4. Jan 26, 2010

### peteratcam

Perfect.
I'd have a preference for saying
$$\langle x_2|\hat{P}|x_1\rangle = -i\hbar\delta(x_2-x_1)\nabla_{x_1}$$
since derivatives of delta functions tend to scare me.

As an aside, I'm also a fan of keeping the final answer like:
$$H = \int dxdx' a(x')^\dagger\delta(x-x')\left[\frac{-\hbar^2}{2m}\nabla^2 + V(x)\right]a(x)$$
since that reminds me that we are really looking at a quadratic form.

FWIW, I mentioned some references on second quantization in this post in this thread:
https://www.physicsforums.com/showpost.php?p=2474928&postcount=14

5. Jan 26, 2010

### Alexios

Thanks a lot.

I'm trying now to derive the second quantized expression for a general two-body operator $$\hat{V}$$.

Diagonal representation: $$\hat{V}=\frac{1}{2}\sum_{ij} v_{ij}\hat{P}_{ij}$$
where $$\hat{P}_{ij} = a_i^\dagger a_j^\dagger a_j a_i$$ counts the number of pairs of particles in states $$|i\rangle, |j\rangle$$.

When following the same procedure as before I get an ingegral over $$dx_1 \ldots dx_4$$ with the terms

$$\langle x_1|v_i\rangle \langle x_2 | v_j \rangle \langle v_j | x_3 \rangle \langle v_i |x_4\rangle$$.

Is it correct to rearrange these terms like

$$\langle x_1|\otimes \langle x_2| |v_i\rangle \otimes |v_j \rangle \langle v_j | \otimes \langle v_i | |x_3\rangle \otimes |x_4 \rangle$$

so as to plug in $$\hat{V} = \sum_{ij} v_{ij} |v_i\rangle \otimes |v_j\rangle \langle v_j|\otimes \langle v_i|$$ ?