- #1
chwu
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- Homework Statement
- Show that, with this definition the total energy of the spin up components reads:
$$U^+ = U_0 (1+x)^{5/3} + \frac{1}{2} N \mu_B B (1+x)$$ where $$U_0=\frac{3}{10} N E_F$$, in terms of the usual Fermi energy of the gas at $$B=0; E_F = \hbar^2 (3\pi^2 N/V)^{2/3}/(2m)$$. Find the corresponding expression for $$U^-$$
- Relevant Equations
- $$N^+ = \frac{1}{2} N (1+x)$$ and $$N^- = \frac{1}{2} N (1-x)$$
Hello everybody, I tried the following approach. I calculated the density of states for the spin up states:
$$g_+(\epsilon + \mu_BB) = (1/2) g(\epsilon) \Rightarrow g_+ \frac{1}{2} g(\epsilon - \mu_B B)$$. As far as I understand it, this comes from the fact that when there is no magnetic field the spin up and spin down make up excactly half of the density of states (and of the occupation number) Now i'd like to perform the integral $$U_+ = \int_0^{E_F(B)} \epsilon g_+(\epsilon) d\epsilon$$, but for this I need the Fermi energy depending on the magnetic field $$E_F(B)$$ and I do not know yet how to obtain it. In fact I would like to obtain $$E_F(B)$$ in terms of x. Any help is appreciated, thanks in advance!
$$g_+(\epsilon + \mu_BB) = (1/2) g(\epsilon) \Rightarrow g_+ \frac{1}{2} g(\epsilon - \mu_B B)$$. As far as I understand it, this comes from the fact that when there is no magnetic field the spin up and spin down make up excactly half of the density of states (and of the occupation number) Now i'd like to perform the integral $$U_+ = \int_0^{E_F(B)} \epsilon g_+(\epsilon) d\epsilon$$, but for this I need the Fermi energy depending on the magnetic field $$E_F(B)$$ and I do not know yet how to obtain it. In fact I would like to obtain $$E_F(B)$$ in terms of x. Any help is appreciated, thanks in advance!
