Shanghai Jiao Tong University Lecture 2 Shanghai Jiao Tong University Lecture 2 2017 Anthony J. Leggett Department of Physics University of Illinois at Urbana-Champaign, USA and Director, Center for Complex Physics Shanghai Jiao Tong University

Reminder: the vector potential 𝑨 𝒓𝑡 in classical mechanics The equation of motion of a charged particle in an electric field 𝑬 and magnetic field 𝑩: 𝑚 𝑑𝒗 𝑑𝑡 =𝑒 𝑬+v×𝑩 + 𝐅 non−em How to find 𝐻 𝒓,𝒑 s.t. 𝑑𝒓 𝑑𝑡 = 𝜕𝐻 𝜕𝒑 𝑑𝒑 𝑑𝑡 =− 𝜕𝐻 𝜕𝒓 , ? Solution: define 𝐴 𝒓,𝑡 s.t. 𝑬 𝒓,𝑡 =− 𝜕𝑨 𝑟𝑡 𝜕𝒕 , 𝑩 𝒓,𝑡 =𝜵×𝑨 𝒓𝑡 (hence 𝜵×𝑬=𝜕B/𝜕t) and put Faraday 𝐻 𝒓,𝒑 = 𝒑−𝑒𝑨 𝑟𝑡 2 /2𝑚 + 𝑉 𝑛𝑜𝑛−𝑒𝑚 This works! [Recall: 𝑑𝑨(𝑟𝑡) 𝑑𝑡 = 𝜕𝑨 𝑟𝑡 𝜕𝑡 +𝒗∙𝛁𝑨, 𝐯× 𝛁×𝑨 𝒊 = 𝐯 𝒋 𝜕 𝒊 𝑨 𝒋 − 𝐯∙𝛁 𝐴 𝒊 ] 𝐯≡ 𝑑𝒓 𝑑𝑡 = 𝜕𝐻 𝜕𝒑 = 1 𝒎 𝒑−𝑒𝐴 𝑟𝑡 note: ≠𝒑/𝑚 𝐻 𝒓,𝒑 = 1 2 𝑚 𝐯 2 = so kinetic energy (only)! (but expressed in terms of p and A).

Quantum mechanics: 𝒑→−𝒊ℏ𝜵 so 𝐾𝐸 is 𝑯 𝑲 = −𝑖ℏ𝜵−𝑒𝑨 𝟐 /2𝑚 and so, including possible V 𝑛𝑜𝑛−𝑒𝑚 ≡V 𝑟 , 𝐻 = 1 2𝑚 −𝑖ℏ𝜵−𝑒𝑨 𝑟𝑡 2 +V 𝒓 In CM (classical mechanics), all effects obtainable from 𝑨(𝒓𝑡) are equally derivable only from 𝑬(𝒓𝑡) and 𝑩(𝒓𝑡)⇒ vector potential redundant. In QM (quantum mechanics) this is not true: 𝑨(𝒓𝑡) has a “life of its own”!

. Single charged particle on thin ring If field 𝑩 through ring is uniform, can take . 𝑩 𝑅 𝑨 𝜽 𝑨≡ 𝐴 𝜃 𝜽 , 𝐴 𝜃 ≡ 1 2 𝐵𝑅 then flux Φ through ring is Φ≡𝜋 𝑅 2 𝐵⇒ 𝐴 𝜃 =Φ/2𝜋𝑅 TISE for 𝜓≡𝜓 𝜃 is 𝐻 𝜓 𝜃 ≡ 1 2𝑚 −𝑖ℏ𝜵−𝑒𝑨 𝒓 2 𝜓 𝜃 =𝐸𝜓 𝜃 𝜵 𝜽 = 1 𝑅 𝜕 𝜕𝜃 only nonzero component of 𝜵 is and only component of 𝑨 is 𝐴 𝜃 , so ℏ 2 2𝑚 𝑅 2 −𝑖 𝜕 𝜕𝜃 − 𝑒 ℏ 𝐴 𝜃 𝑅 2 𝜓 𝜃 =𝐸𝜓 𝜃 or putting 𝐴 𝜃 =Φ/2𝜋𝑅 and defining Φ 𝑜 𝑠𝑝 ≡ℎ/𝑒 (single-particle) flux quantum ℏ 2 2𝑚 𝑅 2 −𝑖 𝜕 𝜕𝜃 −Φ/ Φ 𝑜 𝑠𝑝 2 𝜓 𝜃 =𝐸𝜓 𝜃 = 𝐿 𝑧 (angular momentum in units of ℏ) Formal solution is 𝐸= ℏ 2 2𝑚 𝑅 2 𝑘−Φ/ Φ 𝑜 𝑆𝑝 2 𝜓 𝜃 = exp 𝑖𝑘𝜃 , (𝑘 arbitrary),

However, crucial point: 𝜓 𝜃 must be single-valued, i.e. 𝜓 𝜃+2𝑛𝜋 =𝜓 𝜃 (SVBC) single-valuedness boundary condition Hence, only allowed values of 𝑘 are integers ℓ=0,±1,±2… (i.e. angular momentum 𝐿 𝑧 is quantized in units of ℏ) Thus, 𝜓 ℓ 𝜃 = exp 𝑖ℓ𝜃 , ℓ=0,±1,±2 , 𝐸 ℓ = ℏ 2 2𝑚 𝑅 2 ℓ−Φ/ Φ 𝑜 𝑠𝑝 2 𝑗 ℓ = 𝑒ℏ 𝑚/𝑅 ℓ−Φ/ Φ 𝑜 𝑠𝑝 For Φ< Φ 𝑜 𝑠𝑝 /2 , GS has ℓ=0⇒ 𝑝 𝜃 ≡ 𝐿 𝑧 /𝑅= ℓℏ/𝑅 =0 However, recall: 𝐯= 𝒑−𝒆𝑨 /𝑚⇒ 𝐯 𝜃 =−𝑒 𝐴 𝜃 /𝑚 ⇒ 𝐣 𝜃 ≡𝑒 v 𝜃 =− 𝑒 2 /𝑚 𝐴 𝑜 ≠0 in general, in sense to produce magnetic field opposite to 𝑩 ⇒ GS is diamagnetic, 𝐣 𝜃 =− 𝑒 2 /𝑚 𝐴 𝜃 ≠0

Single charged particle on ring: two notes 1. What is the situation in classical mechanics? We can still formally introduce ℓ≡ 𝐿 𝑧 /ℏ and write 𝐸 ℓ = ℏ 2 2𝑚 𝑅 2 ℓ−Φ/ Φ 𝑜 𝑆𝑝 2 but now there is no restriction on ℓ (SVBC is meaningless since no wave function!) so now GS always corresponds to ℓ=Φ/ Φ 𝑜 𝑠𝑝 , equivalent to 𝐣 𝜃 =0 (no diamagnetism). 𝑨 𝑓 𝑡 𝑜 𝑨 𝜃 𝑡 However, consider time-dependent problem: since motion is restricted to ring, Lorenz force 𝐯×𝑩 is irrelevant and we have by Newton II 𝑚 𝑑𝐯 𝑑𝑡 =𝑬 𝑡 =− 𝜕𝑨 𝜕𝑡 or 𝑚 𝑑 v 𝜃 𝑑𝑡 =− 𝑑 𝐴 𝜃 𝑑𝑡

If at t=0 v 𝜃 =0, solution is simply 𝐴 𝜃 𝑗 𝜃 𝑡 𝑡 𝑜 If at t=0 v 𝜃 =0, solution is simply v 𝜃 𝑡 =− 𝑒/𝑚 𝐴 𝜃 𝑡 and in particular for 𝑡= 𝑡 𝑜 v 𝜃 =− 𝑒/𝑚 𝐴 𝑓 ⇒ j 𝜃 =− 𝑒 2 𝑚 𝐴 𝑓 as in Quantum Mechanics case. However, this is not the lowest-energy state, so scattering by walls, etc. will reduce j 𝜃 to zero. 2. Aharonov-Bohm effect note induced diamagnetic current depends only on total trapped flux Φ, not on details of how it is produced. Hence in particular can get nonzero effect even when 𝑩=0 everywhere on ring! (e.g. 𝑩 produced by “Helmholtz coil”) 𝑩=0 𝑩≠0

Atomic diamagnetism To the extent that argument applies, velocity of electrons at radius r given by 𝐯 𝑟 =−𝑒𝐴 𝑟)/𝑚 but electric current density j 𝑟 =𝑛 𝑟 𝑒𝐯 𝑟 , hence 𝐣 𝒓 = −𝑛 𝑟 𝑒 2 𝑚 𝑨 𝑟 Circulating current produces magnetic field 𝛥𝐵 opposite to original one. Estimate order of magnitude: take 𝐴~𝐵 𝑅 𝑎𝑡 , 𝐽~ 𝑅 𝑎𝑡 2 𝑗 ~𝑅 𝑎𝑡 2 𝑛 𝑒 2 𝐴/𝑚~ 𝑁 𝑒 2 /𝑚 𝑅 𝑎𝑡 𝐴~ 𝑁 𝑒 2 /𝑚 𝐵, Δ𝐵~ 𝜇 𝑜 𝐽 𝑅 𝑎𝑡 ~ 𝑁 𝑒 2 𝜇 𝑜 𝑚 𝑅 𝑎𝑡 𝐵: 𝑁 𝑒 2 𝜇 𝑜 /𝑚 𝑅 𝑎𝑡 ~ 10 −6 , so effect very small.

Superconductors: London phenomenology Basic postulate: as in atomic diamagnetism, 𝐣 𝒓 = −𝑛 𝑒 2 𝑚 𝐴 𝒓 Combine with Maxwell’s equation 𝐣=𝛁×𝑯≡ 1 𝜇 𝑜 𝛁×𝑩= 1 𝜇 𝑜 𝛁× 𝛁×𝑨 =− 1 𝜇 𝑜 𝛻 2 𝑨 (div 𝑨=0) gives 𝛻 2 𝑨=+ 𝑛 𝑒 2 𝑚 𝜇 𝑂 𝑨 or taking curl. 𝛻 2 𝑩= 𝑛 𝑒 2 𝜇 𝑂 𝑚 𝑩≡ 𝝀 𝑳 −𝟐 𝑩 London penetration depth Hence, both in atomic diamagnets and in superconductors, (n(𝑟), hence 𝜆 𝐿 , comparable in two cases) 𝐵~ 𝐵 𝑜 exp−z/ 𝜆 𝐿 Qualitative difference: in both cases 𝜆 𝐿 ~ 10 −5 cm, but: in atomic diamagnets, 𝜆 𝐿 ≫ atomic size ⇒ effect very small in superconductors, 𝜆 𝐿 ≪ size of sample ⇒ effect spectacular: magnetic field totally excluded from bulk (Meissner effect)

Problems with London phenomenology Meissner effect is thermodynamically stable phenomenon, circulating supercurrents on metastable. Hence London argument does not explain stability of supercurrents! (: beware misleading statements in literature) No explanation of vanishing Peltier coefficient. Why do not all metals show Meissner effect? Let’s turn question C around: when does Meissner effect not occur? 1. Classical systems: no restriction on v 𝜃 ≡ 𝑝 𝜃 −𝑒 𝐴 𝜃 𝑚 , and by Maxwellian statistical mechanics 𝑃 𝜐 𝜃 ∝exp 1 2 𝑚 𝜐 𝜃 2 /𝑘𝑇, hence from symmetry 𝜐 =0⇒ no circulating current ⇒ no diamagnetism (Bohr-van Leeuwen theorem)

2. Quantum Mechanics, but noninteracting particles obeying classical statistics: now 𝑝 (or angular momentum 𝐿 ) is quantized 𝐿=ℓℏ , ℓ= …−2,−1, 0, 1, 2… and energy ∝ ℓ−Φ/ Φ 𝑜 𝑠𝑝 2 ℏ 2 /2𝑚 𝑅 2 Φ 𝑜 𝑠𝑝 ≡ℎ/𝑒 so 𝑃 ℓ ∝exp− ℓ−Φ/ Φ 𝑜 𝑠𝑝 2 ∙ ℏ 2 /2𝑚 𝑅 2 𝑘 𝐵 𝑇 Φ≲ Φ 𝑜 𝑠𝑝 Crucial point: under normal circumstances 𝑘 𝐵 𝑇≫ ℏ 2 /2𝑚 𝑅 2 , so can effectively replace discrete values of ℓ by continuum ⇒ back to classical mechanics.* 3. So, will only get Meissner effect if for some reason all or most particles forced to be in same state. Then the probability of angular momentum ℓ for this state is 𝑃 ℓ ∝exp − 𝑁 𝑜 ℓ−Φ/ Φ 𝑜 2 ℏ 2 /2m 𝑅 2 𝑘 𝐵 𝑇 Number of particles in same state and provided 𝑘 𝐵 𝑇, though ≫ ℏ 2 /2𝑚 𝑅 2 , is ≪𝑁 ℏ 2 /2𝑚 𝑅 2 , can get results similar to atomic diamagnetism. Does this ever happen? Yes, e.g. for noninteracting gas of bosons! *Doesn’t work for atomic diamagnetism because ℏ 2 /2𝑚 𝑅 2 is ~𝑒𝑉, hence ≫ 𝑘 𝐵 𝑇. Electron volts

Summary of lecture 2: In presence of electromagnetic vector potential 𝑨 𝒓 , Hamiltonian for single particle of charge 𝑒 is For single particle on ring, in flux Φ< 1 2 ℎ/𝑒, this leads in ground state to 𝑗 𝜃 =− 𝑒 2 /𝑚 𝐴 𝜃 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx For a closed-shell atom, similar argument leads to London phenomenology: assume (*) also describes superconductor ⇒ Meissner effect Difficulty: doesn’t work for classical systems (Bohr – van Leeuwen theorem) nor (for 𝑘𝑇≳ ℏ 2 /𝑚 𝑅 2 ) for quantum systems obeying Maxwell-Boltzmann statistics Difficulty can be overcome if for some reason all particles forced to behave in same way. 𝐻 =𝑝 −𝑖ℏ𝜵− 2𝑚 𝑒𝐴(𝑟) 2 +𝑉(𝑟) 𝑗 𝒓 = −𝑛 𝑟 𝑚 𝑒 2 𝐴 𝒓 (diamagnetism) (*)