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A Brief Introduction to Relativistic

Hsin-Chia Cheng, U.C. Davis

1 Introduction

In 215AB, you learned non-relativistic , e.g., Schr¨odinger , p2 E = + V, 2m ∂ E → i , p → −i ∇, ~∂t ~ ∂ 2 i Ψ = ~ ∇2Ψ + V Ψ. (1) ~∂t 2m Now we would like to extend quantum mechanics to the relativistic domain. The natural thing at first is to search for a relativistic single- equation to replace the Schr¨odinger equation. It turns out that the form of the relativistic equation depends on the of the particle,

spin-0 Klein-Gordon equation spin-1/2 spin-1 Proca equation etc

It is useful to study these one-particle and their solutions for certain problems. However, at certain point these one-particle relativistic encounter fatal inconsistencies and break down. Essentially, this is because while is conserved in but is not. with mass can be created and destroyed in real physical processes. For example, pair annihilation + − − − e e → 2γ, muon decay µ → e ν¯eνµ. They cannot be described by single-particle theory. At that stage we are forced to abandon single-particle relativistic wave equations and go to a many-particle theory in which particles can be created and destroyed, that is, quantum field theory, which is the subject of the course.

1 2 Summary of Special Relativity

An event occurs at a single point in - and is defined by its coordinates xµ, µ = 0, 1, 2, 3, x0 = ct, x1 = x, x2 = y, x3 = z, (2) in any given frame. The interval between 2 events xµ andx ¯µ is called s,

s2 = c2(t − t¯)2 − (x − x¯)2 − (y − y¯)2 − (z − z¯)2 = (x0 − x¯0)2 − (x1 − x¯1)2 − (x2 − x¯2)2 − (x3 − x¯3)2. (3)

We define the metric  1 0 0 0   0 −1 0 0  gµν =   , (4)  0 0 −1 0  0 0 0 −1 then we can write

2 X µ µ ν ν µ ν s = gµν(x − x¯ )(x − x¯ ) = gµν∆x ∆x , (5) µ,ν where we have used the Einstein convention: repeated indices (1 upper + 1 lower) are summed except when otherwise indicated. Lorentz transformations The postulates of Special Relativity tell us that the of is the same in any inertial frame. s2 is invariant under transformations from one inertial frame to any other. Such transformations are called Lorentz transformations. We will only need to discuss the homogeneous Lorentz transformations (under which the origin is not shifted) here, 0µ µ ν x = Λ νx . (6)

µ ν 0µ 0ν gµνx x = gµνx x µ ρ ν σ ρ σ = gµνΛ ρx Λ σx = gρσx x µ ν ⇒ gρσ = gµνΛ ρΛ σ. (7)

It’s convenient to use a notation,

 x0  1 µ  x  x :   = x. (8)  x2  x3

2 s2 = xT gx, x0 = Λx ⇒ g = ΛT gΛ (9)

Take the determinant, det g = det ΛT det g det Λ, (10) so det Λ = ±1 (+1: proper Lorentz transformations, −1: improper Lorentz trans- formations). Example: (proper):

x00 = x0 x01 = x1 cos θ + x2 sin θ x02 = −x1 sin θ + x2 cos θ x03 = x3 (11)  1 0 0 0   0 cos θ sin θ 0  Λ =   (12)  0 − sin θ cos θ 0  0 0 0 1

Example: Boosts (proper): v t0 = γ(t − x1) or x00 = γx0 − γβx1 c2 x01 = γ(x1 − vt) = γx1 − γβx0 x02 = x2 x03 = x3 (13) where v 1 β = , γ = . (14) c p1 − β2 It’s convenient to define a quantity η such that cosh η = γ, sinh η = γβ, then  cosh η − sinh η 0 0   − sinh η cosh η 0 0  Λ =   . (15)  0 0 1 0  0 0 0 1 One can easily check that det Λ = cosh2 η − sinh2 η = 1. Four-vectors, A contravariant vector is a set of 4 quantities which transforms like xµ under a

3 ,

 V 0  1 µ  V  0µ µ ν V =   ,V = Λ V . (16)  V 2  ν V 3

A covariant vector is a set of 4 quantities which transforms as

0 −1ν −1 T Aµ = Aν Λ µ , Λ = gΛ g. (17) An upper index is called a contravariant index and a lower index is called a co- variant index. Indices can be raised or lowered with the metric gµν and its µν µλ µ inverse g = diag(1, −1, −1, −1), g gλν = δ ν. The product of a contravari- µ ant vector and a covariant vector V Aµ is invariant under Lorentz transformations. Examples: Energy and form a contravariant 4-vector, E pµ = ( , p , p , p ). (18) c x y z 4- gradient, ∂ 1 ∂ ∂ ∂ ∂  = , , , ≡ ∂ (19) ∂xµ c ∂t ∂x ∂y ∂z µ is a covariant vector, ∂ ∂xν ∂ ∂ = = Λ−1ν . (20) ∂x0µ ∂x0µ ∂xν µ ∂xν One can generalize the concept to tensors,

0µ0ν0··· µ0 ν0 −1 ρ −1 σ µν··· T ρ0σ0··· = Λ µΛ ν ··· (Λ ) ρ0 (Λ ) σ0 ··· T ρσ···. (21)

Maxwell’s equations in Lorentz covariant from (Heaviside-Lorentz conven- tion)

∇ · E = ρ (22) ∇ · B = 0 (23) 1 ∂B ∇ × E + = 0 (24) c ∂t 1 ∂E 1 ∇ × B − = J (25) c ∂t c From the second equation we can define a vector potential A such that

B = ∇ × A (26)

4 Substituting it into the third equation, we have

 1 ∂A ∇ × E + = 0, (27) c ∂t then we can define a potential φ, such that 1 ∂A E = −∇φ − . (28) c ∂t

Gauge invariance: E, B are not changed under the following transformation,

A → A − ∇χ 1 ∂ φ → φ + χ. (29) c ∂t (cρ, J) form a 4-vector J µ. conservation can be written in the Lorentz µ covariant form, ∂µJ = 0, µ (φ, A) from a 4-vector A (Aµ = (φ, −A)), from which one can derive an antisym- metric electromagnetic field tensor, ∂ F µν = ∂µAν − ∂νAµ (note: ∂i = −∂ = − , i = 1, 2, 3). (30) i ∂xi     0 −Ex −Ey −Ez 0 Ex Ey Ez µν  Ex 0 −Bz By   −Ex 0 −Bz By  F =   ,Fµν =    Ey Bz 0 −Bx   −Ey Bz 0 −Bx  Ez −By Bx 0 −Ez −By Bx 0 (31) Maxwell’s equations in the covariant form: 1 ∂ F µν = J ν (32) µ c ˜µν ∂µF = ∂µFλν + ∂λFνµ + ∂νFµλ = 0 (33) where 1 F˜µν ≡ µνλρF , (34) 2 λρ 0123 and its even = +1, its odd permutation = −1. Gauge invariance: Aµ → Aµ + ∂µχ. One can check F µν is invariant under this transformation.

5 3 Klein-Gordon Equation

In non-, the energy for a is

p2 E = . (35) 2m To get quantum mechanics, we make the following substitutions: ∂ E → i , p → −i ∇, (36) ~∂t ~ and the Schr´odinger equation for a free particle is

2 ∂Ψ − ~ ∇2Ψ = i . (37) 2m ~ ∂t In relativistic mechanics, the energy of a free particle is p E = p2c2 + m2c4. (38)

Making the same substitution we obtain √ ∂Ψ − 2c2∇2 + m2c2Ψ = i . (39) ~ ~ ∂t It’s difficult to interpret the on the left hand side, so instead we try

E2 = p2c2 + m2c4 (40)  ∂ 2 ⇒ i Ψ = − 2c2∇2 + m2c4Ψ, (41) ~∂t ~ 1  ∂ 2 m2c2 or Ψ − ∇2Ψ ≡ 2Ψ = − Ψ, (42) c2 ∂t ~2 where 1  ∂ 2 2 = − ∇2 = ∂ ∂µ. (43) c2 ∂t µ Plane-wave solutions are readily found by inspection,

1  i   i  Ψ = √ exp p · x exp − Et , (44) V ~ ~ where E2 = p2c2 + m2c4 and thus E = ±pp2c2 + m2c4. Note that there is a negative energy solution as well as a positive energy solution for each value of p. Na¨ıvely one should just discard the negative energy solution. For a free particle in a positive energy state, there is no mechanism for it to make a transition to

6 the negative energy state. However, if there is some external potential, the Klein- Gordon equation is then altered by the usual replacements, e E → E − eφ, p → p − A, (45) c e (i ∂ − eφ)2Ψ = c2(−i ∇ − A)2Ψ + m2c4Ψ. (46) ~ t ~ c The solution Ψ can always be expressed as a superposition of free particle solutions, provided that the latter form a complete set. They from a complete set only if the negative energy components are retained, so they cannot be simply discarded. Recall the and current in Schr´odinger equation. If we multiply the Schr´odinger equation by Ψ∗ on the left and multiply the conjugate of the Schr¨odinger equation by Ψ, and then take the difference, we obtain 2 − ~ (Ψ∗∇2Ψ − Ψ∇2Ψ∗) = i (Ψ∗Ψ˙ + ΨΨ˙ ∗) 2m ~ 2 ∂ ⇒ − ~ ∇(Ψ∗∇Ψ − Ψ∇Ψ∗) = i (Ψ∗Ψ) (47) 2m ~∂t ∗ ~ ∗ ∗ Using ρs = Ψ Ψ, js = 2mi (Ψ ∇Ψ − Ψ∇Ψ ), we then obtain the equation of continuity, ∂ρ s + ∇ · j = 0 (48) ∂t s Now we can carry out the same procedure for the free-particle Klein-Gordon equa- tion: m2c2 Ψ∗2Ψ = − Ψ∗Ψ ~ m2c2 Ψ2Ψ∗ = − ΨΨ∗ (49) ~ Taking the difference, we obtain

∗ ∗ ∗ µ µ ∗ Ψ 2Ψ − Ψ2Ψ = ∂µ(Ψ ∂ Ψ − Ψ∂ Ψ ) = 0. (50) This suggests that we can define a probability 4-current, jµ = α(Ψ∂µΨ − Ψ∂µΨ∗), where α is a constant (51)

µ µ 0 and it’s conserved, ∂µj = 0, j = (j , j). To make j agree with js, α is chosen to ~ be α = − 2mi . So, j0 i  ∂Ψ ∂Ψ∗  ρ = = ~ Ψ∗ − Ψ . (52) c 2mc2 ∂t ∂t ∗ ρ does reduce to ρs = Ψ Ψ in the non-relativistic limit. However, ρ is not positive -definite and hence can not describe a probability density for a single particle. Pauli and Weisskopf in 1934 showed that Klein-Gordon equation describes a spin-0 (scalar) field. ρ and j are interpreted as charge and current density of the particles in the field.

7 4 Dirac Equation

To solve the density problem of the Klein-Gordon equation, people were looking for an equation which is first order in ∂/∂t. Such an equation is found by Dirac.

It is difficult to take the square root of −~2c2∇2 + m2c4 for a single . One can take the inspiration from E&M: Maxwell’s equations are first-order but combining them gives the second order wave equations.

Imagining that ψ consists of N components ψl,

3 N N 1 ∂ψl X X ∂ψn imc X + αk + β ψ = 0, (53) c ∂t ln ∂xk ln n k=1 n=1 ~ n=1 where l = 1, 2,...,N, and xk = x, y, z, k = 1 , 2, 3.   ψ1  ψ   2  ψ =  .  , (54)  .  ψN and αk, β are N × N matrices. Using the matrix notation, we can write the equations as 1 ∂ψ imc + α · ∇ψ + βψ = 0, (55) c ∂t ~ where α = α1xˆ + α2yˆ + α3zˆ. N components of ψ describe a new degree of freedom just as the components of the Maxwell field describe the of the light quantum. In this case, the new degree of freedom is the spin of the particle and ψ is called a . We would like to have positive-definite and conserved probability, ρ = ψ†ψ, where ψ† is the hermitian conjugate of ψ (so is a row matrix). Taking the hermitian conjugate of Eq. (55),

1 ∂ψ† imc + ∇ψ† · α − ψ†β† = 0. (56) c ∂t ~ Multiplying the above equation by ψ and then adding it to ψ†× (55), we obtain

1  ∂ψ ∂ψ†  imc ψ† + ψ + ∇ψ† · α†ψ + ψ†α · ∇ψ + (ψ†βψ − ψ†β†ψ) = 0. (57) c ∂t ∂t ~ The ∂ (ψ†ψ) + ∇ · j = 0 (58) ∂t

8 can be obtained if α† = α, β† = β, then 1 ∂ (ψ†ψ) + ∇ · (ψ†αψ) = 0 (59) c ∂t with j = cψ†αψ. (60) From Eq. (55) we can obtain the Hamiltonian,

∂ψ   Hψ = i = c∇ · ~∇ + βmc2 ψ. (61) ~ ∂t i

One can see that H is hermitian if α, β are hermitian. To derive properties of α, β, we multiply Eq. (55) by the conjugate operator,

1 ∂ imc  1 ∂ imc  − α · ∇ − β + α · ∇ + β ψ = 0 c ∂t ~ c ∂t ~  2 2 2  1 ∂ i j m c 2 imc i i ⇒ − α α ∂i∂j + β − (βα + α β)∂i ψ = 0 (62) c2 ∂t2 ~2 ~

i j 1 i j j i We can rewrite α α ∂i∂j as 2 (α α + α α )∂i∂j. Since it’s a relativistic , the second order equation should coincide with the Klein-Gordon equation. Therefore, we must have

αiαj + αjαi = 2δijI (63) βαi + αiβ = 0 (64) β2 = I (65)

Because βαi = −αiβ = (−I)αiβ, (66) if we take the determinant of the above equation,

det β det αi = (−1)N det αi det β, (67) we find that N must be even. Next, we can rewrite the relation as

(αi)−1βαi = −β (no summation). (68)

Taking the trace,

Tr (αi)−1βαi = Tr (αiαi)−1β = Tr[β] = Tr[−β], (69) we obtain Tr[β] = 0. Similarly, one can derive Tr[αi] = 0.

9 Covariant form of the Dirac equation Define

γ0 = β, γj = βαj, j = 1, 2, 3 µ 0 1 2 3 ν γ = (γ , γ , γ , γ ), γµ = gµνγ (70)

Multiply Eq. (55) by iβ,

1 ∂ imc  iβ × + α · ∇ + β ψ = 0 c ∂t ~   0 ∂ j ∂ mc  µ mc ⇒ iγ + iγ − ψ = iγ ∂µ − ψ = 0 (71) ∂x0 ∂xj ~ ~

µ µ Using the short-hand notation: γ ∂µ ≡6∂, γ Aµ ≡6A,  mc i 6∂ − ψ = 0 (72) ~ From the properties of the αj and β matrices, we can derive

γ0† = γ0, (hermitian) (73) γj† = (βαj)† = αj†β† = αjβ = −βαj = −γj, (anti-hermitian)(74) 㵆 = γ0γµγ0, (75) γµγν + γνγµ = 2gµνI. (Clifford ). (76)

Conjugate of the Dirac equation is given by

† µ† mc † −i∂µψ γ − ψ = 0 ~ † 0 µ 0 mc † ⇒ −i∂µψ γ γ γ − ψ = 0 (77) ~ We will define the Dirac adjoint spinor ψ by ψ ≡ ψ†γ0. Then

µ mc i∂µψγ + ψ = 0. (78) ~ The four-current is jµ  j  = ψγµψ = ρ, , ∂ jµ = 0. (79) c c µ

10 Properties of the γµ matrices We may form new matrices by multiplying γ matrices together. Because different γ matrices anticommute, we only need to consider products of different γ’s and the order is not important. We can combine them in 24 −1 ways. Plus the identity we have 16 different matrices, I γ0, iγ1, iγ2, iγ3 γ0γ1, γ0γ2, γ0γ3, iγ2γ3, iγ3γ1, iγ1γ2 iγ0γ2γ3, iγ0γ3γ1, iγ0γ1γ2, γ1γ2γ3 0 1 2 3 5 iγ γ γ γ ≡ γ5(= γ ). (80)

Denoting them by Γl, l = 1, 2, ··· , 16, we can derive the following relations.

(a) ΓlΓm = almΓn, alm = ±1 or ± i.

(b) ΓlΓm = I if and only if l = m.

(c) ΓlΓm = ±ΓmΓl.

(d) If Γl 6= I, there always exists a Γk, such that ΓkΓlΓk = −Γl.

(e) Tr(Γl) = 0 for Γl 6= I. Proof: Tr(−Γl) = Tr(ΓkΓlΓk) = Tr(ΓlΓkΓk) = Tr(Γl) . P16 (f) Γl are linearly independent: k=1 xkΓk = 0 only if xk = 0, k = 1, 2, ··· , 16. Proof: 16 ! X X X xkΓk Γm = xmI + xkΓkΓm = xmI + xkakmΓn = 0 (Γn 6= I) k=1 k6=m k6=m P . Taking the trace, xmTr(I) = − k6=m xkakmTr(Γn) = 0 ⇒ xm = 0. for any m. This implies that Γk’s cannot be represented by matrices smaller than 4 × 4. In fact, the smallest representations of Γk’s are 4 × 4 matrices. (Note that this 4 is not the of the space-time. the equality is accidental.) (g) Corollary: any 4 × 4 matrix X can be written uniquely as a of the Γk’s. 16 X X = xkΓk k=1 X Tr(XΓm) = xmTr(ΓmΓm) + xkTr(ΓkΓm) = xmTr(I) = 4xm k6=m 1 x = Tr(xΓ ) m 4 m 11 (h) Stronger corollary: ΓlΓm = almΓn where Γn is a different Γn for each m, given a fixed l. Proof: If it were not true and one can find two different Γm,Γm0 such that ΓlΓm = almΓn, ΓlΓm0 = alm0 Γn, then we have

alm Γm = almΓlΓn, Γm0 = alm0 ΓlΓn ⇒ Γm = Γm0 , alm0 which contradicts that γk’s are linearly independent. (i) Any matrix X that commutes with γµ (for all µ) is a multiple of the identity. Proof: Assume X is not a multiple of the identity. If X commutes with all γµ then it commutes with all Γl’s, i.e., X = ΓlXΓl. We can express X in terms of the Ga matrices, X X = xmΓm + xkΓk, Γm 6= I. k6=m

There exists a Γi such that ΓiΓmΓi = −Γm. By the hypothesis that X commutes with this Γi, we have X X = xmΓm + xkΓk = ΓiXΓi k6=m X = xmΓiΓmΓi + xkΓiΓkΓi k6=m X = −xmΓm + ±xkΓk. k6=m

Since the expansion is unique, we must have xm = −xm.Γm was arbitrary except that Γm 6= I. This implies that all xm = 0 for Γm 6= I and hence X = aI. (j) Pauli’s fundamental theorem: Given two sets of 4×4 matrices γµ and γ0µ which both satisfy {γµ, γν} = 2gµνI, there exists a nonsingular matrix S such that

γ0µ = SγµS−1.

µ 0 Proof: F is an arbitrary 4 × 4 matrix, set Γi is constructed from γ and Γi is constructed from γ0µ. Let 16 X 0 S = ΓiF Γi. i=1

ΓiΓj = aijΓk 2 2 2 ΓiΓjΓiΓj = aijΓk = aij 2 3 ΓiΓiΓjΓiΓjΓj = ΓjΓi = aijΓiΓj = aijΓk 0 0 0 ΓiΓj = aijΓk

12 For any i,

0 X 0 0 X 4 0 X 0 4 ΓiSΓi = ΓiγjF ΓjΓi = aijΓkF Γk = ΓkF Γk = S, (aij = 1). j j j It remains only to prove that S is nonsingular.

16 0 X 0 S = ΓiGΓi, for G arbitrary. i=1

0 0 0 By the same argument, we have S = ΓiS Γi.

0 0 0 0 0 S S = ΓiS ΓiΓiSΓi = ΓiS SΓi,

0 0 S S commutes with Γi for any i so S S = aI. We can choose a 6= 0 because F,G are arbitrary, then S is nonsingular. Also, S is unique up to a constant. Otherwise µ −1 µ −1 −1 µ µ −1 −1 if we had S1γ S1 = S2γ S2 , then S2 S1γ = γ S2 S1 ⇒ S2 S1 = aI. Specific representations of the γµ matrices

Recall H = (−cα(i~)∇+βmc2). In the non-relativistic limit, mc2 term dominates the total energy, so it’s convenient to represent β = γ0 by a diagonal matrix. Recall Trβ = 0 and β2 = I, so we choose I 0 1 0 β = where I = . (81) 0 I 0 1

αk’s anticommute with β and are hermitian,  0 Ak αk = , (82) (Ak)† 0

Ak: 2 × 2 matrices, anticommute with each other. These properties are satisfied by the , so we have  0 σk 0 1 0 −i 1 0  αk = , σ1 = , σ2 = , σ3 = (83) σk 0 1 0 i 0 0 −1 From these we obtain I 0   0 σi 0 I γ0 = β = , γi = βαi = , γ = iγ0γ1γ2γ3 = . 0 −I −σi 0 5 I 0 (84) This is the “Pauli-Dirac” representation of the γµ matrices. It’s most useful for system with small , e.g., . Let’s consider the simplest possible problem: free particle at rest. ψ is a 4- component wave-function with each component satisfying the Klein-Gordon equa- tion, i (p·x−Et) ψ = χ e ~ , (85)

13 where χ is a 4-component spinor and E2 = p2c2 + m2c4. Free particle at rest: p = 0, ψ is independent of x,

2 0 2 0 Hψ = (−i~cα · ∇ + mc γ )ψ = mc γ ψ = Eψ. (86) In Pauli-Dirac representation,γ0 = diag(1, 1, −1, −1), the 4 fundamental solutions are 1 0 2 χ1 =   ,E = mc , 0 0 0 1 2 χ2 =   ,E = mc , 0 0 0 0 2 χ3 =   ,E = −mc , 1 0 0 0 2 χ4 =   ,E = −mc . 0 1

As we shall see, Dirac wavefunction describes a particle pf spin-1/2. χ1, χ2 repre- 2 sent spin-up and spin-down respectively with E = mc . χ3, χ4 represent spin-up and spin-down respectively with E = −mc2. As in Klein-Gordon equation, we have negative solutions and they can not be discarded. For ultra-relativistic problems (most of this course), the “Weyl” representation is more convenient.   ψ1       ψ2 ψA ψ1 ψ3 ψPD =   = , ψA = , ψb = . (87) ψ3 ψB ψ2 ψ4 ψ4

In terms of ψA and ψB, the Dirac equation is ∂ mc i ψA + iσ · ∇ψB = ψA, ∂x0 ~ ∂ mc −i ψB − iσ · ∇ψA = ψB. (88) ∂x0 ~ Let’s define 1 1 ψA = √ (φ1 + φ2), ψB = √ (φ2 − φ1) (89) 2 2

14 and rewrite the Dirac equation in terms of φ1 and φ2, ∂ mc i φ1 − iσ · ∇φ1 = φ2, ∂x0 ~ ∂ mc i φ2 + iσ · ∇φ2 = φ1. (90) ∂x0 ~

On can see that φ1 and φ2 are coupled only via the mass term. In ultra-relativistic limit (or for nearly such as ), rest mass is negligible, then φ1 and φ2 decouple, ∂ i φ − iσ · ∇φ = 0, ∂x0 1 1 ∂ i φ + iσ · ∇φ = 0, (91) ∂x0 2 2 The 4-component wavefunction in the Weyl representation is written as   φ1 ψWeyl = . (92) φ2

Let’s imagine that a massless spin-1/2 is described by φ1, a state of a definite momentum p with energy E = |p|c,

i (p·x−Et) φ1 ∝ e ~ . (93)

∂ 1 ∂ E i φ1 = i φ1 = φ1, ∂x0 c ∂t ~c 1 iσ · ∇φ1 = − σ · pφ1 ~ σ · p ⇒ Eφ = |p|cφ = −cσ · pφ or φ = −φ . (94) 1 1 1 |p| 1 1 The operator h = σ · p/|p| is called the “helicity.” Physically it refers to the component of spin in the direction of . φ1 describes a neutrino with helicity −1 (“left-handed”). Similarly, σ · p φ = φ , (h = +1, “right-handed”). (95) |p| 2 2 The γµ’s in the Weyl representation are 0 I  0 σi −I 0 γ0 = , γi = , γ = . (96) I 0 −σi 0 5 0 I

Exercise: Find the S matrix which transform between the Pauli-Dirac represen- tation and the Weyl representation and verify that the gaµ matrices in the Weyl representation are correct.

15 5 of the Dirac Equation

We will set ~ = c = 1 from now on. In E&M, we write down Maxwell’s equations in a given inertial frame, x, t, with the electric and magnetic fields E, B. Maxwell’s equations are covariant with respect to Lorentz transformations, i.e., in a new Lorentz frame, x0, t0, the equations have the same form, but the fields E0(x0, t0), B0(x0, t0) are different. Similarly, Dirac equation is Lorentz covariant, but the wavefunction will change when we make a Lorentz transformation. Consider a frame F with an observer O and coordinates xµ. O describes a particle by the wavefunction ψ(xµ) which obeys

 ∂  iγµ − m ψ(xµ). (97) ∂xµ

In another inertial frame F 0 with an observer O0 and coordinates x0ν given by

0ν ν µ x = Λ µx , (98)

O0 describes the same particle by ψ0(x0ν) and ψ0(x0ν) satisfies

 ∂  iγν − m ψ0(x0ν). (99) ∂x0ν

Lorentz covariance of the Dirac equation means that the γ matrices are the same in both frames. What is the transformation matrix S which takes ψ to ψ0 under the Lorentz trans- formation? ψ0(Λx) = Sψ(x). (100) Applying S to Eq. (97),

∂ iSγµS−1 Sψ(xµ) − mSψ(xµ) = 0 ∂xµ ∂ ⇒ iSγµS−1 ψ0(x0ν) − mψ0(x0ν) = 0. (101) ∂xµ Using ∂ ∂ ∂x0ν ∂ = = Λν , (102) ∂xµ ∂x0ν ∂xµ µ ∂x0ν we obtain ∂ iSγµS−1Λν ψ0(x0ν) − mψ0(x0ν) = 0. (103) µ ∂x0ν Comparing it with Eq. (99), we need

µ −1 ν ν µ −1 −1µ ν Sγ S Λ µ = γ or equivalently Sγ S = Λ ν γ . (104)

16 We will write down the form of the S matrix without proof. You are encouraged to read the derivation in Shulten’s notes Chapter 10, p.319-321 and verify it by yourself.

µ µ µ νλ For an infinitesimal Lorentz transformation, Λ ν = δ ν +  ν. Multiplied by g it can be written as Λµλ = gµλ + µλ, (105) where µλ is antisymmetric in µ and λ. Then the corresponding Lorentz transfor- mation on the spinor wavefunction is given by i S(µν) = I − σ µν, (106) 4 µν where i i σ = (γ γ − γ γ ) = [γ , γ ]. (107) µν 2 µ ν ν µ 2 µ ν For finite Lorentz transformation,

 i  S = exp − σ µν . (108) 4 µν

Note that one can use either the active transformation (which transforms the ob- ject) or the passive transformation (which transforms the coordinates), but care should be taken to maintain consistency. We will mostly use passive transforma- tions unless explicitly noted otherwise. Example: about z-axis by θ angle (passive).

−12 = +21 = θ, (109) i σ = [γ , γ ] = iγ γ 12 2 1 2 1 2 −iσ3 0  = i 0 −iσ3 σ3 0  = ≡ Σ (110) 0 σ3 3  i σ3 0  θ σ3 0  θ S = exp + θ = I cos + i sin . (111) 2 0 σ3 2 0 σ3 2

We can see that ψ transforms under rotations like an spin-1/2 . For a rotation around a general direction nˆ, θ θ S = I cos + inˆ · Σ sin . (112) 2 2

17 Example: Boost inx ˆ direction (passive).

01 = −10 = η, (113) i σ = [γ , γ ] = iγ γ , (114) 01 2 0 1 0 1  i   i  S = exp − σ µν = exp − ηiγ γ 4 µν 2 0 1 η   η  = exp γ γ = exp − α1 2 0 1 2 η η = I cosh − α1 sinh . (115) 2 2 For a particle moving in the direction of nˆ in the new frame, we need to boost the frame in the −nˆ direction, η η S = I cosh + α · nˆ sinh . (116) 2 2

6 Free Particle Solutions to the Dirac Equation

The solutions to the Dirac equation for a free particle at rest are

1 r 2m 0 −imt ψ1 =   e ,E = +m, V 0 0 0 r 2m 1 −imt ψ2 =   e ,E = +m, V 0 0 0 r 2m 0 imt ψ3 =   e ,E = −m, V 1 0 0 r 2m 0 imt ψ4 =   e ,E = −m, (117) V 0 1 where we have set ~ = c = 1 and V is the total volume. Note that I have chosen a particular normalization Z d3xψ†ψ = 2m (118)

18 for a particle at rest. This is more convenient when we learn field theory later, because ψ†ψ is not invariant under boosts. Instead, it’s the zeroth component of a 4-vector, similar to E. The solutions for a free particle moving at a constant can be obtained by a Lorentz boost, η η S = I cosh + α · nˆ sinh . (119) 2 2 Using the Pauli-Dirac representation,

 0 σi αi = , σi 0  η η  cosh 2 σ · nˆ sinh 2 S = η η , (120) σ · nˆ sinh 2 cosh 2 and the following relations, E0 cosh η = γ0 = , sinh η = γ0β0, m r r η 1 + cosh η r1 + γ0 m + E0 cosh = = = , 2 2 2 2m r r η cosh η − 1 E0 − m sinh = = 2 2 2m 0 0µ 0 0 0 0 µ −pµx = p+ · x − E t = −pµx = −mt, (121)

0 0 where p+ = |p+|nˆ is the 3-momentum of the positive energy state, we obtain     η 1 r cosh 2 0 0 2m  0  −imt ψ (x ) = Sψ1(x) =   e 1 V 1 σ · nˆ sinh η  2 0  1  √ 1  0  i(p0 ·x0−E0t0) = √ m + E0    e + q 0 1 V  E −m σ · nˆ  E0+m 0  1  √ 1  0  i(p0 ·x0−E0t0) = √ m + E0 e + (122)  0   V  σ·p+ 1  E0+m 0 where we have used √ rE0 − m E02 − m2 |p0 | = = + (123) E0 + m E0 + m E0 + m in the last line.

19 1 0 ψ0 (x0) has the same form except that is replaced by . 2 0 1

0 0 p 2 2 0 0 For the negative energy solutions E− = −E = − |p| + m and p− = vnˆE− = 0 0 −vE nˆ = −p+. So we have

 0   σ·p− 1 − 0 1 √ E +m 0 0 0 0 0 0 0 0   i(p−·x +E t ) ψ3(x ) = √ m + E     e , (124) V  1  0 1 0 and ψ0 (x0) is obtained by the replacement → . 4 0 1 Now we can drop the primes and the ± subscripts, √   1 χ+,− i(p·x−Et) 1 i(p·x−Et) ψ1,2 = √ E + m σ·p e = √ u1,2e , V E+m χ+,− V √  σ·p  1 − E+m χ+,− i(p·x+Et) 1 i(p·x+Et) ψ3,4 = √ E + m e = √ u3,4e , (125) V χ+,− V where 1 0 χ = , χ = (126) + 0 − 1 (V is the proper volume in the frame where the particle is at rest.)

Properties of u1, ··· u4

† urus = 0 for r 6= s. (127)

  †  † † σ·p  χ+ u1u1 = (E + m) χ+ χ+ E+m σ·p E+m χ+  (σ · p)(σ · p) = (E + m)χ† 1 + χ . (128) + (E + m)2 + Using the following identity:

(σ · a)(σ · b) = σiaiσjbj = (δij + iijkσk)aibj = a · b + iσ · (a × b), (129) we have  |p|2  u†u = (E + m)χ† 1 + χ 1 1 + (E + m)2 + E2 + 2Em + m2 + |p|2 = (E + m)χ† χ + (E + m)2 + 2E2 + 2Em = χ χ + E + m + † = 2Eχ+χ+ = 2E. (130)

20 † † Similarly for other ur we have urus = δrs2E, which reflects that ρ = ψ ψ is the zeroth component of a 4-vector. One can also check that urus = ±2mδrs (131) where + for r = 1, 2 and − for r = 3, 4.

I 0  u u = u†γ0u γ0 = 1 1 1 1 0 −I  |p|2  = (E + m)χ† 1 − χ + (E + m)2 + E2 + 2Em + m2 − |p|2 = (E + m)χ† χ + (E + m)2 + 2m2 + 2Em = χ χ + E + m + † = 2mχ+χ+ = 2m (132) is invariant under Lorentz transformation. Orbital and spin Orbital angular momentum

L = r × p or

Li = ijkrjpk. (133)

(We don’t distinguish upper and lower indices when dealing with space only.)

dL i = i[H,L ] dt i 2 = i[cα · p + βmc ,Li]

= icαn[pn, ijkrjpk]

= icαnijk[pn, rj]pk = icαnijk(−iδnj~)pk = c~ijkαjpk = c~(α × p)i 6= 0. (134) We find that the orbital angular momentum of a free particle is not a constant of the motion.

21   1 1 σi 0 Consider the spin 2 Σ = 2 , 0 σi dΣ i = i[H, Σ ] dt i 2 = i[cαjpj + βmc , Σi]         σi 0 0 I 0 σi = ic[αj, Σi]pj using Σiγ5 = = = αi = γ5Σi 0 σi I 0 σi 0

= ic[γ5Σj, Σi]pj

= icγ5[Σj, Σi]pj

= icγ5(−2iijkΣk)pj

= 2cγ5ijkΣkpj

= 2cijkαkpj

= −2c(α × p)i. (135) Comparing it with Eq. (134), we find 1 d(Li + ~Σi) 2 = 0, (136) dt 1 so the total angular momentum J = L + 2 ~Σ is conserved.

7 Interactions of a Relativistic with an External Electromagnetic

We make the usual replacement in the presence of external potential: ∂ E → E − eφ = i − eφ, e < 0 for electron ~∂t e e p → p − A = −i ∇ − A. (137) c ~ c In covariant form, ie ∂µ → ∂µ + Aµ → ∂µ + ieAµ ~ = c = 1. (138) ~c Dirac equation in external potential: µ iγ (∂µ + ieAµ)ψ − mψ = 0. (139) Two component reduction of Dirac equation in Pauli-Dirac basis: I 0  ψ  0 σ ψ  ψ  (E − eφ) A − (p − eA) A − m A = 0, 0 −I ψB σ 0 ψB ψB

⇒ (E − eφ)ψA − σ · (p − eA)ψB − mψA = 0 (140)

−(E − eφ)ψB + σ · (p − eA)ψA − mψB = 0 (141)

22 where E and p represent the operators i∂t and −i∇ respectively. Define W = E − m, π = p − eA, then we have

σ · πψB = (W − eφ)ψA (142)

σ · πψA = (2m + W − eφ)ψB (143)

From Eq. (143), −1 ψB = (2m + W − eφ) σ · πψA. (144) Substitute it into Eq. (142),

(σ · π)(σ · π) ψ = (W − eφ)ψ . (145) 2m + W − eφ A A In non-relativistic limit, W − eφ  m,

1 1  W − eφ  = 1 − + ··· . (146) 2m + W − eφ 2m 2m

1 In the lowest order approximation we can keep only the leading term 2m , 1 (σ · π)(σ · π)ψ ' (W − eφ)ψ . (147) 2m A A Using Eq. (129),

(σ · π)(σ · π)ψA = [π · π + iσ · (π × π)]ψA. (148)

(π × π)ψA = [(p − eA) × (p − eA)]ψA

= [−eA × p − ep × A]ψA

= [+ieA × ∇ + ie∇ × A]ψA

= ieψA(∇ × A)

= ieBψA, (149) so 1 e (p − eA)2ψ − σ · Bψ + eφψ = W ψ . (150) 2m A 2m A A A Restoring ~, c, 1 e e (p − A)2ψ − ~ σ · Bψ + eφψ = W ψ . (151) 2m c A 2mc A A A This is the “Pauli-Schr¨odinger equation” for a particle with the spin-magnetic , e e µ = ~ σ = 2 S. (152) 2mc 2mc

23 In comparison, the relation between the angular momentum and the of a classical charged object is given by Iπr2 ω πr2 eωr2 e e µ = = e = = mωr2 = L. (153) c 2π c 2c 2mc 2mc

We can write e µ = g S (154) s 2mc in general. In Dirac theory, gs = 2. Experimentally,

− −13 gs(e ) = 2 × (1.0011596521859 ± 38 × 10 ). (155)

α The deviation from 2 is due to radiative corrections in QED, (g − 2)/2 = 2π + ··· . The predicted value for gs − 2 using α from the quantum Hall effect is

−13 (gs − 2)qH /2 = 0.0011596521564 ± 229 × 10 . (156)

They agree down to the 10−11 level. There are also spin-1/2 particles with anomalous magnetic moments, e.g., |e| |e| µproton = 2.79 , µneutron = −1.91 . (157) 2mpc 2mnc This can be described by adding the Pauli moment term to the Dirac equation,

µ µν iγ (∂µ + iqAµ)ψ − mψ + kσµνF ψ = 0. (158)

Recall i σ = (γ γ − γ γ ), µν 2 µ ν ν µ I 0   0 −σi  0 −σi σ = iγ γ = i = i = −iαi, 0i 0 i 0 −I σi 0 −σi 0 σk 0  σ = iγ γ =  Σk =  , ij i j ijk ijk 0 σk F 0i = −Ei, F ij = −ijkBk. (159)

Then the Pauli moment term can be written as

µ iγ (∂µ + iqAµ)ψ − mψ + 2ikα · Eψ − 2kΣ · Bψ = 0. (160)

The two component reduction gives

(E − qφ)ψA − σ · πψB − mψA + 2ikσ · EψB − 2kσ · BψA = 0, (161)

−(E − qφ)ψB + σ · πψA − mψB + 2ikσ · EψA − 2kσ · BψB = 0. (162)

24 (σ · π − 2ikσ · E)ψB = (W − qφ − 2kσ · B)ψA, (163)

(σ · π + 2ikσ · E)ψA = (2m + W − qφ + 2kσ · B)ψB. (164)

Again taking the non-relativistic limit, 1 ψ ' (σ · π + 2ikσ · E)ψ , (165) B 2m A we obtain 1 (W − qφ − 2kσ · B)ψ = (σ · π − 2ikσ · E)(σ · π + 2ikσ · E)ψ . (166) A 2m A Let’s consider two special cases. (a) φ = 0, E = 0 1 (W − 2kσ · B)ψ = (σ · π)2ψ A 2m A 1 q ⇒ W ψ = π2ψ − σ · Bψ + 2kσ · Bψ A 2m A 2m A A q ⇒ µ = − 2k. (167) 2m

(b) B = 0, E 6= 0 for the (q = 0) 1 W ψ = σ · (p + iµ E) σ · (p − iµ E)ψ A 2m n n A 1 = [(p + iµ E) · (p + iµ E) + iσ · (p + iµ E) × (p − iµ E)] ψ 2m n n n n A 1 = p2 + µ2 E2 + iµ E · p − iµ p · E + iσ · (iµ p × E − iµ E × p) ψ 2m n n n n n A 1 = p2 + µ2 E2 − µ (∇ · E) + 2µ σ · (E × p) + iµ σ · (∇ × E) ψ 2m n n n n A 1 = p2 + µ2 E2 − µ ρ + 2µ σ · (E × p) ψ . (168) 2m n n n A The last term is the spin- interaction, 1 dφ 1 dφ σ · (E × p) = − σ · (r × p) = − σ · L. (169) r dr r dr The second to last term gives an effective potential for a slow neutron moving in the electric field of an electron, µ ρ µ V = − n = n (−e)δ3(r). (170) 2m 2m It’s called “Foldy” potential and does exist experimentally.

25 8 Foldy-Wouthuysen Transformation

We now have the Dirac equation with interactions. For a given problem we can solve for the and wavefunctions (ignoring the negative energy solutions for a moment), for instance, the , We can compare the solutions to those of the schr¨odinger equation and find out the relativistic corrections to the spectrum and the wavefunctions. In fact, the problem of can be solved exactly. However, the exact solutions are problem-specific and involve unfamiliar special functions, hence they not very illuminating. You can find the exact solutions in many textbooks and also in Shulten’s notes. Instead, in this section we will develop a systematic approximation method to solve a system in the non-relativistic regime (E−m  m). It corresponds to take the approximation we discussed in the previous section to higher orders in a systematic way. This allows a physical interpretation for each term in the approximation and tells us the relative importance of various effects. Such a method has more general applications for different problems.

In Foldy-Wouthuysen transformation, we look for a unitary transformation UF removes operators which the large to the small components.

i i Odd operators (off-diagonal in Pauli-Dirac basis): α , γ , γ5, ··· Even operators (diagonal in Pauli-Dirac basis): 1, β, Σ, ···

0 iS ψ = UF ψ = e ψ, S = hermitian (171) First consider the case of a free particle, H = α · p + βm not time-dependent. ∂ψ0 i = eiSHψ = eiSHe−iSψ0 = H0ψ0 (172) ∂t We want to find S such that H0 contains no odd operators. We can try eiS = eβα·pˆθ = cos θ + βα · pˆsin θ, where pˆ = p/|p|. (173)

H0 = (cos θ + βα · pˆsin θ)(α · p + βm) (cos θ − βα · pˆsin θ) = (α · p + βm) (cos θ − βα · pˆsin θ)2 = (α · p + βm) exp (−2βα · pˆθ)  m  = (α · p) cos 2θ − sin 2θ + β (m cos 2θ + |p| sin 2θ) . (174) |p| To eliminate (α · p) term we choose tan 2θ = |p|/m, then

H0 = βpm2 + |p|2. (175) This is the same as the first Hamiltonian we tried except for the β factor which also gives rise to negative energy solutions. In practice, we need to expand the Hamiltonian for |p|  m.

26 General case: H = α · (p − eA) + βm + eΦ = βm + O + E, (176) O = α · (p − eA), E = eΦ, βO = −Oβ, βE = Eβ (177) H time-dependent ⇒ S time-dependent We can only construct S with a non-relativistic expansion of the transformed Hamiltonian H0 in a series in 1/m.

p4 p×(E,B) We’ll expand to m3 and m2 . ∂ ∂ψ0  ∂  Hψ = i e−iSψ0 = e−iSi + i e−iS ψ0 ∂t ∂t ∂t ∂ψ0   ∂   ⇒ i = eiS H − i e−iS ψ0 = H0ψ0 (178) ∂t ∂t S is expanded in powers of 1/m and is “small” in the non-relativistic limit. i2 in eiSHe−iS = H + i[S,H] + [S, [S,H]] + ··· + [S, [S, ··· [S,H]]]. (179) 2! n! 1 S = O( m ) to the desired order of accuracy 1 i 1 H0 = H + i[S,H] − [S, [S,H]] − [S, [S, [S,H]]] + [S, [S, [S, [S, βm]]]] 2 6 24 i 1 −S˙ − [S, S˙] + [S, [S, S˙]] (180) 2 6 We will eliminate the odd operators order by order in 1/m and repeat until the desired order is reached. First order [O(1)]: H0 = βm + E + O + i[S, β]m. (181) iβO To cancel O, we choose S = − 2m , β 1 i[S,H] = −O + [O, E] + βO2 (182) 2m m i2 βO2 1 1 [S, [S,H]] = − − [O, [O, E]] − O3 (183) 2 2m 8m2 2m2 i3 O3 1 [S, [S, [S,H]]] = − βO4 (184) 3! 6m2 6m3 i4 βO4 [S, [S, [S, [S,H]]]] = (185) 4! 24m3 iβO˙ −S˙ = (186) 2m i i − [S, S˙] = − [O, O˙ ] (187) 2 8m2 27 Collecting ,  O2 O4  1 i H0 = β m + − + E − [O, [O, E]] − [O, O˙ ] (188) 2m 8m3 8m2 8m2 β O3 iβO˙ + [O, E] − + 2m 3m2 2m = βm + E 0 + O0 (189)

0 1 0 0 0 Now O is O( m ), we can transform H by S to cancel O , ! −iβ −iβ β O3 iβO˙ S0 = O0 = [O, E] − + (190) 2m 2m 2m 3m2 2m

After transformation with S0,

  0 0 ∂ 0 β iβO˙ H00 = eiS H0 − i e−iS = βm + E 0 + [O0, E 0] + (191) ∂t 2m 2m = βm + E 0 + O00, (192)

00 1 00 −iβO00 where O is O( m2 ), which can be cancelled by a third transformation, S = 2m   00 ∂ 00 H000 = eiS H00 − i e−iS = βm + E 0 (193) ∂t  O2 O4  1 i = β m + − + E − [O, [O, E]] − [O, O˙ ] (194) 2m 8m3 8m2 8m2 Evaluating the operator products to the desired order of accuracy, O2 (α · (p − eA))2 (p − eA)2 e = = − Σ · B (195) 2m 2m 2m 2m 1   e ie [O, E] + iO˙ = (−iα · ∇Φ − iα · A˙ ) = α · E (196) 8m2 8m2 8m2  ie  ie O, α · E = [α · p, α · E] 8m2 8m2 ie X  ∂Ej  e = αiαj −i + Σ · E × p (197) 8m2 ∂xi 4m2 i,j e ie e = (∇ · E) + Σ · (∇ × E) + Σ · E × p 8m2 8m2 4m2 So, the effective Hamiltonian to the desired order is  (p − eA)2 p4  e H000 = β m + − + eΦ − βΣ · B 2m 8m3 2m ie e e − Σ · (∇ × E) − Σ · E × p − (∇ · E) (198) 8m2 4m2 8m2 28 The individual terms have a direct physical interpretation. The first term in the parentheses is the expansion of

p(p − eA)2 + m2 (199) and −p4/(8m3) is the leading relativistic corrections to the kinetic energy. The two terms ie e − Σ · (∇ × E) − Σ · E × p (200) 8m2 4m2 together are the spin-orbit energy. In a spherically symmetric static potential, they take a very familiar form. In this case ∇ × E = 0, 1 ∂Φ 1 ∂Φ Σ · E × p = − Σ · r × p = − Σ · L, (201) r ∂r r ∂r and this term reduces to e 1 ∂Φ H = Σ · L. (202) spin−orbit 4m2 r ∂r The last term is known as the Darwin term. In a coulomb potential of a nucleus with charge Z|e|, it takes the form

e e Ze2 Zαπ − (∇ · E) = − Z|e|δ3(r) = δ3(r) = δ3(r), (203) 8m2 8m2 8m2 2m2 so it can only affect the S (l = 0) states whose wavefunctions are nonzero at the origin. For a Hydrogen-like (single electron) atom,

Ze2 eΦ = − , A = 0. (204) 4πr The shifts in of various states due to these correction terms can be com- puted by taking the expectation values of these terms with the corresponding wavefunctions. Darwin term (only for S (l = 0) states):

  4 4 Zαπ 3 Zαπ 2 Z α m ψns δ (r) ψns = |ψns(0)| = . (205) 2m2 2m2 2n3

Spin-orbit term (nonzero only for l 6= 0):

 Zα 1  Z4α4m [j(j + 1) − l(l + 1) − s(s + 1)] 2 3 σ · r × p = 3 1 . (206) 4m r 4n l(l + 1)(l + 2 )

29 Relativistic corrections:  p4  Z4α4m 3 n  − 3 = 4 − 1 . (207) 8m 2n 4 l + 2 We find Z4α4m 3  ∆E(l = 0) = − n (208) 2n4 4 1 = ∆E(l = 1, j = ), (209) 2 so 2S1/2 and 2P1/2 remain degenerate at this level. They are split by (2S1/2 > 2P1/2) which can be calculated after you learn radiative corrections in QED. The 2P1/2 and 2P3/2 are split by the spin-orbit interaction (fine structure) which you should have seen before.

3 1 Z4α4m ∆E(l = 1, j = ) − ∆E(l = 1, j = ) = (210) 2 2 4n3

9 Klein Paradox and the Hole Theory

So far we have ignored the negative solutions. However, the negative energy solu- tions are required together with the positive energy solutions to form a complete set. If we try to localize an electron by forming a , the wavefunc- tion will be composed of some negative energy components. There will be more negative energy components if the electron is more localized by the relation ∆x∆p ∼ ~. The negative energy components can not be ignored if the electron is localized to comparable to its Compton ~/mc, and we will encounter many paradoxes and dilemmas. An example is the Klein paradox described below. In order to localize , we must introduce strong external confining them to the desired region. Let’s consider a simplified situation that we want to confine a free electron of energy E to the region z < 0 by a one-dimensional step- function potential of height V as shown in Fig. 1. Now in the z < 0 half space there is an incident positive energy plan wave of momentum k > 0 along the z axis,  1 

ikz  0  ψinc(z) = e  k  , (spin-up). (211)  E+m  0

30 Figure 1: Electrostatic potential idealized with a sharp boundary, with an incident free electron wave moving to the right in region I.

The reflected wave in z < 0 region has the form

 1   0 

−ikz  0  −ikz  1  ψref (z) = a e  −k  + b e   , (212)  E+m   0  k 0 E+m and the transmitted wave in the z > 0 region (in the presence of the constant potential V ) has a similar form

 1   0 

iqz  0  iqz  1  ψtrans(z) = c e  q  + d e   , (213)  E−V +m   0  −q 0 E−V +m with an effective momentum q of

q = p(E − V )2 − m2. (214)

The total wavefunction is

ψ(z) = θ(−z)[ψinc(z) + ψref (z)] + θ(z)ψtrans(z). (215)

Requiring the continuity of ψ(z) at z = 0, ψinc(0) + ψref (0) = ψtrans(0), we obtain

1 + a = c (216) b = d (217) k q (1 − a) = c (218) E + m E − V + m k −q b = d (219) E + m E − V + m

31 From these equations we can see

b = d = 0 (no spin-flip) (220) 1 + a = c (221) q E + m 1 − a = rc where r = (222) k E − V + m 2 1 − r ⇒ c = , a = . (223) 1 + r 1 + r As long as |E − V | < m, q is imaginary and the transmitted wave decays exponen- tially. However, when V ≥ E +m the transmitted wave becomes oscillatory again. The probability currents j = ψ†αψ = ψ†α3ψzˆ, for the incident, transmitted, and reflected are k j = 2 , inc E + M q j = 2c2 , trans E − V + m k j = 2a2 . (224) ref E + m we find

jtrans 2 4r = c r = 2 (< 0 for V ≥ E + m), jinc (1 + r) j 1 − r2 ref = a2 = (> 1 for V ≥ E + m). (225) jinc 1 + r

Although the conservation of the looks satisfied: jinc = jtrans + jref , but we get the paradox that the reflected flux is larger than the incident one! There is also a problem of violation of the single particle theory which you can read in Prof. Gunion’s notes, p.14–p.15. Hole Theory In spite of the success of the Dirac equation, we must face the difficulties from the negative energy solutions. By their very they require a massive reinterpretation of the Dirac theory in order to prevent atomic electrons from making radiative transitions into negative-energy states. The transition rate for an electron in the of a hydrogen atom to fall into a negative-energy state may be calculated by applying semi-classical theory. The rate for the electron to make a transition into the energy interval −mc2 to −2mc2 is 2α6 mc2 ∼ ' 108sec−1 (226) π ~ and it blows up if all the negative-energy states are included, which clearly makes no .

32 A solution was proposed by Dirac as early as 1930 in terms of a many-particle theory. (This shall not be the final standpoint as it does not apply to scalar particle, for instance.) He assumed that all negative energy levels are filled up in the state. According to the Pauli exclusion , this prevents any electron from falling into these negative energy states, and thereby insures the stability of positive energy physical states. In turn, an electron of the negative energy sea may be excited to a positive energy state. It then leaves a hole in the sea. This hole in the negative energy, negatively charged states appears as a positive energy positively —the positron. Besides the properties of the positron, its charge |e| = −e > 0 and its rest mass me, this theory also predicts new phenomena: —The annihilation of an electron-positron pair. A positive energy electron falls into a hole in the negative energy sea with the emission of radiation. From energy momentum conservation at least two are emitted, unless a nucleus is to absorb energy and momentum. —Conversely, an electron-positron pair may be created from the vacuum by an incident beam in the presence of a target to balance energy and momentum. This is the process mentioned above: a hole is created while the excited electron acquires a positive energy. Thus the theory predicts the existence of positrons which were in fact observed in 1932. Since positrons and electrons may annihilate, we must abandon the interpretation of the Dirac equation as a . Also, the reason for discarding the Klein-Gordon equation no longer hold and it actually describes spin-0 particles, such as pions. However, the hole interpretation is not satisfactory for , since there is no Pauli exclusion principle for bosons. Even for fermions, the concept of an infinitely charged sea looks rather queer. We have instead to construct a true many-body theory to accom- modate particles and antiparticles in a consistent way. This is achieved in the quantum theory of fields which will be the subject of the rest of this course.

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