Elementary Par,cles Rohlf Ch , p474 et seq.

Transcription

1 Elementary Par,cles Rohlf Ch , p474 et seq.

2 The Schroedinger equa,on is non- rela,vis,c. Rela,vis,c wave equa,on (Klein- Gordon eq.) Rela,vis,c equa,on connec,ng the energy and momentum of a free electron. E 2 = p 2 c 2 + m 2 c 4 The rela,vis,c form of the Schroedinger equa,on, which is known as the Klein- Gordon eq. 2 2 Ψ t 2 = 2 c 2 2 Ψ + m 2 c 4 Ψ

3 Nega,ve energy solu,ons. Non-relativistic free electron: E = p2 2m (positive) Relativistic electron: E 2 = p 2 c 2 + m 2 c 4 E ± = ± p 2 c 2 + m 2 c 4 = ±E 0 (positive and negative)

4 Meaning of a negative energy electron? The negative energy electron is equivalent to a positive energy electron moving backward in time. It behaves like a positive energy particle moving forward in time with charge - q.

5 The Schroedinger equa,on is non- rela,vis,c and doesn t account for the internal angular momentum of elementary par,cles Dirac equa,on for electron - electron proper,es. Dirac relativistic electron equation: Ĥ is a 4 4 matrix,. ψ = i c α x x + α y ψ y + α z ψ z + βmc2 ψ = i ψ t ψ is a 1 4 colum matrix called a spinor Using p x = i ψ x etc. (c αi p + βmc 2 )ψ = Êψ

6 Dirac relativistic electron equation: i ψ t = ( cαi p + βmc 2 )ψ β = α 1 = i 0 0 i 0 α 2 = 0 i 0 0 i α 3 = ψ = ψ 1 (x µ,t) ψ 2 (x µ,t) ψ 3 (x µ,t) ψ 4 (x µ,t) ψ 1 = ψ E,1/2 ψ 2 = ψ E, 1/2 ψ 3 = ψ E,1/2 ψ 4 = ψ E, 1/2 = positive energy electron with spin up. = positive energy electron with spin down. = negative energy electron with spin up. = negative energy electron with spin down. Spin: S = 1 / 2 m s = ±1 / 2. Magnetic moment: µ = g e 2m s = g e 2m 1 2 = e 2m (g = 2)

7 1927 Dirac equa,on - covariant nota,on Dirac relativistic electron equation: i c γ µ ψ mc 2 ψ = 0 x µ 3 µ=0 γ µ = γ 0, γ 1, γ 2, γ 3 each 4 4 matrices. Simplify notation: = 1 c = 1 ψ x µ µ and summation convention on repeated incices. iγ µ µ ψ mψ = 0

8 Relativistic equation for a photon field. E 2 p 2 c 2 = Ψ γ t c 2 2 Ψ γ = Ψ γ c 2 t 2 2 Ψ γ = 0 Time independence-steady state 2 Ψ γ t 2 = 0 2 Ψ γ = 0 Add source term -ρ/ε 0 = qδ ( r 0) / ε 0 ε 0 2 Ψ γ = qδ ( r 0) Homework: verify Ψ γ is a solu,on.

9 2 Ψ γ = q ε 0 δ r 0 Divergence theorem. Poisson s equa,on ( ) i Ψγ dv = q δ ( r 0) ε 0 dv = q 1 ε 0 ( ) i( Ψγ ) dv = Ψ γ V S ids = S Ψ γ ids Spherical symmetry. = S Ψ γ r ˆriˆrr 2 dω = Ψ γ r S r 2 dω = 4πr 2 Ψ γ r Ψ γ r = q 1 4πε 0 r 2 Ψ γ = q 1 4πε 0 r

10 Away from r = 0 ε 0 2 Ψ γ = ε 0 1 r 2 r r2 Ψ γ r = 2 Ψ γ r r Ψ γ r = 0 q Exercise: Verify Ψ γ = 4πε 0 r is a solution to the Laplace equation.

11 Quantum Electrodynamics - QED Feynman graphs of an electron and positron t +E 0 -E 0 x

12 Basic Electromagne,c Interac,on e e e H e α Rate H 2 α H e 2 α Rate H 2 α 2

13 Quantum Electrodynamics (Rohlf P23) Fine structure constant. Velocity of electron in the lowest quantum state depends on the electronic charge strength. α v(bohr) c = β(bohr) = e 2 4πε 0 c = ke2 c α is a dimentionless quantity which is a measure of the basic electronic coupling strength.

14 Vacuum polariza,on. Self energy. Running coupling constant

15 Running Coupling Constant What is e or α? It depends on distance from the charge. Distance of penetration corresponds to momentum of the probe. Momentum of probe is related to wavelength and resolution of probe p is related to the resolution of the probe since p= h λ = k So, distance r ~ λ = 2π k = h p There are an infinite number of vacuum polarization graphs to sum. 1 Leading order: α = α 0 1 α 0 3π Λ2 ln k 2 Λ is a parameter which is necessary to keep the entire procedure finite. α 0 cannot be directly measured.

16 e = e 0 1+ α 0 3π Λ2 ln k 2 What is e 0 or α 0? It is unmeasurable! But, we do know e at some µ 2 = - k 2. ( ) = α 0 1 α µ 1 + α 0 3π Λ2 ln µ 2 α ( k 2 ) = 1 + α 0 3π α 0 µ 2 k 2 µ 2 ln Λ2 = 1+ α 0 3π Λ2 ln µ 2 α 0 + α 0 3π ln µ 2 k α 0 3π Λ2 ln µ 2 = α 0 α µ ( ) α ( k 2 ) = α ( k 2 ) = α 0 α 0 α ( µ ) + α 0 3π ln µ 2 k 2 ( ) α µ 2 1 α(µ 2 ) 3π ln k 2 µ 2 α ( k 2 ) = α ( µ ) ( ) 1+ α µ 3π ln µ 2 k 2

17 Vacuum polariza,on. Self energy. Running coupling constant α ( k 2 ) = ( ) α µ 2 1 α(µ2 ) 3π k2 ln µ 2 k = λ ~ r k is "momentum transfer" α ( r) α(r µ ) 1 α(r µ ) 3π ln r 2 µ r 2

18 Exercise The charge on an electron as seen from a distance 1 nm is e = C, and is α= = The charge of an electron is related to α = e 2 4πε 0 hc Find α and then e seen at a distance from the electron r = fm = nm. α( nm) α( 1 nm) ~ 1.03 e( nm) e(1 nm) In reality the effect is much larger since at such small distances lots of other intermeduate states can also contribute.

19 Gauge Invariance Symmetry This is the basis of the Standard Model Equations of physics should be invariant with respect to a phase shift applied to the wave function. Global Unitary Phase Transformation: Let ψ (x,t) ψ (x,t)e ia = ψ (x,t). Show Schroedinger's equation in 1-dim. for a free particle: 2 2m 2 ψ (x,t) x 2 = i ψ (x,t) t gives the same for when ψ (x,t) is substituted for ψ (x,t)

20 Show that Schroedinger s equa,on in one dimension is invariant under a global phase transforma,on.

21 Local Unitary Phase Transformation: Let ψ (x, t) ψ (x, t)e iaf (x,t). Schroedinger's equation for a free particle: 2 2m 2 ψ (x, t) ψ (x, t) x 2 = i t 2 ( iaf (x,t) ψ (x, t)e ) x 2 = i 2m ( iaf (x,t) ψ (x, t)e ) t This no longer has the same form of the Schroedinger eq. Carry out the differentiation. It is straight forward but messy.

22 Modify Schroedinger s equa,on m 2 ψ (x,t) x 2 = i ψ (x,t) t 1 2m i x e c A 2 ψ (x,t) = i t ev ψ (x,t) Perform the local phase transformation: ψ (x,t) ψ (x,t) = ψ (x,t)e iε(x,t) = ψ (x,t)e iaf (x,t), a = e c ( e iε(x,t) is called a local U(1)transformation, or a guage transformation of the first kind. ) with the gauge conditions: A A = A + f x V V = V 1 c f t. The Schroedinger equation is now phase invariant.

23 What are the new fields? B = A E = V A t

24 Special homework exercise. Show that the modified Schroedinger equation in one dimension: 1 2m i x e c A 2 ψ (x,t) = i t ev ψ (x,t) remains invariant under a U(1) phase transformation: iaf (x,t) ψ (x,t) ψ (x,t) = ψ (x,t)e with the gauge conditions: A A = A + f x V V = V 1 c f t.