2nd quantization (1 Viewer)

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

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

[itex]\frac{N(N-1)}{2\mathcal{V}}\widetilde{V}(0)[/itex]
where
[itex]\widetilde{V}(q) = \int d^3 \vec{x} e^{i \vec{q} \cdot \vec{x}}(\vec{x})[/itex]
and [itex]\mathcal{V}[/itex] 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

[itex]\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}')[/itex]

Set V = 0 and then

[itex]\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}) [/itex]

Now use the definition [itex]\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}[/itex]

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

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

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

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

[itex]\hat{H} = \sum_{\vec{k}} \frac{\hbar^2 \vec{k}^2}{2m}\hat{a}^\dag_{\vec{k}}\hat{a}_\vec{k}[/itex]

so the eigenstates of the non-interacting Hamiltonian are the occupation number states [itex]| n_{\vec{k}_1},n_{\vec{k}_2},\ldots\rangle[/itex]
 
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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.
 
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Avodyne

Science Advisor
1,384
81
Yes, correct so far.

But for the ground state, what is the value of the momentum [itex]\vec k[/itex]?
 
Yes, correct so far.

But for the ground state, what is the value of the momentum [itex]\vec k[/itex]?
I'm not sure, I basically left it as [itex]\vec{k}_1[/itex] 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
[itex]\vec{k} = \frac{2\pi}{\mathcal{V}^{1/3}}\vec{n}[/itex] so [itex]\frac{2\pi}{\mathcal{V}^{1/3}}[/itex] is the lowest wavenumber?
 

reilly

Science Advisor
1,070
0
For a system of non-interacting bosons, the ground state consists of only zero momentum particles. (See Bogulubov's(sp?) work on superfluidity, which is strongly based on this idea.) Your V(0) is odd: the potential does not appear in the defining integral -- if it does then you are on the right track. If there were non-zero momentum particles, then your N(N-1) factor would be incorrect.

Regards,
Reilly Atkinson
 

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