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Inclusion of the Electromagnetic Field in Quantum Mechanics Similar to Classical Mechanics but with Interesting Consequences

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Inclusion of the Electromagnetic Field in Quantum Mechanics Similar to Classical Mechanics but with Interesting Consequences 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….
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