noospace
Apr8-08, 05:13 AM
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
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