LLeeccttuurree 1100:: CCoonnddeennsseedd mmaatttteerr ssyysstteemmss

Ideal quantum, condensed system of fermions 66 Aims: ? Degenerate systems: non-interacting quantum particles at high density ? Fermions: / Fermi energy and chemical potential µ(T). / Electrons in metals: / Heat capacity / Pauli paramagnetism / Degeneracy pressure (introduction)

May 05 Lecture 10 1 DDeeggeenneerraattee ssyysstteemmss

66 Summary of previous results: 66 Particles in a box (cube of side a) ? Single-particle energies ≈ >2 ’ >2 2 ∆ ÷ 2 ≈π ’ 2 2 2 ε = ∆ ÷k = ∆ ÷ (l + m + n ) « 2m ◊ 2m « a ◊ ? Density of states in 3-dimensions ≈ a3 ’≈ 2m ’3/ 2 g (ε )dε = ∆ ÷∆ ÷ ε 1/ 2 dε = Aε 1/ 2 dε o « 4π 2 ◊« >2 ◊ to this we need to add any spin-degeneracy. For example, electrons have spin 1/2 and hence a degeneracy of two. g(ε) from g(ε) from g (ε )dε = 2Aε 1/ 2 dε nnooww oonn electron ? Occupation numbers for SSppinin uupp aanndd spin down FD (+ sign) and BE (- sign) spin down distributions:

pi (ε i ) =1 [exp((ε i − µ(T )) kT )±1]

? Classical limit when p<<1 (high T, low density) ? Quantum limit when p>~1 (low T, high density) Then it is called a degenerate system

May 05 Lecture 10 2 DDeeggeenneerraattee FFeerrmmii ggaass

66 At T = 0. ? Low temperature limit of the Fermi-Dirac distribution is a step function: ? States fill up to energy µ starting at ε = 0. All higher energy states are empty. ? The value of µ(T) at T = 0K is known as the

Fermi energy, εF. All states within a sphere of radius kF, the Fermi wavevector, are filled, where >2 2 ε F = kF 2m

FFeerrmmii eenneerrggyy

May 05 Lecture 10 3 FFeerrmmii eenneerrggyy aanndd µµ((TT==00))

66 T = 0 continued… ? Determine εF from the requirement that we have N electrons (i.e. N filled states)

∞ ε F N = pF (ε )g(ε )dε = g(ε )dε —0 —0 ≈ >2 ’ Ω ∆ ÷ 2 2/3 ε F = (3π n) « 2m ◊ DDeennssiittyy ooff ppaarrttiicclleess nn== NN// aa33 == NN// VV 66 Fermi function for T>0K. 1 pF (ε ) = exp((εi − µ(T )) kT )+1

May 05 Lecture 10 4 CChheemmiiccaall ppootteennttiiaall,, µµ((TT)),, ffoorr TT>>00

66 Occupied states in 3-D Fermi gas

66 Chemical potential µ(T) for T>0K. ? Result follows using the previous approach: ∞ g(ε ) N = — dε 0 exp[(ε − µ) kT ]+1 µµ~~ ccoonnssttaanntt an implicit equation for µ(T) ? Numerical solution: Key point / µ(T) ≈ εF at low T (true for most practical situations)

May 05 Lecture 10 5 EElleeccttrroonnss iinn mmeettaallss

66 Free electron metals ? Valence electrons move freely and independently. / Best example: Alkali metals (Li, Na, K, …); also noble metals (Cu, Ag, Au) ? A surprisingly good approximation / for subtle reasons to do with e-e interactions ? Typical data T = /k v =

/m TFF = εεFF/k vFF = κκFF/m

-3 Metal n/(m ) εF/(eV) TF/(K) vF/ (m/s)

Na 2.65x1028 3.2 3.75x104 1.07x106

Cu 8.45x1028 7.0 8.12x104 1.57x106

? Metals (at room temperature) are degenerate Fermi systems ? The original free-electron model (Drude model) assumed the electrons were a classical perfect gas. It largely fails to account for metallic properties. ? Eg: Heat capacity ~3NkT/2 was far too large.

May 05 Lecture 10 6 TThheerrmmaall pprrooppeerrttiieess ooff mmeettaallss

66 Thermal capacity ? Electronic contribution to the heat capacity follows from differentiating the electronic energy w.r.t. T. ∞ ε g(ε ) U el = — dε 0 exp[(ε − µ) kT ]+1 RReessuulltt ffrroomm aa ssttrraaiigghhtt-- forward, if lengthy, 2 forward, if lengthy, ∂U π T manipulation. See C = el ≈ Nk manipulation. See el ∂T 2 T AAsshhccrroofftt aanndd F MMeerrmmiinnpp4422--77

? A similar result follows from a qualitative argument – emphasising the essential physics / Compare occupied states at T=0K with those at T≠0K

May 05 Lecture 10 7 EElleeccttrroonniicc ccoonnttrriibbuuttiioonn ttoo tthhee hheeaatt ccaappaacciittyy ooff aa mmeettaall

? cont….. / only electrons within ~kT are active thermally. A direct consequence of degeneracy. / There are n e x = g ( ε F ) k T such electrons / If we treat these excited electrons like classical electrons. That is, having kinetic energy 3kT/2 per electron. 2 2 U el = nex 3kT 2 = g(ε F )3k T 2 ∂U C = el = g(ε )3k 2T AA el ∂T F / Recall 3 pp.. 22 ≈ V ’≈ 2m ’2 g(ε ) = 2∆ ÷∆ ÷ ε 1 2 F « 4π 2 ◊« >2 ◊ F 2 pp.. 44 ≈ >2 ’≈ 3π 2 N ’3 ε = ∆ ÷∆ ÷ F « 2m ◊« V ◊ Ω g(ε F ) = 3N 2ε F = 3N 2kTF BB / Combining A and B gives NNoottee TT ddeeppeennddeennccee 3N 2 T Cel ≈ 3k T = 4.5Nk 2kTF TF

May 05 Lecture 10 8 EElleeccttrroonniicc hheeaatt ccaappaacciittyy ccoonntt……....

66 Notes: ? Absolute magnitude is much less than the -2 classical result (3Nk/2). T/TF~10 at room temperature for typical metals.

? Observed values are in good agreement. E.g. Sodium (Na) a “classic”, free-electron metal: -1 -1 (Cel)meas = 15.T J mol K -1 -1 (Cel)f.e. = 11.T J mol K

? The result is one of the most important consequences of Fermi-Dirac statistics.

? T dependence arises because, as the Fermi- edge broadens, more electrons get excited.

? N.B. the electronic contribution to the heat capacity of a metal is masked by a much larger contribution from vibration (see future lectures) at all but the lowest temperatures.

May 05 Lecture 10 9 PPaauullii ppaarraammaaggnneettiissmm

66 Paramagnetism in metals ? Metals are weakly paramagnetic but, unlike molecular-paramagnets, do not obey Curie’s Law (χ∼1/T) ? An external field, B, shifts spin-up and spin-

down electrons in energy by ±µBB gg((εε))ddεε//22== g( ) B/2 g(εεFF)µµBBB/2

? Net spin = 2 x no. switching-spin x spin-moment 2 M = 2(g(ε F )µB B 2)µB V = g(ε F )µB µ0 H V 2 χ P = g(ε F )µB µ0 V ? Small and independent of temperature. ? Again, only the electrons near εF are involved. ? The result is accurate but needs reducing by 1/3 due to Landau diamagnetism (Part II Quantum).

May 05 Lecture 10 10 DDeeggeenneerraaccyy pprreessssuurree

66 Pressure due to an electron gas: ? Consider the energy levels in a box under compression. It takes energy to compress the box, since the single-particles energies rise. Hence there must be an outwards pressure. n n 4 3 ≈ >2 ’ >2 2 ∆ ÷ 2 ≈π ’ 2 2 2 3 2 ε = ∆ ÷k = ∆ ÷ (l + m + n ) 2 « 2m ◊ 2m « a ◊ 1 1 a = l a = l/√2 66 Astrophysical examples, typical star: ? Gravity tends to compress the star. ? Nuclear energy keeps the star hot and inflated. ? Mainly composed of ionised hydrogen M ~ 3x1030 Kg, TT ~~ 33xx110088KK R ~ 3x107 m FF TT<

May 05 Lecture 10 11