# Bogoliubov Transformation

• CMJ96
In summary: To do this, one defines a matrix ##K_{BT}## which diagonalizes the BT:$$K_{BT} = \left[ \begin{array}{cccc} u_i & v_i \\ u_i & -v_i \\ \vdots & -v_i \end{array} \right]$$The first two columns of this matrix are the canonical commutation relations for the original operators, and the last two columns are the same as the original except that the negation of each term is replaced by its opposite.
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)

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

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
$$\hat{h}_0 = -\frac{\hbar^2}{2m} \nabla^2 + V_{ext}$$

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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.

## What is a Bogoliubov Transformation?

A Bogoliubov Transformation is a mathematical technique used in quantum mechanics to transform a system of interacting particles into a system of non-interacting particles. It is commonly used in the study of superfluidity and superconductivity.

## Who developed the Bogoliubov Transformation?

The Bogoliubov Transformation was developed by Russian physicist Nikolay Bogoliubov in the 1950s. He used this technique to study the properties of Bose-Einstein condensates, a state of matter in which a large number of bosons (particles with integer spin) occupy the same quantum state.

## What are the applications of the Bogoliubov Transformation?

The Bogoliubov Transformation has many applications in physics, including the study of superfluidity, superconductivity, and quantum field theory. It is also used in condensed matter physics to study the behavior of particles in a material.

## How does the Bogoliubov Transformation work?

The Bogoliubov Transformation involves diagonalizing the Hamiltonian (the operator that describes the total energy of a system) by introducing new creation and annihilation operators. This allows for the separation of the system into a non-interacting part and an interacting part, making it easier to solve complex quantum mechanical problems.

## What are the limitations of the Bogoliubov Transformation?

The Bogoliubov Transformation is only applicable to systems with weak interactions, as it assumes that the interactions between particles are small compared to their kinetic energy. It also does not take into account the effects of temperature, which can be important in certain systems.

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