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Draft Forbes Group Website (Build by Nikola). The official site is hosted at:

https://labs.wsu.edu/forbes

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License: GPL3
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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):

ParseError: KaTeX parse error: Undefined control sequence: \d at position 51: …^{k_F} 1 \frac{\̲d̲^3 k}{(2\pi)^3}…

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

ETF[n]=ξEFG[n]=ξℏ2m3(3π2)2/310n5/3,μUFG=ξϵF.\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:

ParseError: KaTeX parse error: Undefined control sequence: \d at position 560: …} n(r) 4\pi r^2\̲d̲{r} = \fra…
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])
3ℏ433N43ωξ\frac{3 \hbar}{4} \sqrt[3]{3} N^{\frac{4}{3}} \omega \sqrt{\xi}
n_HO
m3ω3(R2−r2)323π2ℏ3ξ32\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:

ParseError: KaTeX parse error: Undefined control sequence: \d at position 431: … N = \int n(r) \̲d̲^3{x} = \frac{4…

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

ParseError: KaTeX parse error: Undefined control sequence: \d at position 264: …nt n(\tilde{r})\̲d̲^3{\tilde{x}} =…

We can obtain effective 2D and 1D densities which would be obtained by integrating along the line-of-sight (along the zz axis here) or in the y−zy-z plane. For reference, we fix the chemical potential by fixing RxR_x, the Thomas-Fermi radius along the xx axis μ=mωx2Rx2/2\mu = m\omega_x^2R_x^2/2:

n2D(x,y)=2m(ωx2(Rx2−x2)−ωy2y2)3/23ωzgn1D(x)=mπωx44gωyωz(Rx2−x2)2n_{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: ParseError: KaTeX parse error: Undefined control sequence: \abs at position 35: … \frac{\hbar^2\̲a̲b̲s̲{\nabla_z\psi}^…

ParseError: KaTeX parse error: Undefined control sequence: \abs at position 1: \̲a̲b̲s̲{\psi(z)}^2 = \…

The Thomas-Fermi approximation gives

ωx4(Rx2−x2)24ω⊥2(ℏ2m2+gn(x)2πm)=(ℏ2m2+3gn(x)4πm)2.\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 n1D(x)n_{1D}(x) above, we can write this as:

gn1D(x)mπ(ℏ2m2+gn(x)2πm)=(ℏ2m2+3gn(x)4πm)2,\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,
94n(x)=(n1D(x)−3ℏ2πmg)+n1D(x)(n1D(x)+3πℏ2mg)\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 ℏω⊥\hbar \omega_\perp:

n(R)=0,mωx22R2=mωx22Rx2−ω⊥ℏ.n(R) = 0, \qquad \frac{m\omega_x^2}{2} R^2 = \frac{m\omega_x^2}{2} R_x^2 - \omega_\perp\hbar.