Bogoliubov Hamiltonian and elementary excitations

In the absence of the external potential $V({\bf r})$, the Hamiltonian (1.1) can be conveniently expressed in momentum space

$$\hat H = \sum \frac{p^2}{2 m}\,\hat a_{\bf p}^\dagger \hat a_{\bf... ...^\dagger \hat a_{\bf p_1'}\hat a_{\bf p_2'}\, \delta_{\bf p_1+p_2,\,p_1'+p_2'},$$ (1.19)

where the summation is carried out over all indices that appear twice. By assuming that the relevant scattering processes involve particles at low momenta, the matrix elements in the Hamiltonian (1.19) can be replaced by their values at zero momenta, then

$$\hat H = \sum\frac{p^2}{2 m}\,\hat a_{\bf p}^\dagger \hat a_{\bf ... ...{\bf p_1}}^\dagger{\hat a_{\bf p_2}}^\dagger \hat a_{\bf p_1'}\hat a_{\bf p_2'}$$ (1.20)

In a dilute gas almost all particles are found in the condensed state $N~\approx~N_{\bf0}~=~{\hat a_{\bf0}}^\dagger \hat a_{\bf0}$, then, as it was already discussed above (see eq. (1.7)) the operators $\hat a_{\bf0}$ and ${\hat a_{\bf0}}^\dagger$ can be treated as ordinary numbers. The application of perturbation theory means that the last term in (1.20) should be decomposed in powers of the small quantities ${\hat a_{\bf p}}^\dagger$ and $\hat a_{\bf p}$, with ${\bf p\ne 0}$. The zeroth term is

$${\hat a_{0}}^\dagger{\hat a_{0}}^\dagger \hat a_{0}\hat a_{0} = a_0^4$$ (1.21)

The first order terms are absent because they do not satisfy the law of momentum conservation. The second order terms are

$$a_{\bf0}^2 \sum\limits_{\bf p\ne 0} (\hat a_{\bf p}\hat a_{\bf -p... ... p}}^\dagger{\hat a_{\bf -p}}^\dagger+4{\hat a_{\bf p}}^\dagger \hat a_{\bf p})$$ (1.22)

Here the $a_{\bf0}^2 = N_{\bf0}$ factor can be substituted with the total number of particles $N$, although in equation () it is necessary to use the more precise formula

$$a_{\bf0}^2 + \sum\limits_{\bf p\ne 0} {\hat a_{\bf p}}^\dagger \hat a_{\bf p} = N$$ (1.23)

As a result the sum of equations (1.21) and (1.22) becomes equal to

$$N^2 + N \sum\limits_{\bf p\ne 0} (\hat a_{\bf p}\hat a_{\bf -p}+ ... ... p}}^\dagger{\hat a_{\bf -p}}^\dagger+2{\hat a_{\bf p}}^\dagger \hat a_{\bf p})$$ (1.24)

and substitution into Hamiltonian (1.20) gives

$$\hat H = \frac{N^2}{2V}U_0 + \sum\limits_{\bf p}\frac{p^2}{2m} {\... ...f p}}^\dagger{\hat a_{\bf -p}}^\dagger+2{\hat a_{\bf p}}^\dagger\hat a_{\bf p})$$ (1.25)

The matrix element $U_0$ has to be expressed in terms of the scattering length $a$. In the second order terms this can be done using the first Born approximation $U_0 = 4\pi\hbar^2a/m$, although in the zeroth order term one should use the second Born approximation for collisions of two particles from the condensate

$$U_0 = \frac{4\pi\hbar^2a}{m} \left(1+\frac{4\pi\hbar^2a}{V}\sum\limits_{\bf p\ne 0}\frac{1}{p^2}\right)$$ (1.26)

or by introducing the speed of sound (see equation (1.18))

$$c = \sqrt{\frac{4\pi \hbar^2 a N}{m^2 V}}$$ (1.27)

one obtains

$$U_0 = \frac{Vmc^2}{N}\left(1+\frac{1}{N}\sum\limits_{\bf p\ne 0} \biggl(\frac{mc}{p}\biggr)^2\right)$$ (1.28)

Substitution of this formula into (1.25) gives

$$\hat H = \frac{N}{2} mc^2 \left(1+\frac{1}{N}\sum\limits_{\bf p\n... ... p}}^\dagger{\hat a_{\bf -p}}^\dagger+2{\hat a_{\bf p}}^\dagger \hat a_{\bf p})$$ (1.29)

In order to calculate the energy levels of the system one has to diagonalize the Hamiltonian (1.29). This can be accomplished by using the Bogoliubov canonical transformation of the field operators [15]. The operators $\hat a_{\bf p}^\dagger$ and $\hat a_{\bf p}$ should be expressed as a linear superposition of the qusiparticle operators $\hat b_{\bf p}^\dagger$ and $\hat b_{\bf p}$

$$\left\{ \begin{array}{rcl} \hat a_{\bf p}&=&u_p \hat b_{\bf p} + ... ...dagger&=&u_p \hat b_{\bf p}^\dagger + v_p \hat b_{\bf -p}\\ \end{array}\right.$$ (1.30)

which have to satisfy the same commutation rules as the operators $\hat a_{\bf p}^\dagger,\hat a_{\bf p}$ (see eq.(1.6))

$$[\hat b_{\bf p}, \hat b_{\bf p'}^\dagger] = \delta_{\bf pp'},\qua... ...at b_{\bf p'}] = 0,\quad [\hat b_{\bf p}^\dagger, \hat b_{\bf p'}^\dagger] = 0.$$ (1.31)

From the commutation rules (1.31) one can show that the coefficients must satisfy the condition $u_p^2 - v_p^2 = 1$. The transformation (1.30) can be rewritten as

$$\left\{ \begin{array}{rcl} \displaystyle\hat a_{\bf p}&=&\display... ...hat b_{\bf p}^\dagger + L_p \hat b_{\bf -p}}{\sqrt{1-L_p^2}} \end{array}\right.$$ (1.32)

Let us substitute (1.32) into the Hamiltonian (1.25) and set to zero the coefficient of the term proportional to $\hat b_{\bf p}\hat b_{\bf -p}$. This gives an equation for $L_p$

$$L_p^2 + 2 \frac{\frac{p^2}{2m}+ mc^2}{mc^2}L_p +1 = 0,$$ (1.33)

which has two solutions

$$L_p = \frac{1}{mc^2} \left(-\frac{p^2}{2m} - mc^2 \pm \sqrt{\left(\frac{p^2}{2m}\right)^2+(pc)^2}\,\right)$$ (1.34)

The solution with negative sign is unphysical, because the $1-L_p^2$ term in the square root in (1.32) becomes negative. Thus, the solution is

$$L_p = \frac{1}{mc^2} \left(E(p) - \frac{p^2}{2m} - mc^2\right),$$ (1.35)

where $E(p)$ stands for

$$E(p) = \sqrt{\left(\frac{p^2}{2m}\right)^2+(pc)^2}$$ (1.36)

The condition that the coefficient of the term proportional to $\hat b_{\bf p}^\dagger \hat b_{\bf -p}^\dagger$ be zero gives the same equation (1.33). Thus, if condition (1.35) is satisfied, the Hamiltonian has been diagonalized and has the form

$$\hat H = E_0 + \sum\limits_{\bf p}E(p)~\hat b_{\bf p}^\dagger \hat b_{\bf p},$$ (1.37)

where

$$E_0 = \frac{N}{2} mc^2 + \frac{1}{2} \sum\limits_{\bf p\ne 0} \le... ...(p) - \frac{p^2}{2m} - mc^2\left(1 - \biggl(\frac{mc}{p}\biggr)^2\right)\right]$$ (1.38)

From the Hamiltonian (1.37) and the commutation rules (1.31) one can identify $\hat b_{\bf p}$ and $\hat b_{\bf p}^\dagger$ as the creation and annihilation operators of quasiparticles with energy $E(p)$. The ground state energy is given by $E_0$ which is the energy of the ``vacuum'' of quasiparticles $\hat H \vert\rangle = E_0 \vert\rangle$, where the ``vacuum'' state is defined as $\hat b_{\bf p}\vert\rangle = 0$ for any value of ${\bf p\ne 0}$ The excited states are given by $\vert{\bf p}\rangle = \hat b_{\bf p}\vert\rangle$ and have energy $E(p)$ and momentum ${\bf p}$. It is interesting to note that the spectrum (1.36) of the elementary excitations was already obtained in () from the Gross-Pitaevskii equations by considering small oscillations of the order parameter around the stationary solution.