Exercises — Fermi-Dirac statistics — fermions, Fermi energy

Sirf exponent ka sign aur size matter karta hai $x=(\epsilon -\mu )/k_{B}T$. (a) $x=0$: $f=\frac{1}{e^0+1}=\frac{1}{2}$. Chemical potential hamesha half-filled energy hoti hai. (b) $x=-10$: $e^{-10}\approx 4.5\times 10^{-5}$, isliye $f=\frac{1}{1.000045}\approx 0.99995\approx 1$. Deep states essentially full hoti hain. (c) $x=+10$: $e^{10}\approx 2.2\times 10^4$, isliye $f=\frac{1}{22027}\approx 4.5\times 10^{-5}\approx 0$. High states essentially empty hoti hain.

$k_F=(3{\pi }^{2}n{)}^{1/3}$ — yeh one-third power hai. $E_F=\frac{{\hslash }^2}{2m}(3{\pi }^{2}n{)}^{2/3}$ mein density two-thirds power se aati hai, kyunki $E_F\propto k_F^2$ aur $k_F\propto n^{1/3}$, isliye $k_F^2\propto n^{2/3}$.

(i) Yes — fermions mein $+1$ hota hai, bosons mein $-1$. (ii) Yes — fermion $f\le 1$ (exclusion isko cap karta hai); boson $f$ diverge ho sakta hai (condensation, dekho Bose-Einstein Statistics). (iii) No — $\mu$ dono cases mein chemical potential hai; yeh woh reservoir property hai jo average particle number fix karti hai (dekho Chemical Potential).

Step 1 — argument banao. $3{\pi }^{2}n=3(9.8696)(8.5\times 10^{28})=2.517\times 10^{30}{\text{m}}^{-3}$. Kyun: yahi woh quantity hai jise hum $2/3$ tak raise karenge. Step 2 — $2/3$ tak raise karo. $(2.517\times 10^{30}{)}^{2/3}=1.855\times 10^{20}{\text{m}}^{-2}$. (Yeh $k_F^2$ hai.) Step 3 — prefactor se multiply karo. $E_F=\frac{(1.055\times 10^{-34}{)}^2}{2(9.11\times 10^{-31})}\times 1.855\times 10^{20}=1.13\times 10^{-18}\text{J}.$ Yeh step kyun? Humne jo bracket evaluate ki abhi, woh $k_F^2$ hai; Fermi energy ${\hslash }^2{k}_F^2/2m$ hai, isliye prefactor ${\hslash }^2/2m$ se multiply karne par squared wavenumber energy mein convert ho jaata hai — yeh literally free-particle relation $\epsilon ={\hslash }^2{k}^2/2m$ hai jo Fermi sea ke top par evaluate ki gayi hai. $1.602\times 10^{-19}$ se divide karo: $E_F\approx 7.05\text{eV}$. Step 4 — Fermi temperature. $T_F=\frac{1.13\times 10^{-18}}{1.381\times 10^{-23}}\approx 8.2\times 10^4\text{K}$. $k_B$ se kyun divide karein? $T_F$ is tarah define hai ki $k_B{T}_F=E_F$ — yeh Fermi energy ko us temperature ke roop mein re-express karta hai jis par ek classical gas ko wohi energy scale milti. Energy ko $k_B$ se divide karna simply joules ko kelvin mein convert karta hai. Kyunki room temperature $300\text{K}\ll T_F$ hai, copper ka electron gas strongly degenerate hai.

$f=0.25$ set karo: $e^x+1=4\Rightarrow e^x=3\Rightarrow x=\ln 3\approx 1.0986$. Isliye $\epsilon -\mu =(\ln 3)k_{B}T\approx +1.10k_{B}T$: ek quarter-full state $\mu$ ke thoda upar hoti hai.

$f=0.9\Rightarrow x=\ln (1/9)=-2.197$; $f=0.1\Rightarrow x=\ln 9=+2.197$. Width $=\Delta x=2\ln 9=4.394$, isliye $\Delta \epsilon \approx 4.39k_{B}T$. $300\text{K}$ par: $k_{B}T=(1.381\times 10^{-23})(300)=4.14\times 10^{-21}\text{J}=0.02585\text{eV}$. $\Delta \epsilon \approx 4.39\times 0.02585\approx 0.114\text{eV}$. $E_F\approx 7\text{eV}$ se compare karo: smear $E_F$ ka $\sim 1.6\%$ hai — ek bahut patli fuzzy band.

Pehle $\delta$ define karo: maano $\delta$ ek fixed energy offset hai chemical potential se, isliye $\epsilon =\mu +\delta$, $\mu$ se $\delta$ upar hai aur $\epsilon =\mu -\delta$, $\delta$ neeche hai. Iske units energy ke hain, bilkul $\epsilon$ aur $\mu$ ki tarah. Maano $x=\delta /k_{B}T$ (dimensionless). Tab $f(\mu +\delta )=\frac{1}{e^x+1},f(\mu -\delta )=\frac{1}{e^{-x}+1}.$ Doosre ko $\frac{e^x}{e^x}$ se multiply karo: $f(\mu -\delta )=\frac{e^x}{1+e^x}$. Add karo: $\frac{1}{e^x+1}+\frac{e^x}{e^x+1}=\frac{1+e^x}{e^x+1}=1.$ ∎ Interpretation: $\mu$ se $\delta$ upar kisi state ke empty hone ki probability, $\mu$ se $\delta$ neeche kisi state ke occupied hone ki probability ke barabar hai. Excitations symmetrically particle-hole pairs create karti hain — $\mu$ ke neeche jo "hole" banta hai woh $\mu$ ke upar promote hue "electron" ko mirror karta hai. Yahi electron-hole symmetry hai jo Sommerfeld expansion ke peeche hai.

Yeh tool (derivative) kyun? Metal ki thermal aur electrical response un states se aati hai jo apna occupation change karti hain jab tum energy ya temperature nudge karte ho. Slope $-\partial f/\partial \epsilon$ exactly measure karta hai ki $f$ kitna sharply change karta hai — jahan yeh bada hai, states "active" hain. $x=(\epsilon -\mu )/k_{B}T$ ke saath: $-\frac{\partial f}{\partial \epsilon }=\frac{1}{k_{B}T}\cdot \frac{e^x}{(e^x+1{)}^2}.$ $\epsilon =\mu$ par, $x=0$: $\frac{e^0}{(1+1{)}^2}=\frac{1}{4}$, isliye $-{\frac{\partial f}{\partial \epsilon }|}_{\mu }=\frac{1}{4k_{B}T}.$ Yeh ek bump hai jo $\mu$ par peaked hai, height $\sim 1/(4k_{B}T)$ aur width $\sim k_{B}T$ — ek "window" jo sirf Fermi surface ke paas ke electrons ko select karti hai. Baki sab exclusion se frozen hain.

Step 1 — $k_F$. $k_F=(3{\pi }^{2}n{)}^{1/3}=(2.517\times 10^{30}{)}^{1/3}=1.362\times 10^{10}{\text{m}}^{-1}$. Step 2 — $v_F$. $v_F=\frac{(1.055\times 10^{-34})(1.362\times 10^{10})}{9.11\times 10^{-31}}=1.58\times 10^6\text{m/s}$. Step 3 — thermal speed. $\sqrt{k_{B}T/m}=\sqrt{\frac{4.14\times 10^{-21}}{9.11\times 10^{-31}}}=6.74\times 10^4\text{m/s}$. Ratio: $v_F/v_{\text{thermal}}\approx 23$. $300\text{K}$ par bhi Fermi surface ke electrons classical gas se $\sim 23\times$ zyaada fast move karte hain — kyunki Pauli exclusion, temperature nahi, pushing kar raha hai. Yahi degeneracy pressure ka microscopic origin hai.

Integrate kyun karein? $T=0$ par, $f=1$ $E_F$ ke neeche aur $0$ upar hai, isliye occupied states ka sum matlab $g(\epsilon )$ ko $0$ se $E_F$ tak integrate karna hai. $N={\int }_0^{E_F}g(\epsilon )d\epsilon =\frac{V}{2{\pi }^2}{\left(\frac{2m}{{\hslash }^2}\right)}^{3/2}{\int }_0^{E_F}\sqrt{\epsilon }d\epsilon .$ Integral $\frac{2}{3}{E}_F^{3/2}$ deta hai: $N=\frac{V}{3{\pi }^2}{\left(\frac{2m}{{\hslash }^2}\right)}^{3/2}{E}_F^{3/2}.$ $E_F$ ke liye solve karo. $n=N/V$ likho: $E_F^{3/2}=3{\pi }^{2}n{\left(\frac{{\hslash }^2}{2m}\right)}^{3/2}\Rightarrow E_F={\left(3{\pi }^{2}n\right)}^{2/3}\frac{{\hslash }^2}{2m}.$ Exactly parent formula. ∎

Kyun: mean energy har occupied state ko $\epsilon$ se weight karti hai, matlab $⟨\epsilon ⟩=\frac{{\int }_0^{E_F}\epsilon gd\epsilon }{{\int }_0^{E_F}gd\epsilon }$. Kyunki $g\propto \sqrt{\epsilon }$: $⟨\epsilon ⟩=\frac{{\int }_0^{E_F}{\epsilon }^{3/2}d\epsilon }{{\int }_0^{E_F}{\epsilon }^{1/2}d\epsilon }=\frac{\frac{2}{5}{E}_F^{5/2}}{\frac{2}{3}{E}_F^{3/2}}=\frac{2/5}{2/3}{E}_F=\frac{3}{5}{E}_F.$ Isliye total energy $U_0=N⟨\epsilon ⟩=\frac{3}{5}N{E}_F$. Matlab: absolute zero par bhi gas enormous kinetic energy carry karti hai — copper ke liye, $\frac{3}{5}(7\text{eV})=4.2\text{eV}$ per electron. Yahi stored energy degeneracy pressure ke roop mein compression resist karti hai.

Pehle, $E_F\propto V^{-2/3}$ kyun? Hum electron number $N$ fixed rakhte hain aur box shrink karte hain. Kyunki $n=N/V$, density $n\propto V^{-1}$ scale karti hai; isse $E_F\propto n^{2/3}$ mein dalne par $E_F\propto (V^{-1}{)}^{2/3}=V^{-2/3}$ milta hai. Box ko squeeze karne par har electron ki Fermi energy badhti hai — woh rising energy hi pressure hai. $E_F=C{V}^{-2/3}$ likho jahan $C=\frac{{\hslash }^2}{2m}(3{\pi }^{2}N{)}^{2/3}$ ($N$ fixed rakhe). Tab $U_0=\frac{3}{5}NC{V}^{-2/3}$. $P=-\frac{\partial U_0}{\partial V}=-\frac{3}{5}NC\left(-\frac{2}{3}\right)V^{-5/3}=\frac{2}{5}\cdot \frac{NC}{V}{V}^{-2/3}=\frac{2}{5}\frac{N}{V}{E}_F.$ Isliye $P=\frac{2}{5}n{E}_F$. Copper ke liye: $P=\frac{2}{5}(8.5\times 10^{28})(1.13\times 10^{-18})\approx 3.8\times 10^{10}\text{Pa}\approx 3.8\times 10^5\text{atm}$. Yeh vast internal pressure — sirf exclusion se — woh hai jo white dwarf stars ko support karta hai.

$\frac{E_F-\mu }{E_F}=\frac{{\pi }^2}{12}{\left(\frac{k_{B}T}{E_F}\right)}^2=\frac{{\pi }^2}{12}{\left(\frac{1}{T_F/T}\right)}^2$. $T=300\text{K}$, $T_F=8.2\times 10^4\text{K}$ ke saath: $\frac{T}{T_F}=3.66\times 10^{-3}$. $\frac{{\pi }^2}{12}(3.66\times 10^{-3}{)}^2=0.8225\times (1.34\times 10^{-5})=1.10\times 10^{-5}$. Isliye $\mu$ $E_F$ se 11 parts per million neeche hai. Copper ke liye room temperature par shorthand $\mu \approx E_F$ extremely safe hai — lekin yeh ek $(T/T_F{)}^2$ effect hai jo badhta hai aur hot ya low-density systems mein ignore nahi kiya ja sakta.

Step 1 — argument. $3{\pi }^{2}n=(29.608)(10^{36})=2.961\times 10^{37}{\text{m}}^{-3}$. Step 2 — $2/3$ power. $(2.961\times 10^{37}{)}^{2/3}$. Base ko $29.608\times 10^{36}$ likho; tab $(10^{36}{)}^{2/3}=10^{24}$ aur $(29.608{)}^{2/3}=9.63$, isliye result $9.63\times 10^{24}{\text{m}}^{-2}$ hai. (Yeh $k_F^2$ hai.) Step 3 — energy. $E_F=\frac{(1.055\times 10^{-34}{)}^2}{2(9.11\times 10^{-31})}\times 9.63\times 10^{24}=6.11\times 10^{-8}\times 9.63\times 10^{24}\times \dots$ Zyaada cleanly, prefactor $\frac{{\hslash }^2}{2m}=\frac{(1.055\times 10^{-34}{)}^2}{2(9.11\times 10^{-31})}=6.11\times 10^{-39}{\text{J⋅m}}^2$, isliye $E_F=6.11\times 10^{-39}\times 9.63\times 10^{24}=5.88\times 10^{-14}\text{J}.$ eV mein: $5.88\times 10^{-14}/1.602\times 10^{-19}=3.67\times 10^5\text{eV}\approx 0.37\text{MeV}$. Step 4 — rest energy se compare karo. $\frac{E_F}{m_e{c}^2}=\frac{5.88\times 10^{-14}}{8.19\times 10^{-14}}=0.72$. $E_F$ electron rest energy ka lagbhag $72\%$ hai — Fermi-surface electrons firmly relativistic hain. Non-relativistic dispersion $\epsilon ={\hslash }^2{k}^2/2m$ yahan trustworthy nahi raha; tumhe relativistic energy-momentum relation use karni hogi, jo degeneracy pressure ko soften karti hai aur aakhirkar white dwarfs ke liye Chandrasekhar mass limit produce karti hai.

Yeh limit kyun? Jab states bahut rarely occupied hoti hain ($f\ll 1$), toh koi bhi do particles ek hi state ke liye compete nahi karte, isliye exclusion irrelevant ho jaata hai aur quantum statistics ko classical answer par collapse karna chahiye. Agar $\mu \to -\infty$ to kisi bhi finite $\epsilon$ ke liye, $(\epsilon -\mu )/k_{B}T\to +\infty$, isliye $e^{(\epsilon -\mu )/k_{B}T}\gg 1$ aur $+1$ negligible hai: $f(\epsilon )=\frac{1}{e^{(\epsilon -\mu )/k_{B}T}+1}⟶e^{-(\epsilon -\mu )/k_{B}T}=e^{\mu /k_{B}T}{e}^{-\epsilon /k_{B}T}.$ Yeh $f\propto e^{-\epsilon /k_{B}T}$ hai — Maxwell-Boltzmann factor. Kab: high temperature aur/ya low density, matlab $T\gg T_F$, jahan occupation sparse hai aur "$+1$" (fermion signature) matter karna band kar deta hai. Dono compare karta figure dekho.