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0

$$ n(x)=|\phi(x)|^{2}=\frac{\mu-V_{ext}(x)}{g}=\frac{2\mu-m\omega^{2}x^{2}}{2g} $$ As you can see when $|x|$ goes to $\infty$, $n(x)$ becomes $-ve$, that is unphysical. We can define the Thomas-Fermi radius, $R_{TF}$ such that $n(x=R_{TF})=0$, $$ \mu=\frac{m\omega^{2}R_{TF}^{2}}{2} $$ So the corrected Thomas-Fermi density profile is, $$ ...


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If operators $a_j$, $a^\dagger_j$ correspond to the system's orthonormal natural orbitals $\phi_j({\bf x})$, such that $$ \int{d{\bf x}\; \phi^*_j({\bf x})\phi_k({\bf x})} = \delta_{jk}, \;\;\;\sum_j{\phi^*_j({\bf x})\phi_j({\bf x'})} = \delta({\bf x} - {\bf x'})\\ \hat\psi({\bf x}) = \sum_j{\phi_j({\bf x})\;a_j},\;\;\; a_j = \int{d{\bf x} \;\phi^*_j({\bf ...


0

See the question Laws and theories and the answers to it. The terms law and theory are somewhat vaguely defined so your question doesn't make sense. Relativity, both special and general, is well enough tested that physicists regard it as an excellent working description of the universe. However it can only be an effective theory since it does not take into ...


5

The 1st order density matrix, $$ \rho({\bf x}, {\bf x'}) = \langle \hat\psi^\dagger({\bf x'})\hat\psi({\bf x}) \rangle $$ is a Hermitian operator since $\rho({\bf x}, {\bf x'}) = \rho^*({\bf x'}, {\bf x})$ and its diagonal entries, $$ \rho({\bf x}, {\bf x}) = \langle \hat\psi^\dagger({\bf x})\hat\psi({\bf x}) \rangle = n({\bf x}) $$ give the number density ...


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You can think of the chemical potential as the amount of free energy needed to add one additional particle to the system. Because the ground state of a BEC is degenerate and can hold an infinite number of particles, there's no energy cost to add another particle to that state. So, $\mu = 0$.


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To determine the upper limit on chemical potential for a gas of $\mathcal N$ bosons, look at the form of the Bose distribution in the grand canonical ensemble with $\langle N \rangle = \mathcal N$. When using the GCE, it's easiest to work at chemical potential $\mu$ and to then choose $\mu(\mathcal N)$ so that $\langle N\rangle(\mu)=\mathcal N$. Each state ...



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