Self consistency and mean field theory

[Hazz]

Homework Statement

I am a little confused about the how self consistency conditions work and I was wondering if in the following case I have correctly understood the details?

Homework Equations

[/B]
Say we have a harmonic oscillator with a perturbation
$H=\frac{\hat{p}}{2m}+\frac{1}{2}m\omega^2\hat{q}^2+\gamma \hat{q}^4$Say we want to perform something like a Hartree-Fock approximation. We replace the $(a+a^\dagger)^4$ in $\hat{q}^4$ with $\langle (a+a^\dagger)^2 \rangle (a+a^\dagger)^2$. Say our new Hamiltonian is then

$H_{HF}=\frac{\hat{p}}{2m}+\frac{1}{2}m\omega^2\hat{q}^2+\delta (a+a^\dagger)^2$

with $\delta=\gamma(\frac{\hbar}{2m\omega})^2\langle (a+a^\dagger)^2 \rangle$

The Attempt at a Solution

[/B]
This is now a quadratic Hamiltonian and can be diagonalised as

$H= \sqrt{\frac{\hbar\omega}{2}(\frac{\hbar\omega}{2}+2\delta)}\hat{b}^\dagger \hat{b}$

Using this $\langle (a+a^\dagger)^2 \rangle$ can be calculated and found to be$\langle (a+a^\dagger)^2 \rangle = \frac{2n_B(\hbar\omega)-1}{\sqrt{\frac{\hbar\omega}{2}(\frac{\hbar\omega}{2}+2\gamma(\frac{\hbar}{2m\omega})^2\langle (a+a^\dagger)^2 \rangle)}}$

Am I correct in thinking this equation would be the self-consistency condition we are looking for? You could then plot the RHS and LHS with respect to $\langle (a+a^\dagger)^2 \rangle$ and see where the lines cross, if they meet anywhere then the approximation is consistent?

 

[mark726]

Hi there,

Your understanding of self-consistency conditions seems to be correct. In this case, the self-consistency condition is the equation that relates the expectation value of the operator $(a+a^\dagger)^2$ to itself. This equation is then solved to find the self-consistent value of $\langle (a+a^\dagger)^2 \rangle$.

Your approach of plotting the RHS and LHS with respect to $\langle (a+a^\dagger)^2 \rangle$ is a good way to visualize the self-consistency condition and see if there is a consistent solution. If the lines do not cross, then there is no consistent solution and the approximation may not be valid.

Overall, it seems like you have a good understanding of how self-consistency conditions work in this particular case. Keep up the good work!