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Superfluid fraction $\rho _s/\rho $ in the presence of disorder

Let us apply the theory developed in the previous section to a dilute Bose system. As we will see, the system without disorder is fully superfluid at zero temperature while the presence of impurities create a depletion of the superfluid density. According to (2.33) and (2.34) the normal component is given by the limit of the transverse current-current response function. Let us consider, for example, $\chi_T = \chi_{z}$, i.e. $z$ response and in the $z$ direction (see Fig. 2.2). First one has to take the limit $k_z \to 0$ and after $k_x
\to 0$ and $k_y \to 0$. This can be accomplished by considering ${\bf k} =(k,0,0)$ and letting $k$ decrease toward zero. The $k$-component of the current operator $\hat j$ in second quantization is given by the formula
$\displaystyle \hat j_{\bf k} = \frac{\hbar}{\sqrt{V}}
\sum\limits_{\bf q} \left({\bf q}+\frac{\bf k}{2}\right)
\hat a^\dagger_{\bf q}\hat a_{\bf q+k},$     (2.36)

where in the absence of disorder the particle creation and annihilation operators can be expressed in terms of quasiparticle operators by means of the Bogoliubov transformation (1.32). Once the current (2.36) is calculated, the transverse response function can be obtained by averaging the commutator
$\displaystyle \chi_T({\bf k},t) = -i\Theta(t)\,
\langle[j^z_{\bf k}(t), j^z_{\bf -k}(0)]\rangle$     (2.37)

By taking the Fourier transform
$\displaystyle \chi_T({\bf k},\omega)=\int\limits_{-\infty}^{+\infty}\exp(i\omega t)\chi_T({\bf k},t)\,dt$     (2.38)

the limiting procedure
$\displaystyle \rho_n = \lim\limits_{k\to 0} \lim\limits_{\omega \to 0} \chi_T(k,\omega)$     (2.39)

yields the density of the normal fluid. It is easy to check that in the absence of disorder $\chi_T({\bf
k}, \omega)$ goes to zero as $\bf k \to 0$ and $\omega \to 0$ that the commutator (2.37) of $\chi_{zz}$ goes to zero in the limit $k\to 0$. This corresponds to the fact that a homogeneous dilute Bose gas is completely superfluid at $T = 0$. A useful check consists in the calculation of the longitudinal response (see Fig. 2.3). This means taking first the limit $k_x
\to 0$, $k_y \to 0$ in $\chi_{zz}({\bf k})$. We consider ${\bf k} = (0,0,k)$ and then let $k\to 0$. It is easy to calculate $\chi_{zz}$ in Bogloliubov approximation. For the $z$ component of the current operator one has
$\displaystyle \hat j^z_{\bf k} = \frac{\hbar{\bf k}\sqrt{N_0}}{\sqrt{V}}
\frac{...
...{\bf k}\sqrt{N_0}}{\sqrt{V}}
\frac{(\hat a_{\bf k}+\hat a_{\bf -k}^\dagger)}{2}$     (2.40)

Within this level of accuracy the limit (2.35) of the longitudinal component is given by
$\displaystyle \lim\limits_{\bf k\to 0} \lim\limits_{\omega \to 0}
\chi_L({\bf k},\omega) = n_0m \approx nm$     (2.41)

Let us now study the system in the presence of the random external potential.Starting from the transformation ([*]) between the particles operators $\hat a_{\bf
k}$, $\hat a^\dagger_{\bf k}$ and the corresponding quasiparticle operators $\hat c_{\bf k}$, $\hat c^\dagger_{\bf k}$ one can write the contribution to the current proportional to the external potential (there is no need to consider the contribution independent of the external potential, because as calculated before it is equal to zero).
$\displaystyle \hat j_0^{z}(t)
= \sum\limits_{\bf q} \frac{\hbar q_z Z_q}{\sqrt{...
...\frac{1-L_q}{1+L_q}}
(\hat c^\dagger_{\bf q} V_{\bf -q}-\hat c V_{\bf q}) + ...$     (2.42)

Notice that in order to make calculations simpler we take $k = 0$ from the very begging. Result (2.34) is independent of the order of the two limits. The response function is then given by
$\displaystyle \chi_T(0, t) = -i \Theta(t)
\frac{1}{V}\sum\limits_{\bf q}\hbar^2...
...\bf -q}\rangle
\frac{1-L_q}{1+L_q}
(e^{-i\omega_{\bf q}t}-e^{i\omega_{\bf q}t})$     (2.43)

its Fourier transformation being equal to
$\displaystyle \chi_T(0, w) =
\frac{1}{V}\sum\limits_{\bf q}\hbar^2 q^2_x \vert ...
...1-L_q}{1+L_q}
\left(\frac{1}{\omega-\omega_q}-\frac{1}{\omega+\omega_q}\right),$     (2.44)

and, finally, setting $\omega = 0$ one obtains
$\displaystyle \rho_n
=\chi_T(0, 0) =
\frac{4m\hbar}{3V}
\sum\limits_q q^2_x \ve...
...q}\rangle
= \frac{2\sqrt{\pi}}{3}(na^3)^{1/2}
\chi\left(\frac{b}{a}\right)^2 nm$     (2.45)

This result gives the depletion of the superfluid density due to the presence of impurities.
next up previous contents
Next: calculation of the superfluid Up: Superfluid density Previous: Connection between and the   Contents