How to Find the Ground State of a System of Identical Bosons?

In summary, the conversation discusses finding the expectation of the Hamiltonian in the ground state of a system of N identical bosons interacting via a pairwise potential. The expectation is shown to be equal to \frac{N(N-1)}{2\mathcal{V}}\widetilde{V}(0), where \widetilde{V}(q) is the integral of the potential and \mathcal{V} is the volume of the system. The conversation also touches on finding the ground state in the absence of potential and the value of the momentum for the ground state.
  • #1
noospace
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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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  • #2
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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  • #3
Yes, correct so far.

But for the ground state, what is the value of the momentum [itex]\vec k[/itex]?
 
  • #4
Avodyne said:
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?
 
  • #5
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
 

1. What is 2nd quantization?

2nd quantization is a mathematical framework used to describe the behavior of particles in quantum mechanics. It is based on the principles of quantum field theory and allows for the treatment of many-particle systems.

2. How is 2nd quantization different from 1st quantization?

1st quantization describes the behavior of individual particles, while 2nd quantization takes into account the collective behavior of many particles. In 2nd quantization, particles are represented as excitations of quantum fields, rather than individual point particles.

3. What is the significance of 2nd quantization in quantum mechanics?

2nd quantization allows for the treatment of systems with an arbitrary number of particles, making it a useful tool in studying many-body systems. It also provides a more intuitive and efficient way of calculating quantities such as energy and momentum in quantum systems.

4. How is 2nd quantization used in condensed matter physics?

2nd quantization is commonly used in condensed matter physics to study the behavior of electrons in materials. It allows for the description of collective phenomena, such as superconductivity and superfluidity, which cannot be explained using 1st quantization methods.

5. Are there any limitations to 2nd quantization?

2nd quantization is a powerful tool, but it does have some limitations. It assumes that particles are indistinguishable and that interactions between particles are instantaneous, which may not always be the case in real systems. Additionally, it is not applicable to systems with high energies or strong interactions.

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