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 ⇤ 4 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 = c t r 1 A or E = 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 ⇤ 2 + ( A)= 4 ⇥ r c ⇤t r · 1 ⇤2A 1 ⇤ 4 2A A + = j r c2 ⇤t2 r r · c ⇤t c ✓ ◆
To decouple one could choose (gauge freedom) 1 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 + 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 = q v A r dt v c ·
E&M Check 1 q Yields Lorentz from L = T U = mv2 q + 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 + q · 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 x y r ~ c m a† = (vx ivy) 2~ c r qB • with cyclotron frequency = c mc
• Straightforward to check a, a† =1 • So Hamiltonian becomes ⇥ ⇤
1 H = ~ c a†a + 2 • and consequently spectrum is (called Landau levels)
1 En = ~ c (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 = B (r ⇤)da = B⇥r2 r ⇥ · 1 1 ZS ZS – for any circular loop outside inner cylinder (and centered) – Vector potential in the direction of ˆ and line integral --> 2 r – Resulting in Br 2 modifying the Hamiltonian and the spectrum!! A = 1 ˆ 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 ⇤ 4 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 = 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 ⇤ 2 + ( A)= 4 ⇥ r c ⇤t r · 1 ⇤2A 1 ⇤ 4 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 + ) 1r ⇥ ⇥0 = ⇥ ) c t 1 A • With E = --> same E&M fields r c t B = A 1 • Can always find potentialsr ⇥ that satisfy A + =0 r · c t • If not: choose ⇤ such that 2 1 ⇥0 1 ⇥ 2 1 0= A0 + = A + + r · c t r · c t r c2 2t
E&M Employ this gauge freedom
1 ⇤ 2 + ( A)= 4 ⇥ r c ⇤t r · 1 ⇤2A 1 ⇤ 4 2A A + = j r c2 ⇤t2 r r · c ⇤t c ✓ ◆ 1 Can choose A + =0 (Lorentz gauge) r · c t
Leads to wave equations 1 ⇤2 2 = 4 ⇥ r c2 ⇤t2 1 ⇤2A 2A = 4 j r c2 ⇤t2 E&M Radiation gauge 1 ⇤ 2 + ( A)= 4 ⇥ r c ⇤t r · 1 ⇤2A 1 ⇤ 4 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 2 = 4 ⇥ r 1 ⇤2A 1 ⇤ 4 2A = j r c2 ⇤t2 c r ⇤t c
E&M Instantaneous Coulomb
3 (x0,t) Yields instantaneous Coulomb potential (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 2 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 · = 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 A k (polarizations) =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 ⇒ k = ck
i kt 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 = i k Ak(t)+i k A⇥ k(t) e · p 2c V k X i ik x = k 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 8 · · ZV
i ik x Use E(x,t)= k 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 k Ak(t) A⇥ k(t) e · V 4c V Z k k0 X X ik0 x k0 (Ak⇥ (t) A k0 (t)) e · ⇥ · 0 1 2 2 Orthogonality = k 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 8 · · ZV 1 1 2 2 2 = 2k Ak(t) + A k(t) 8 4 | | | | Xk ⇣ ⌘ 1 1 = k2 A (t) 2 = k2 A 2 8 | k | 8 | k| Xk Xk
Note: no time dependence! E&M Expand Fourier coefficients along polarization vectors
Use Ak = ek Ak =1,2 X 1 --> H = k2 A 2 em 8 | k | Xk
E&M Preparation for QUANTIZATION
In order to quantize, introduce real canonical variables i Qk(t)= [Ak(t) A⇤ (t)] 2cp4 k k Pk(t)= [Ak(t)+A⇤ (t)] 2p4 k i Invert --> Ak(t)= icp4 Qk(t)+ Pk(t) ⇥ k 2 2 2 2 2 Pk (t) 2 2 Pk So Ak(t) = c 4 Qk(t)+ = c 4 Qk + | | ⇥2 ⇥2 k k
And thus …..(what else)
E&M Oscillators of course 1 H = k2 A (t) 2 em 8 | k | k X 2 1 2 2 2 Pk = k c 4 Qk + 2 8 ⇥k Xk ✓ ◆ 1 = P 2 + ⇥2Q2 2 k k k k X Expand in polarizations Pk = ek Pk Qk = ek Qk =1,2 =1,2 X X 1 2 2 2 then Hem = P + Q 2 k k k k X E&M True canonical variables
Q k ,P k are canonical variables
i kt Check Ak(t)=e Ak
So A˙ (t)= i A (t) k k k i and from Qk(t)= [Ak(t) A⇤ (t)] 2cp4 k i it follows Q˙ k(t)= [ i⇥kAk(t) (i⇥k)A⇤ (t)] = Pk(t) 2cp4 k
Hem But also: Q˙ k = Pk = Pk Hem Similarly for generalized momentum P˙k = Qk E&M And now…. • Back to Hamiltonian • Looks like a sum of oscillators --> treat as such!
• From canonical classical variables in classical mechanics
• Quantize by introducing commutation relations between operators!!! (Dirac)
[Pk ,Pk0 0 ]=0
[Qk ,Qk0 0 ]=0
[Qk ,Pk0 0 ]=i~ k,k0 , 0
E&M Photons Introduce the usual operators 1 ak = (Pk i kQk ) p2~ k 1 ak† = (Pk + i kQk ) p2~ k with commutators [ak ,ak0 0 ]=0
a† ,a† =0 k k0 0 h i a ,a† = k k0 0 k,k0 , 0 h i E&M Each mode HO
ˆ Number operator for each mode Nk↵ = ak† ↵ak↵
Then ak↵, Nˆk ↵ = ak↵a† ak ↵ a† ak ↵ ak↵ 0 0 k0↵0 0 0 k0↵0 0 0 h i = ak↵a† ak ↵ a† ak↵ak ↵ k0↵0 0 0 k0↵0 0 0 = ak↵a† a† ak↵ ak ↵ k0↵0 k0↵0 0 0 h i = kk0 ↵↵0 ak↵ and a† , Nˆ = a† k↵ k0↵0 kk0 ↵↵0 k↵ h i So enough to work with one mode Nˆ = a†a Eigenkets of this Hermitian operator Nˆ n = n n | | Consider Naˆ † n = a†Nˆ + a† n =(n + 1)a† n | | | also eigenket with eigenvalueh n +1i E&M More • Similarly Naˆ n = aNˆ a n =(n 1)a n | ⇥ | ⇥ | ⇥ • So h i a† n = c n +1 | ⇥ + | ⇥ a n = c n 1 | ⇥ | ⇥ • Normalization from n = n Nˆ n = n a†a n 0 ⇥ | | ⇤ ⇥ | | ⇤ • Phase choice a n = ⇤n n 1 | ⇥ | ⇥ • Also a† n = ⇥n +1 n +1 | | • Integers otherwise negative norm appears a n = ⇤n n 1 | ⇥ | ⇥ a n 1 = ⇤n 1 n 2 | ⇥ | ⇥ ... a 2 = ⇤2 1 | ⇥ | ⇥ a 1 = ⇤1 0 | ⇥ | ⇥ a 0 =0 E&M | ⇥ Photon states • Operator that adds a photon with momentum ~ k and polarization ↵ ak† ↵ • Single photon state a† 0 = 0, 0,...,0, 1 , 0...... = 1 k↵ | | k↵ | k↵ • No quantum: vacuum state 0 | • Normalized two-photon state (same mode) 1 a† a† 0 = 0, 0,...,0, 2k↵, 0...... = 2k↵ ⇥2 k↵ k↵ | | |
• Different modes
a† a† 0 = 0, 0,...,0, 1k↵, 0..., 0, 1k ↵ , 0...... = 1k↵1k ↵ = a† a† 0 k↵ k0↵0 | | 0 0 | 0 0 k0↵0 k↵ |
E&M Development
• General state nki i a† ki i n n n ... = 0 k1 1 k2 2 k3 3 ⇣ ⌘ | nki i ! | kYi i • So that p
a† nk ↵ ...nk ↵ ... = nk ↵ +1 nk ↵ ...(nk ↵ + 1)... ki↵i | 1 1 i i i i | 1 1 i i p • Photons: quantum excitations of the radiation field since classical vector potential has been replaced by quantum operator acting on photon states! i cp4 1 Ak icp4 Qk + Pk = [ i⇥kQk + Pk ] 2~⇥k ) ⇥k ⇥k p2~⇥k ⇥ p 8 ~ = c a ⇥ k r k 8 ~ • also Ak c ak† ) ⇥k r E&M Vector potential operator
2 1/2 2 ~c i(k x ⇥kt) i(k x ⇥kt) A(x,t)= ak ek e · + ak† ek e · ⇥kV Xk ✓ ◆ n o
Acts on photon states: adds or removes one!
Acts on charged particle at x and t (first quantization)
First rewrite Hamiltonian of free field for further interpretation No work...
E&M Hamiltonian free field
ˆ Number operator for each mode Nk↵ = ak† ↵ak↵
ˆ ˆ 1 ˆ Hamiltonian operator Hem = ~ k Nk + 2 ~ kNk ) Xk ⇣ ⌘ Xk Momentum operator from Poynting vector (exercise) 1 Pˆ = d3x (E B B E) em 8 c ⇥ ⇥ ZV ˆ 1 ˆ = ~k Nk + 2 = ~kNk Xk ⇣ ⌘ Xk ˆ Single photon state Hema† 0 = ~ k a† 0 k | k | ˆ Pema† 0 = ~k a† 0 k | k | So massless! 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 m c = E p c = ~ k ~ k c = ~ k c ~ k c =0 E&M More on photon states
• Characterized also by polarization vector ek↵ • Transforms as vector --> interpret as 1 unit of intrinsic angular momentum or spin of the photon • Consider circular polarization vectors ( ) 1 e ± = (ek,1 iek,2) k ⇥2 ± • Rotate by angle about propagation axis
e0 = cos e +sin e e + e k,1 k,1 k,2 ) k,1 k,2 e0 = sin e + cos e e + e k,2 k,1 k,2 ) k,1 k,2 1 • New circular polarization vectors e± = (ek,1 iek,2 ) k ⇥⇤2 0 ± 0 ( ) 1 = e ± ⇥(ek,2 ( )iek,1) k ⇥ ⇤2 ± ( ) ( ) = e ± i ⇥ e ± k ⇥ k ( ) =(1 i ⇥) ek± ⇥ E&M Angular momentum ( ) • Compare e± =(1 i ⇥) e ± k ⌥ k
i Jz im e ~ 1m = e 1m • With | ⇤ | ⇤ (1 im ⇥) 1m ⇥ | ⇤ (+) • Interpret m =1 e ) k ( ) m = 1 e ) k • Quantization axis along k so photons can have helicity 1 or -1 but not 0 --> no longitudinal photons • No contradiction (no rest frame where photon is at rest) • Photons with good helicity 1 ak† = ak† ,1 iak† ,2 ± ⇥2 ± ⇣ ⌘ E&M