index.ipynb

Kernel: Python [conda env:work]

import mmf_setup;mmf_setup.nbinit()

<IPython.core.display.Javascript object>

The Unitary Fermi Gas

Here we summarize some features of the unitary Fermi gas (UFG) in harmonic traps.

We start with the properties of a free Fermi gas with two components (spin states):

$$\begin{gather} n_+ = n = 2\int_0^{k_F} 1 \frac{\d^3 k}{(2\pi)^3} = \frac{k_F^3}{3\pi^2} = \frac{k_F^3}{3\pi^2} = \frac{(2m\mu)^{3/2}}{3\pi^2\hbar^3} \tag{density}\\ \mu = E_F = \epsilon_F = \frac{\hbar^2k_F^2}{2m} = \frac{\hbar^2(3\pi^2 n)^{2/3}}{2m} \tag{chemical potential/Fermi energy}\\ \mathcal{E} = 2\int_0^{k_F} \frac{\hbar^2k^2}{2m} \frac{\d^3 k}{(2\pi)^3} = \frac{\hbar^2 k_F^5}{10m\pi^2} = \mu \frac{k_F^3}{5\pi^2} = \frac{3}{5}n \mu \tag{energy density}\\ E_{FG} = \frac{\mathcal{E}}{n} = \frac{3}{5}E_F \tag{energy per particle}. \end{gather}$$

The energy density for the UFG is expressed in terms of these and the Bertsch parameter $\xi = 0.370(5)$:

$$\mathcal{E}_{TF}[n] = \xi \mathcal{E}_{FG}[n] = \xi\frac{\hbar^2}{m}\frac{3(3\pi^2)^{2/3}}{10}n^{5/3}, \qquad \mu_{UFG} = \xi \epsilon_F.$$

Harmonic Traps

Here we derive the results for a harmonically trapped gas in a spherical trap using the local density approximation (LDA) or Thomas-Fermi (TF) approximation:

$$\begin{gather} V(r) = \frac{m\omega^2 r^2}{2}, \tag{HO traping potential}\\ a^2 = \frac{\hbar}{m\omega}, \tag{HO length}\\ R_{TF}^2 = \frac{2\mu}{m\omega^2} = a^2 \frac{2\mu}{\hbar\omega}, \tag{Thomas-Fermi or cloud radius}\\ \xi \epsilon_F[n] = \mu - V(r) = \frac{m\omega^2(R_{TF}^2 - r^2)}{2}, \tag{local chemical potential}\\ n(r) = \frac{m^3\omega^3}{3\pi^2\hbar^3 \xi^{3/2}}\left(R_{TF}^2 - r^2\right)^{3/2} = \frac{(R_{TF}^2 - r^2)^{3/2}}{3\pi^2 a^6 \xi^{3/2}} \tag{density}\\ N = \int_0^{R_{TF}} n(r) 4\pi r^2\d{r} = \frac{4}{3\pi a^6 \xi^{3/2}} \int_0^{R_{TF}}r^2( R_{TF}^2 - r^2)^{3/2}\d{r} = \frac{R_{TF}^6}{24 a^6 \xi^{3/2}} = \frac{m^3\omega^3 R_{TF}^6}{24 \hbar^3 \xi^{3/2}} \tag{paticle number}\\ E = \int_0^{R_{TF}} \left(\mathcal{E}_{TF}[n(r)] + \frac{m\omega^2}{2}n(r)\right) 4\pi r^2\d{r} = \frac{\hbar^2 R_{TF}^8}{ 64 m a^{10} \xi^{3/2}} = \frac{m^4 \omega^5}{64 \hbar^3 \xi^{3/2}R_{TF}^8} \tag{energy}\\ E(N) = \hbar\omega\sqrt{\xi}\frac{(3N)^{4/3}}{4}. \end{gather}$$

import sympy; sympy.init_session(quiet=True)
h, xi, m, w, r, R, n, mu, R2_r2 = var('hbar, xi, m, omega, r, R, n, mu, R^{2}-r^{2}', positive=True)
pi = sympy.pi
k_ = (3*pi**2*n)**(S(1)/3)
mu_TF = xi*h**2*k_**2/2/m
E_TF = S(3)/5*mu_*n
V = m*w**2*R2_r2/2
n_HO = solve(mu_-V, n)[0]
E_HO = (E_ + m*w**2*r**2/2*n_HO).subs(n,n_HO).simplify().subs(R2_r2, R**2-r**2).simplify()
n_HO = n_HO.subs(R2_r2, R**2-r**2)
#N = (n_HO*4*pi*r**2).integrate((r, (0, R)))
#E = (E_HO*4*pi*r**2).integrate((r, (0, R)))
N = R**6*m**3*w**3/24/h**3/xi**(S(3)/2)
E = R**8*m**4*w**5/64/h**3/xi**(S(3)/2)
l, N_ = symbols('l, N', positive=True)
N = R**6*m**3*w**3/24/h**3/xi**(S(3)/2)
E = R**8*m**4*w**5/64/h**3/xi**(S(3)/2)
E.subs(R, solve(N-N_, R)[0])

IPython console for SymPy 1.0 (Python 2.7.13-64-bit) (ground types: python)

l, N_ = symbols('l, N', positive=True)
N = R**6*m**3*w**3/24/h**3/xi**(S(3)/2)
E = R**8*m**4*w**5/64/h**3/xi**(S(3)/2)
E.subs(R, solve(N-N_, R)[0])

$$\frac{3 \hbar}{4} \sqrt[3]{3} N^{\frac{4}{3}} \omega \sqrt{\xi}$$

n_HO

$$\frac{m^{3} \omega^{3} \left(R^{2} - r^{2}\right)^{\frac{3}{2}}}{3 \pi^{2} \hbar^{3} \xi^{\frac{3}{2}}}$$

GPE Bose Gas

For comparison, here we summarize some features of a trapped Bose gas:

Here we derive the results for a harmonically trapped gas in a spherical trap using the local density approximation (LDA) or Thomas-Fermi (TF) approximation:

$$\begin{gather} V(r) = \frac{m\omega^2 r^2}{2}, \tag{HO traping potential}\\ a^2 = \frac{\hbar}{m\omega}, \tag{HO length}\\ R_{TF}^2 = \frac{2\mu}{m\omega^2} = a^2 \frac{2m\mu}{\hbar\omega}, \tag{Thomas-Fermi or cloud radius}\\ \mu_e(r) = \mu - V(r) = \frac{m\omega^2(R_{TF}^2 - r^2)}{2}, \tag{local chemical potential}\\ n(r) = \frac{m\omega^2(R_{TF}^2 - r^2)}{2g} \tag{density}\\ N = \int n(r) \d^3{x} = \frac{4 m \omega^2 R^5 \pi}{15 g} = \frac{16\pi}{15 g\omega^3}\sqrt{\frac{2\mu^5}{m^3}} \tag{particle number} \end{gather}$$

To generalize this to 3D with different trapping frequencies note that the TF radius in each direction $R_i \propto 1/\omega_i$ so we introduce dimensionless coordinates $\tilde{x}_i = x_i/R_i$:

$$\mu = \frac{m\omega_i^2R_i^2}{2}, \qquad \frac{m\omega_i^2}{2} = \frac{\mu}{R_i^2}, \qquad R_i = \sqrt{\frac{2\mu}{m}}\frac{1}{\omega_i},\\ n = \frac{\mu - m\sum_i \omega_i^2x_i^2/2}{g} = \mu\frac{1 - \tilde{r}^2}{g},\\ N = R_xR_yR_z \int n(\tilde{r})\d^3{\tilde{x}} = \frac{1}{\omega_x\omega_y\omega_z}\sqrt{\frac{2\mu}{m}}^3\frac{8\pi \mu}{15 g} = \frac{16\pi}{15 g \omega_x\omega_y\omega_z}\sqrt{\frac{2\mu^5}{m^3}}$$

We can obtain effective 2D and 1D densities which would be obtained by integrating along the line-of-sight (along the $z$ axis here) or in the $y-z$ plane. For reference, we fix the chemical potential by fixing $R_x$, the Thomas-Fermi radius along the $x$ axis $\mu = m\omega_x^2R_x^2/2$:

$$n_{2D}(x, y) = \frac{2m\left(\omega_x^2(R_x^2 - x^2) - \omega_y^2y^2\right)^{3/2}}{3\omega_z g}\\ n_{1D}(x) = \frac{m\pi \omega_x^4}{4g\omega_y\omega_z}(R_x^2-x^2)^2$$

NPSEQ

Here we consider the NPSEQ, which provides a model for the integrated 1D density: $$\mathcal{E} = \frac{\hbar^2\abs{\nabla_z\psi}^2}{2m} + \Biggl( \overbrace{ \frac{\hbar^2\sigma^2_{,z}(z, t)}{2m\sigma^2} }^{\text{neglected}} + \frac{\hbar^2}{2m\sigma^2} + \frac{m\omega_\perp^2\sigma^2}{2} + \frac{g\abs{\psi}^2}{4\pi\sigma^2} \Biggr)\psi^\dagger\psi,\\ \I\hbar \dot{\psi} = \left( \frac{-\hbar^2\nabla_z^2}{2m} + \frac{\hbar^2}{2m\sigma^2} + \frac{m\omega_\perp^2\sigma^2}{2} + \frac{g\abs{\psi}^2}{2\pi\sigma^2} \right)\psi,\qquad \frac{m\omega_\perp^2}{2}\sigma^4 = \frac{g\abs{\psi}^2}{4\pi}+\frac{\hbar^2}{2m}.$$

$$\abs{\psi(z)}^2 = \int\d{x}\d{y}n(x, y, z) = \frac{\mu R_xR_y}{g} \int (1-\tilde{r}_\perp^2-\tilde{z}^2)2\pi\tilde{r}_\perp \d{\tilde{r}_\perp^2} = \frac{\pi \mu R_xR_y (1-\tilde{z}^2)}{2 g} = \frac{\pi \mu (1-\tilde{z}^2)}{m\omega_x\omega_y g}$$

The Thomas-Fermi approximation gives

$$\frac{\omega_x^4(R_x^2 - x^2)^2}{4\omega_\perp^2}\left(\frac{\hbar^2}{m^2} + \frac{gn(x)}{2\pi m}\right) = \left(\frac{\hbar^2}{m^2} + \frac{3gn(x)}{4\pi m}\right)^2.$$

To compare with $n_{1D}(x)$ above, we can write this as:

$$\frac{gn_{1D}(x)}{m\pi}\left(\frac{\hbar^2}{m^2} + \frac{gn(x)}{2\pi m}\right) = \left(\frac{\hbar^2}{m^2} + \frac{3gn(x)}{4\pi m}\right)^2,$$

$$\frac{9}{4}n(x) = \left( n_{1D}(x) - \frac{3\hbar^2 \pi}{mg} \right) + \sqrt{ n_{1D}(x)\left( n_{1D}(x) + \frac{3\pi\hbar^2}{mg} \right) }$$

Note that the Thomas-Fermi radius is also shifted in this approximation because the transverse modes change the chemical potential by the transverse zero-point energy $\hbar \omega_\perp$:

$$n(R) = 0, \qquad \frac{m\omega_x^2}{2} R^2 = \frac{m\omega_x^2}{2} R_x^2 - \omega_\perp\hbar.$$