How to Diagonalize a Hamiltonian Using Bogoliubov Transformation?

[CMJ96]

Homework Statement

I'd like to diagonalise the following Hamiltonian for quasiparticle excitations in a Bose Einstein Condensate
$$H= K_0 + \hat{K}_1 + \hat{K}_2 $$
where
$$K_0 = \int d^3 r \left[ \phi_0 ^* (\hat{h}_0- \mu) \phi_0 + \frac{g}{2} |\phi_0| ^4 \right]$$
$$\hat{K}_1= \int d^3 r \left[ \hat{\delta}^{\dagger} (\hat{h}_0 + g| \phi_0 |^2 - mu) \phi_0 +\phi_0 ^* (\hat{h}_0 + g| \phi_0 | ^2 - \mu ) \hat{\delta} \right] $$
$$\hat{K}_2= \int d^3 r \left[ \hat{\delta}^{\dagger} (\hat{h}_0 + 2g| \phi_0 |^2 - \mu ) \hat{\delta} + \frac{g}{2} (( \phi_0 ^*)^2 \hat{\delta} \hat{\delta} + \phi_0 ^2 \hat{\delta} ^{\dagger} \hat{\delta} ^{\dagger} ) \right] $$
Eventually reaching the following form for $K_2$
$$K_2= \sum_i \epsilon_i \beta_i ^{\dagger} \beta_i - \sum_i \epsilon _i \int d^3 r |v_i |^2 $$

Homework Equations

I'd like to do this using the following Bogoliubov transformation
$$ \hat{\delta} = \sum_i \left[ u_i \hat{\beta} _i + v_i ^* \hat{\beta}_i ^{\dagger} \right] $$

The Attempt at a Solution

Subbing in the expression for $\delta$ into the Hamiltonian I got the following focusing on the K1 and K2 terms.
$$K_1= \int d^3 r \sum_i (u_i^* \beta_i ^{\dagger} + v_i \beta_i ) (h_0 + g| \phi_0 |^2 - \mu ) \phi_0 + \phi_0 ^* ( \hat{h}_0 +g | \phi_0 |^2 - \mu) (u_i \beta_i + v_i ^* \beta_i ^{\dagger} )$$
$$K_2 = ( h_0 + 2g | \phi_0 |^2 - \mu ) \int d^3 r \sum_{i j} \left[u_i ^* u_j \beta_i ^{\dagger} \beta_j + u_i ^* v_j ^* \beta_i ^{\dagger} \beta_j ^{\dagger} + v_i u_j \beta_i \beta_j + v_i v_j ^* \beta_i \beta_j ^{\dagger} \right] + \frac{g}{2} (\phi_0 ^* )^2 \sum_{i j} \int d^3 r \left[ u_i u_j \beta_i \beta_j +u_i v_j ^* \beta_i \beta_j ^{\dagger} + v_i ^* u_j \beta_i ^{\dagger} \beta_j + v_i ^* v_j ^* \beta_i ^{\dagger} \beta_j ^{\dagger}\right]+ \frac{g}{2} (\phi_0)^2 \sum_{i j} \int d^3 r \left[ u_i ^* u_j ^* \beta_i^{\dagger} \beta_j ^{\dagger} +u_i ^* v_j \beta_i ^{\dagger} \beta_j + v_i u_j ^* \beta_i \beta_j ^{\dagger} + v_i v_j \beta_i \beta_j \right] $$
Defining $L= h_0+2g | \phi_0 |^2 - \mu$ and grouping the terms based upon what coefficient of $\beta$ they have I got the following terms (omitting the integral and sums)
for $\beta_i \beta_j$
$$\beta_i \beta_j \left[ L v_i u_j + \frac{g}{2} | \phi_0| ^2 (u_i u_j +v_i v_j ) \right] $$
for $\beta_i ^{\dagger} \beta_j$
$$\beta_i ^{\dagger} \beta_j \left[L u_i ^* u_j + \frac{g}{2} | \phi_0 |^2 (v_i ^* u_j + u_i ^* v_j ) \right] $$
for $\beta_i \beta_j ^{\dagger}$
$$\beta_i \beta_j ^{\dagger} \left[ L v_i v_j ^* + \frac{g}{2} | \phi_0 | ^2 (u_i v_j ^* + v_i u_j ^* ) \right] $$
for $\beta_i ^{\dagger} \beta_j ^{\dagger}$
$$\beta_i ^{\dagger} \beta_j ^{\dagger} \left[L u_i ^* v_j ^* + \frac{g}{2} | \phi_0 |^2 (v_i ^* v_j ^* + u_i ^* u_j ^* ) \right] $$
I'm not sure how to proceed, for reference I'm using Alexander Fetters 1972 paper "Nonuniform States of an Imperfect Bose Gas" to help me work through this (I'm using slightly different notation though)

 

[strangerep]

Hmm... I need more information. What are $\phi_0$ and $\hat h_0$ ?

 

[CMJ96]

Hi, a field operator has been used and split up into two parts, the condensate part and the non-condensate part $\hat{\psi}= \hat{\phi} + \hat{\delta}$ , $\hat{ \phi}$ is the condensate and $\hat{\delta}$ is the non-condensate. A Bogoliubov approximation was made so $\hat{\phi}$ becomes $\sqrt{N_0} \phi_0$ because were talking about fluctuations this condensate wavefunction is denoted by the subscript $0$ to show it's in the ground state
$$\hat{h}_0 = -\frac{\hbar^2}{2m} \nabla^2 + V_{ext} $$

 

[strangerep]

That's still insufficiently-detailed information, but anyway...

With a BT, one gets constraints on the $u_i, v_i$ by requiring that the new operators satisfy the same canonical commutation relations as the original operators.