In this chapter we showed that for temperatures $k_{\mathrm{B}} T\ll E_{\mathrm{F}}$ the chemical potential for carriers in three dimensions may be approximated to second order in $k_{\mathrm {{B}}}T$ as
$\mu\sim E_{\mathrm{F}}-\frac{\pi^{2}}{12}\frac{(k_{\mathrm {B}}T)^{2}}{E_{\mathrm{F}}}$
Derive an expression for the chemical potential to fourth order in $k_{\mathrm{{B}}}T.$

Question

In this chapter we showed that for temperatures [latex]k_{\mathrm{B}} T\ll E_{\mathrm{F}}[/latex] the chemical potential for carriers in three dimensions may be approximated to second order in [latex]k_{\mathrm {{B}}}T[/latex] as

[latex]\mu\sim E_{\mathrm{F}}-\frac{\pi^{2}}{12}\frac{(k_{\mathrm {B}}T)^{2}}{E_{\mathrm{F}}}[/latex]

Derive an expression for the chemical potential to fourth order in [latex] k_{\mathrm{{B}}}T.[/latex]

Accepted Answer

We wish to derive an expression for the chemical potential similar to Eqn (7.47), but to fourth order in [latex]k_{\mathrm{B}}T[/latex]. We proceed by noting that Eqn (7.36) is of the form

[latex]n=\int\frac{d^{3}k}{(2\pi)^{3}}\cdot2\cdot f_{\bf k}(E_{\bf k})\qquad\qquad\qquad\qquad(7.36)[/latex]

[latex]\mu\sim E_{\mathrm{F}}-\frac{\pi^{2}}{12}\frac {(k_{ \mathrm {B}}T)^{2}}{E_{\mathrm{F}}}-\frac{7\pi^{4}}{960}\frac {(k_{\mathrm {B}}T)^{4}}{E_{\mathrm{F}}^{3}} \qquad\qquad \qquad \qquad(7.47)[/latex]

[latex]n=\int\limits_{E=-\infty}^{E=\infty}h(E)f(E)d E[/latex]

where f (E) is the Fermi–Dirac function given by Eqn (7.25). Our approach is to solve this using a Sommerfeld expansion.We follow the standard approach found in the book by N. Ashcroft and N. D. Mermin, Solid State Physics, Saunders College, Philadelphia, 1976 (ISBN 0 03 049346 3). We let

[latex]|n_{1},n_{2},n_{3},\dots,n_{N} angle\qquad\qquad\qquad\qquad(7.25)[/latex]

[latex]\kappa(E)=\int\limits_{E=-\infty}^{E=\infty}h(E^{\prime})d E^{\prime}[/latex]

so that

[latex]h(E)=\frac{d}{d E}\kappa(E)[/latex]

Integration by parts gives

[latex]n=\int\limits_{E=-\infty}^{E=\infty}h(E)f(E)d E=f(E)\kappa (E)\big|_{E=-\infty}^{E=\infty}-\int\limits_{E=-\infty}^{E=\infty} \kappa(E^{\prime})\frac{\partial}{\partial E^{\prime}}f(E^{\prime}) d E^{\prime}[/latex]

[latex]n=-\int\limits_{E=-\infty}^{E=\infty}\kappa (E^{\prime})\frac {\partial}{\partial E^{\prime}}f(E^{\prime})d E^{\prime}[/latex]

If [latex]h(E)[/latex] is not too rapidly varying in E around [latex]{\mu}[/latex], then [latex]\kappa(E)[/latex] may be expanded in a Taylor series about [latex]E\sim\mu.[/latex]

[latex]\kappa(E)=\kappa(\mu)+(E-\mu)\frac{d}{d E}\kappa(E) \bigg| _{E=\mu}+\frac{(E-\mu)^{2}}{2!}\frac{d^{2}}{d E^{2}}\kappa (E) \bigg|_{E=\mu}+\cdots[/latex][latex]+\frac{(E-\mu)^{n}}{n!}\frac {d^{n}}{d E^{n}}κ(E)[/latex]

Hence,

[latex]-\displaystyle\int\limits_{E=-\infty}^{E=\infty}\kappa (E^{\prime})\displaystyle\frac{\partial}{\partial E^{\prime}}f(E^{ \prime})d E^{\prime}=[/latex][latex]-\int\limits_{E=-\infty}^{E= \infty}\left(\kappa(E) +\sum \limits_{n=1}^{n=\infty}\frac {(E^{ \prime}-\mu)^{n}}{n!}\frac {d^{n}}{d E^{\prime n}}\kappa (E^{ \prime})\Bigg|_{E =\mu} ight )\frac{\partial}{\partial E^{\prime} }f(E^{\prime})d E^{\prime}[/latex]

[latex]-\displaystyle\int\limits_{E=-\infty}^{E=\infty}\kappa(E^ {\prime})\displaystyle{\frac{\partial}{\partial E^{\prime}}}f(E^{ \prime})d E^{\prime}=-\displaystyle\int\limits_{E=-\infty}^{E= \infty}\kappa(E^{\prime}){\frac{\partial}{\partial E^{\prime}}}f (E^{ \prime})d E^{\prime}[/latex] [latex]-\int\sum\limits_{n=1}^{n =\infty}{\frac{(E^{\prime}-\mu)^{ n}}{n!}}{\frac{d^{n}}{d E^{n}}}κ (E^{\prime})\bigg\vert_{E=\mu}{\frac{\partial}{\partial E^{\prime} }}f(E^{\prime})d E^{\prime}[/latex]

[latex]\int\limits_{E=-\infty}^{E=\infty}h(E^{\prime}) f(E^{\prime} )d E^{\prime}=\int\limits_{E=-\infty}^{E=\mu}h(E^{\prime})d E^{ \prime}[/latex][latex]+\sum\limits_{n=1}^{n=\infty}\int\limits_{E=-\infty}^{E=\infty}{\frac{(E^{\prime}-\mu)^{2n}}{(2n)!}}[/latex][latex]\left(-{\frac{\partial}{\partial E^{\prime}}}f(E^{\prime}) ight){\frac{d^{2n-1}}{d E^{\prime 2n-1}}}h(E^{\prime})|_{E^ {\prime}=\mu}d E^{\prime}[/latex]

Let [latex](E-\mu)/k_{\mathrm{B}}T=x[/latex] so that [latex]d E=k_{ \mathrm{{B}}}T d x.[/latex] We may now write

[latex]\int\limits_{E=-\infty}^{E=\infty}h(E^{\prime}) f(E^{\prime}) d E^{\prime}=\int\limits_{E=-\infty}^{E=\mu}h(E^{\prime})d E^{\prime}+\sum\limits_{n=1}^{n=-\infty}a_{n}(k_{\mathrm{B}}T)^{2n}{\frac{d^{2n-1}}{d E^{\prime 2n-1}}}h(E^{\prime})|_{E^{\prime} =\mu}[/latex]

where

[latex]a_{n}=\int\limits_{E=-\infty}^{E=\infty}{\frac{x^{2n}}{(2n) !}}\biggl(-{\frac{d}{d x}}\cdot{\frac{1}{e^{x}+1}}\biggr)d x[/latex][latex]=2{\Biggl(}1-{\frac{1}{2^{2n}}}+{\frac{1}{3^{2n}}}-{\frac{1}{4^{2n}}}+{\frac{1}{5^{2n}}}-\cdots{\Biggr)}[/latex]

[latex]a_{n}=\left(2-{\frac{1}{2^{2(n-1)}}} ight)\xi(2n)[/latex]

[latex]\xi(n)=1+{\frac{1}{2^{n}}}+{\frac{1}{3^{n}}}+{\frac{1}{4^{n} }}+\cdots[/latex]

[latex]\xi(2n)=2^{2n-1}{\frac{\pi^{2n}}{(2n)!}}B_{n}[/latex]

The first few values of [latex] extstyle B_{n}[/latex] are

[latex]B_{1}={\frac{1}{6}},\ B_{2}={\frac{1}{30}},\ B_{3}={\frac{1}{42}},B_{4}={\frac{1}{30}},B_{5}={\frac{5}{66}}[/latex]

so that

[latex]\xi(2)={\frac{\pi^{2}}{6}}\operatorname {and}\xi(4)={\frac {\pi^{4}}{90}}[/latex]

[latex]\int\limits_{E=-\infty}^{E=\infty}h(E^{\prime})f (E^{\prime} )d E^{\prime}=[/latex][latex]\int\limits_{E=-\infty}^{E=\mu}h (E^{ \prime})d E^{\prime}+\frac{\pi^{2}}{6}(k_{\mathrm{B}}T)^{2} h^{ \prime}(\mu)+\frac{7\pi^{4}}{360}(k_{\mathrm{B}}T) ^{4}h^{ \prime \prime\prime}(\mu)+0(k_{\mathrm{B}}T)^{6}[/latex]

[latex]\int\limits_{E=-\infty}^{E=\mu}h(E^{\prime})d E^{\prime} =\int\limits_{0}^{E_{\mathrm {F}}}h(E^{ \prime})d E^{\prime}+(\mu-E_{\mathrm{F}})h( E_{\mathrm{F}})[/latex]

[latex]n=\int\limits_{0}^{E_{\mathrm{F}}}\mathrm g (E^{\prime})d E^{\prime}+(\mu-E_{\mathrm{F}} ) \mathrm g(E_{\mathrm{F}})+{\frac{\pi^{2}}{6}}(k_{ \mathrm{B}}T)^{2}\mathrm g^{\prime}(\mu)+{\frac {7\pi^{4}}{360}}(k_{\mathrm{B}}T)^{4}\mathrm g^{\prime\prime\prime}(\mu)+0(k_{\mathrm{B} }T)^{6}\ .[/latex]

[latex](\mu-E_{\mathrm{F}})\mathrm g(E_{\mathrm{F}})={\frac{-\pi^{2}}{6}}(k_{\mathrm{B}}T)^{2}\mathrm g^{\prime}(\mu)-\left({\frac{7\pi^{4}}{360}} ight)(k_{\mathrm{B}}T)^{4}\mathrm g^{\prime\prime\prime}(\mu)-0(k_{\mathrm{B}}T)^{6}[/latex]

[latex]g(E_{\mathrm{F}})={\frac{m k_{\mathrm{F}}}{\hbar^{2} \pi^{2}}}={\frac{m}{\hbar^{3}\pi^{2}}}(2m)^{1/2}E_{\mathrm {F}}^{1/2}[/latex]

[latex]g^{\prime}(E_{\mathrm{F}})={\frac{m}{\hbar^{2} \pi^{2} k_{\mathrm{F}}^{2}}}{\left({\frac{2m}{\hbar^{2}}} ight) }^{3/2} E_{\mathrm{F}}^{1/2}={\frac{m}{\hbar^{3}\pi^{2}}}{\frac{(2m )^{1/2}}{2}}E_{\mathrm{F}}^{-1/2}[/latex]

[latex]g^{\prime\prime}(E_{\mathrm{F}})={\frac{m}{2\hbar^{2} \pi^{2}k_{\mathrm{F}}^{2}}}[/latex][latex]\left(\frac{2m}{\hbar {}^{2}} ight)^{3/2}F_{\mathrm{F}}^{-1/2}=\frac{-m}{\hbar{}^{3} \pi^{2}}\frac{(2m)^{1/2}}{4}E_{\mathrm{F}}^{-3/2}[/latex]

[latex]g^{\prime\prime\prime}(E_{\mathrm{F}})=\frac{m}{-4 \hbar ^{2}\pi^{2}k_{\mathrm{F}}^{2}}\biggl(\frac{2m}{\hbar^{2}}\biggr )^{3/2}{E}_{\mathrm{F}}^{-3/2}=\frac{m}{\hbar^{3}\pi^{2}}\frac {3(2m)^{1/2}}{8}{E}_{\mathrm{F}}^{-5/2}[/latex]

[latex]\mu=E_{\mathrm{F}}-\frac{\pi^{2}}{6}(k_{\mathrm{B}}T )^{2}\frac{\mathrm g^{\prime}(E_{\mathrm{F}})}{\mathrm g(E_{ \mathrm{F}})}-\frac{7\pi^{2}}{360}(k_{\mathrm{B}}T)^{4}\frac {\mathrm g^{\prime\prime\prime}E_{\mathrm{F}}}{\mathrm g(E_{ \mathrm{F}})}-0{\biggl(}\frac{k_{\mathrm{B}}T}{E_{\mathrm{F}}}{\biggr)}^{6}[/latex]

Hence,

[latex]\mu=E_{\mathrm{F}}-\frac{\pi^{2}}{12}(k_{\mathrm{B} }T)^{2}\frac{1}{E_{\mathrm{F}}}-\frac{7\pi^{2}}{120}(k_{\mathrm {B}}T)^{4}\frac{1}{8}\frac{1}{E_{\mathrm{F}}^{3}}-0 \biggl(\frac {k_{\mathrm{B}}T}{E_{\mathrm{F}}}\biggr)^{6}[/latex]

and, finally,

[latex]\mu=E_{\mathrm{F}}-\frac{\pi^{2}}{12}\frac{(k_{\mathrm {B}}T)^{2}}{E_{\mathrm{F}}}-\frac{7\pi^{4}}{960}\frac{(k_{ \mathrm{B}}T)^{4}}{E_{\mathrm{F}}^{3}}-0\biggl (\frac{k_{ \mathrm{B}}T}{E_{\mathrm{F}}}\biggr)^{6}[/latex]

Question 7.2: In this chapter we showed that for temperatures kBT « EF the......

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