Inclusion of the electromagnetic field in Quantum Mechanics similar to Classical Mechanics but with interesting consequences • Maxwell’s equations • Scalar and vector potentials • Lorentz force • Transform to Lagrangian • Then Hamiltonian • Minimal coupling to charged particles E&M Maxwell’s equations Gaussian units E(x,t)=4⇥(x,t) r · B(x,t)=0 r · 1 ⇤ E(x,t)= B(x,t) r ⇥ − c ⇤t 1 ⇤ 4π B(x,t)= E(x,t)+ j(x,t) r ⇥ c ⇤t c E&M Scalar and Vector potential Quantum applications require replacing electric and magnetic fields! B = A implies B =0 r r · ⇥ 1 ∂ From Faraday E + A =0 r ⇥ c ∂t ✓ ◆ 1 ∂ so E + A = Φ c ∂t −r 1 ∂A or E = Φ −r − c ∂t in terms of vector and scalar potentials. Homogeneous equations are automatically solved. E&M Gauge freedom Remaining equations using ( A)= ( A) 2A r ⇥ r ⇥ r r · − r 1 ⇤ 2Φ + ( A)= 4⇥ r c ⇤t r · − 1 ⇤2A 1 ⇤Φ 4π 2A A + = j r − c2 ⇤t2 − r r · c ⇤t − c ✓ ◆ To decouple one could choose (gauge freedom) 1 ∂Φ A + =0 r · c ∂t more later… first E&M Coupling to charged particles 1 Lorentz F = q E + v B c ⇥ ⇢ 1 ∂A 1 Rewrite F = q Φ + v ( A) −r − c ∂t c ⇥ r ⇥ ⇢ v ( A)= (v A) (v ) A Note ⇥ r ⇥ r · − · r ∂A dA +(v ) A = and ∂t · r dt d ∂U q So that F = U + with U = qΦ v A −r dt ∂v − c · E&M Check 1 q Yields Lorentz from L = T U = mv2 qΦ + v A − 2 − c · d ∂L ∂L =0 Equations of motion dt ∂v − ∂x ∂L q Generalized momentum p = = mv + A ∂v c Solve for v and substitute in Hamiltonian --> Hamiltonian for a charged particle 2 p q A H = p v L = − c + qΦ · − 2m E&M Include external electromagnetic field in QM • Static electric field: nothing new (position --> operator) • Include static magnetic field with momentum and position 2 operators p q A(x) H = − c 2m 1 q • Note velocity operator v = p A m − c ⇣ ⌘ • Note Hamiltonian not “free” particle one • Use ~ ∂Aj [pi,Aj]= i ∂xi q~ • to show that [vi,vj]=i ijkBk m2c 1 • Gauge independent! So think in terms of H = m v 2 2 | | E&M Include external electromagnetic field • Include uniform magnetic field • For example by B(x)=Bzˆ q~B • Only nonvanishing commutator [v ,v ]=i x y m2c • Write Hamiltonian as 1 2 2 2 H = m vx + vy + vz p 2 v = z • but now z m so this corresponds to free particle motion parallel to magnetic field (true classically too) • Only consider 1 H = m v2 + v2 2 x y • Operators don’t commute but commutator is a complex number! • So... E&M Harmonic oscillator again... • Motion perpendicular to magnetic field --> harmonic oscillator m • Introduce a = (v + iv ) 2 ω x y r ~ c m a† = (vx ivy) 2~ωc − r qB • with cyclotron frequency ω = c mc • Straightforward to check a, a† =1 • So Hamiltonian becomes ⇥ ⇤ 1 H = ~ωc a†a + 2 • and consequently spectrum is (called Landau levels) 1 En = ~ωc (n + 2 ) n =0, 1,.... E&M Aharanov-Bohm effect • Consider hollow cylindrical shell • Magnetic field inside inner cylinder either on or off • Charged particle confined between inner and outer radius as well as top and bottom E&M Discussion • Without field: – Wave function vanishes at the radii of the cylinders as well as top and bottom --> discrete energies • With field (think of solenoid) – No magnetic field where the particle moves; inside in z-direction and constant – Spectrum changes because the vector potential is needed in the Hamiltonian – Use Stokes theorem ( A) nˆ da = A d r × · · ZS IC – Only z-component of magnetic field so left-hand side becomes ( A) nˆ da = Bθ(r ⇤)da = B⇥r2 r ⇥ · 1 − 1 ZS ZS – for any circular loop outside inner cylinder (and centered) – Vector potential in the direction of φ ˆ and line integral --> 2πr – Resulting in Br 2 modifying the Hamiltonian and the spectrum!! A = 1 φˆ 2r E&M Example • No field • Example of radial wave function • Problem solved in cylindrical coordinates • Also with field --> E&M Quantize electromagnetic field •Classical free field equations •Quantize •Photons •Coupling to charged particles •One-body operator acting on charged particles and photons E&M Maxwell’s equations Gaussian units E(x,t)=4⇥(x,t) r · B(x,t)=0 r · 1 ⇤ E(x,t)= B(x,t) r ⇥ − c ⇤t 1 ⇤ 4π B(x,t)= E(x,t)+ j(x,t) r ⇥ c ⇤t c E&M Scalar and Vector potential Quantum applications require replacing electric and magnetic fields! 1 ∂A E = Φ −r − c ∂t B = A r ⇥ in terms of vector and scalar potentials. Homogeneous equations are automatically solved. E&M Gauge freedom Remaining equations 1 ⇤ 2Φ + ( A)= 4⇥ r c ⇤t r · − 1 ⇤2A 1 ⇤Φ 4π 2A A + = j r − c2 ⇤t2 − r r · c ⇤t − c ✓ ◆ To decouple employ gauge freedom. Observe: adding gradient of scalar function to vector potential yields same magnetic field To keep electric field the same: change scalar potential accordingly! E&M Gauge transformation • Explicitly A A0 = A + Λ ) 1r∂Λ ⇥ ⇥0 = ⇥ ) − c ∂t 1 ∂A • With E = Φ --> same E&M fields −r − c ∂t B = A 1 ∂Φ • Can always find potentialsr ⇥ that satisfy A + =0 r · c ∂t • If not: choose ⇤ such that 2 1 ∂⇥0 1 ∂⇥ 2 1 ∂ Λ 0= A0 + = A + + Λ r · c ∂t r · c ∂t r − c2 ∂2t E&M Employ this gauge freedom 1 ⇤ 2Φ + ( A)= 4⇥ r c ⇤t r · − 1 ⇤2A 1 ⇤Φ 4π 2A A + = j r − c2 ⇤t2 − r r · c ⇤t − c ✓ ◆ 1 ∂Φ Can choose A + =0 (Lorentz gauge) r · c ∂t Leads to wave equations 1 ⇤2Φ 2Φ = 4⇥ r − c2 ⇤t2 − 1 ⇤2A 2A = 4πj r − c2 ⇤t2 − E&M Radiation gauge 1 ⇤ 2Φ + ( A)= 4⇥ r c ⇤t r · − 1 ⇤2A 1 ⇤Φ 4π 2A A + = j r − c2 ⇤t2 − r r · c ⇤t − c ✓ ◆ Alternative: radiation gauge (Coulomb, or transverse gauge)--> useful for quantizing free field A =0 r · yields 2Φ = 4⇥ r − 1 ⇤2A 1 ⇤Φ 4π 2A = j r − c2 ⇤t2 c r ⇤t − c E&M Instantaneous Coulomb 3 ρ(x0,t) Yields instantaneous Coulomb potential Φ(x,t)= d x0 V x x0 Vector potential --> inhomogeneous wave equation Z | − | rhs can be calculated from instantaneous Coulomb potential 1 ∂A Now no sources ⇒ free field E = − c ∂t B = A 1 ∂2A r ⇥ and 2A =0 ⇒ solve in r − c2 ∂t2 large box with volume V = L3 E&M Free field solutions Use periodic BC so expand in plane waves to avoid standing ones 2π Allowed values k x = n x n x =0 , 1 , 2 ,... also for y and z L ± ± Normalization 1 i(k k0) x dx e − · = δ V kk0 ZV 1 ik x So solution can be written as A(x,t)= Ak(t) e · pV Xk Gauge choice ⇒ k A =0 · k So for future reference: A k = e k α A k α (polarizations) α=1,2 X 2 ∂ Ak(t) From wave equation + c 2 k 2 A ( t )=0 for each mode ∂t2 k E&M Harmonic solutions Fourier coefficients oscillate harmonically ⇒ ωk = ck iωkt So time dependence: Ak(t)=e− Ak Given initial distribution of A k ( t = 0) --> problem solved! E&M fields real so make vector potential explicitly real 1 ik x ik x A(x,t)= Ak(t)e · + Ak⇥ (t)e− · 2pV ! Xk Xk 1 ik x = Ak(t)+A⇥ k(t) e · p − 2 V k X ⇥ ⇤ E&M Fields 1 ik x Use A(x,t)= Ak(t)+A⇥ k(t) e · p − 2 V k X ⇥ ⇤ Then electric field 1 ⇥A E(x,t)= − c ⇥t 1 ik x = iωk Ak(t)+iωk A⇥ k(t) e · − p − − 2c V k X i ik x = ωk Ak(t) A⇥ k(t) e · p − − 2c V k and magnetic field X ⇥ ⇤ B(x,t)= A r ⇥ i ik x = k Ak(t)+A⇥ k(t) e · p ⇥ − 2 V k X ⇥ ⇤ E&M Energy in field 1 General H = dx (E E + B B) em 8π · · ZV i ik x Use E(x,t)= ωk Ak(t) A⇥ k(t) e · p − − 2c V k Note fields are real so X ⇥ ⇤ dxE E = dxE E⇥ · · ZV ZV 1 ik x = dx 2 ωk Ak(t) A⇥ k(t) e · V 4c V − − Z k k0 X X ik0 x ωk0 (Ak⇥ (t) A k0 (t)) e− · ⇥ · 0 − − 1 2 2 Orthogonality = ωk Ak(t) A⇥ k(t) 4c2 − − k X 1 2 2 = k Ak(t) A⇥ k(t) 4 − − k E&M X Energy in field continued dxB B = dxB B⇤ Similarly · · ZV ZV (exercise) 1 2 2 = k Ak(t)+A⇤ k(t) 4 − k X 1 2 2 So with dxE E = k Ak(t) A⇤ k(t) · 4 − − ZV k X 1 Energy becomes H = dx (E E + B B) em 8π · · ZV 1 1 2 2 2 = 2k Ak(t) + A k(t) 8π 4 | | | − | Xk ⇣ ⌘ 1 1 = k2 A (t) 2 = k2 A 2 8π | k | 8π | k| Xk Xk Note: no time dependence! E&M Expand Fourier coefficients along polarization vectors Use Ak = ekαAkα α=1,2 X 1 --> H = k2 A 2 em 8π | kα| Xkα E&M Preparation for QUANTIZATION In order to quantize, introduce real canonical variables i Qk(t)= [Ak(t) A⇤ (t)] 2cp4π − k k Pk(t)= [Ak(t)+A⇤ (t)] 2p4π k i Invert --> Ak(t)= icp4π Qk(t)+ Pk(t) − ⇥ k 2 2 2 2 2 Pk (t) 2 2 Pk So Ak(t) = c 4π Qk(t)+ = c 4π Qk + | | ⇥2 ⇥2 k k And thus …..(what else) E&M Oscillators of course 1 H = k2 A (t) 2 em 8π | k | k X 2 1 2 2 2 Pk = k c 4π Qk + 2 8π ⇥k Xk ✓ ◆ 1 = P 2 + ⇥2Q2 2 k k k k X Expand in polarizations Pk = ekαPkα Qk = ekαQkα α=1,2 α=1,2 X X 1 2 2 2 then Hem = P + ω Q 2 kα k kα kα X E&M True canonical variables Q k ,P k are canonical variables iωkt Check Ak(t)=e− Ak So A˙ (t)= iω A (t) k − k k i and from Qk(t)= [Ak(t) A⇤ (t)] 2cp4π − k i it follows Q˙ k(t)= [ i⇥kAk(t) (i⇥k)A⇤ (t)] = Pk(t) 2cp4π − − k ∂Hem But also: Q˙ k = Pk = ∂Pk ∂Hem Similarly for generalized momentum P˙k = − ∂Qk E&M And now….