I try to derive the Hamiltonian for a weakly-interacting Bose gas. I am stucked right now at this step, could someone explain it?

Hamiltonian weakly-interacting Bose gas

Thank you! Best Michael

  • $\begingroup$ Could you perhaps explain what it is about this equation that you are actually stuck on? As it stands, it is kind of a guessing game what it is you need help with. $\endgroup$
    – Kyle Kanos
    Oct 14, 2015 at 10:33
  • $\begingroup$ Thanks for your answer. I have a Hamiltonian with field operators. Now I try to put explicit field operators (plane waves) into the Hamiltonian. The kinetic term works fine, but I run in some trouble by trying to simplify this interaction term. I have somehow to make one of the integrations and get a delta-function, but I don't know how, because the integration runs also over the (unkown) potential. Best $\endgroup$ Oct 14, 2015 at 10:55

1 Answer 1


What is to be understood I guess is why the equality holds. (and maybe what is q, r being a dummy variable anyway.)

I'm a little bit rusty on the exact relations and numerical factors, but I'll give the idea: one does a change of variable in order to use the following identity: $$ \delta (\mathbf{p}) = \int_V e^{i\mathbf{p}\cdot \mathbf{x}}\ d \mathbf{x}$$ A change of variable is necessary since the "U" in the integral depends on both $\mathbf{r}_1$ and $\mathbf{r}_2$: define $$ \mathbf{R} := \mathbf{r}_1 + \mathbf{r}_2\quad \text{and} \quad \mathbf{S} := \mathbf{r}_1 - \mathbf{r}_2 $$ Here one should be careful about the determinant of the jacobian in the change of variable formula, it seems to me that there should be a 2: (determinant of the following matrix) $$ \left(\frac{d \text{ new variables}}{d \text{ old variables}}\right) = \begin{pmatrix} Id & Id \\ -Id & Id\end{pmatrix} $$ By laziness, denote $\mathbf{K}_2:= \mathbf{k}_2'- \mathbf{k}_2$ and $\mathbf{K}_1:= \mathbf{k}_1- \mathbf{k}_1'$, one obtains $$ \frac{1}{2V^2} \int_{V\times V}\ e^{-i(\mathbf{K}_1-\mathbf{K}_2)\cdot \mathbf{S}} U(|\mathbf{R}|) e^{-i(\mathbf{K}_1+\mathbf{K}_2)\cdot \mathbf{R}}\ d \mathbf{R} d \mathbf{S}$$ Use the first relation for the intergral in $\mathbf{S}$, you get the result up to a factor...

  • 1
    $\begingroup$ The factor is a delta function – $\delta\big((\mathbf k_1 + \mathbf k_2) - (\mathbf k_1' + \mathbf k_2')\big)$ – which is momentum conservation (and this seems to be simply missing in the formulas from the question). $\mathbf q$ is the momentum transfer in the interaction. $\endgroup$ Oct 14, 2015 at 14:35
  • $\begingroup$ Oh yeah, more comments: $\mathbf{q}$ is not uniquely determined (neither are the k's in the l.h.s.) but it is implicity the variable $\mathbf{K}_1 + \mathbf{K}_2$ but with the constraint that $\mathbf{K}_1 = \mathbf{K}_2$ (coming from the $\delta$) $\endgroup$
    – Noix07
    Oct 14, 2015 at 15:36

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