# 2nd quantization

## Main Question or Discussion Point

Suppose I have a system of N identical bosons interacting via pairwise potential $V(\vec{x} - \vec{x}')$.

I want to show that the expectation of the Hamiltonian in the non-interacting ground state is

$\frac{N(N-1)}{2\mathcal{V}}\widetilde{V}(0)$
where
$\widetilde{V}(q) = \int d^3 \vec{x} e^{i \vec{q} \cdot \vec{x}}(\vec{x})$
and $\mathcal{V}$ is the volume of the `box'.

My attempt:

First I need to find the ground state in the absence of potential.

The second-quantized Hamiltonian is

$\hat{H} = \int d^{3}\vec{x} \hat{\psi}^\dag (\vec{x})\left( -\frac{\hbar^2}{2m}\nabla_{\vec{x}}^2 \right) \hat{\psi}(\vec{x}) \quad +\quad \int\int d^3\vec{x}d^3\vec{x}'\hat{\psi}^\dag(\vec{x}')\hat{\psi}^\dag(\vec{x})V(\vec{x},\vec{x}')\hat{\psi}(\vec{x})\hat{\psi}(\vec{x}')$

Set V = 0 and then

$\hat{H} = \int d^{3}\vec{x} \hat{\psi}^\dag (\vec{x})\left( -\frac{\hbar^2}{2m}\nabla_{\vec{x}}^2 \right) \hat{\psi}(\vec{x})$

Now use the definition $\hat{\psi}(x) \equiv \sum_{\lambda} \langle \vec{x} | a^{(\lambda)} \rangle \hat{a}_{\lambda} , \quad \hat{\psi}(x) \equiv \sum_{\lambda} \langle a^{(\lambda)} | \vec{x}} \rangle \hat{a}^\dag_{\lambda}$

where $a_\lambda,a^\dag_\lambda$ are the annihilation and creation operators that subtract or add a particle to the single-particle state $|a^{(\lambda)}\rangle$.

Now I'm going to let the single-particle states be momentum eigenstates so

$\hat{\psi}(\vec{x}) = \frac{1}{\sqrt{\mathcal{V}}}\sum_{\vec{k}} e^{i\vec{k} \cdot\vec{x}} \hat{a}_{\vec{k}}$

Plugging this in and using the fact that $\int d^3\vec{x} e^{i(\vec{k}' - \vec{k})\cdot\vec{x}} = \delta_{\vec{k},\vec{k}'}$ gives

$\hat{H} = \sum_{\vec{k}} \frac{\hbar^2 \vec{k}^2}{2m}\hat{a}^\dag_{\vec{k}}\hat{a}_\vec{k}$

so the eigenstates of the non-interacting Hamiltonian are the occupation number states $| n_{\vec{k}_1},n_{\vec{k}_2},\ldots\rangle$

Last edited:

Related Quantum Physics News on Phys.org
Have I done this much right? I think I might have made a mistake because I was expecting the contribution to the expectation coming from the kinetic energy to vanish.

Last edited:
Avodyne
Yes, correct so far.

But for the ground state, what is the value of the momentum $\vec k$?

Yes, correct so far.

But for the ground state, what is the value of the momentum $\vec k$?
I'm not sure, I basically left it as $\vec{k}_1$ and moved on. I managed to figure out the second part. It comes down to using the fact that everything is in the ground state and using a change of variables in the integration.

I suppose that if the system is in a box then the momenta are quantized according to
$\vec{k} = \frac{2\pi}{\mathcal{V}^{1/3}}\vec{n}$ so $\frac{2\pi}{\mathcal{V}^{1/3}}$ is the lowest wavenumber?

reilly