sign in sign up
<<
Home , Dirac equation, Dirac fermion, Dirac operator, Dirac spinor, Momentum operator, Propagator

5. Quantizing the Dirac

We would now like to quantize the Dirac Lagrangian,

= ¯(x) i @/ m (x)(5.1) L We will proceed naively and treat as we did the field. But we’ll see that things go wrong and we will have to reconsider how to quantize this theory.

5.1 A Glimpse at the -Statistics Theorem We start in the usual way and define the ,

@ 0 ⇡ = L = i ¯ = i † (5.2) @ ˙

For the Dirac Lagrangian, the momentum conjugate to is i †.Itdoesnotinvolve the time derivative of .Thisisasitshouldbeforanequationofmotionthatisfirst order in time, rather than second order. This is because we need only specify and

† on an initial time slice to determine the full evolution.

To quantize the theory, we promote the field and its momentum † to operators, satisfying the canonical commutation relations, which read

[ ↵(~x ), (~y )] = [ ↵† (~x ), † (~y )] = 0 (3) [ (~x ), † (~y )] = (~x ~y )(5.3) ↵ ↵ It’s this step that we’ll soon have to reconsider.

Since we’re dealing with a free theory, where any classical solution is a sum of plane , we may write the operators as

2 3 d p 1 s s +ip~ ~x s s ip~ ~x (~x )= b u (p~ )e · + c † v (p~ )e · (2⇡)3 p~ p~ s=1 2Ep~ X Z h i 2 3 d p p1 s s ip~ ~x s s +ip~ ~x †(~x )= b † u (p~ )†e · + c v (p~ )†e · (5.4) (2⇡)3 p~ p~ s=1 2Ep~ X Z h i s p s s where the operators bp~ † create associated to the u (p~ ), while cp~ † create particles associated to vs(p~ ). As with the scalars, the commutation relations of the fields imply commutation relations for the and creation operators

–106– Claim: The field commutation relations (5.3)areequivalentto

r s 3 rs (3) [b ,b †]=(2⇡) (p~ ~q ) p~ q~ r s 3 rs (3) [c ,c †]= (2⇡) (p~ ~q )(5.5) p~ q~ with all other vanishing.

Note the strange minus sign in the [c, c†]term.Thismeansthatwecan’tdefinethe r 0 as something annihilated by cp~ 0 =0,becausethentheexcitedstates s | i | i cp~ † 0 would have negative norm. To avoid this, we will have to flip the interpretation | i s r of c and c†,withthevacuumdefinedbyc † 0 =0andtheexcitedstatesbyc 0 . p~ | i p~ | i This, as we will see, will be our undoing.

Proof: Let’s show that the [b, b†]and[c, c†]commutatorsreproducethefieldcom- mutators (5.3),

3 3 d pd q 1 r s r s i(~x p~ ~y q~ ) [ (~x ), †(~y )] = [b ,b †]u (p~ )u (~q )†e · · (2⇡)6 p~ q~ r,s 4Ep~ Eq~ X Z ⇣ p r s r s i(~x p~ ~y q~ ) +[cp~ †,cq~ ]v (p~ )v (~q )†e · · 3 ⌘ d p 1 s s 0 ip~ (~x ~y ) s s 0 ip~ (~x ~y ) = u (p~ )¯u (p~ ) e · + v (p~ )¯v (p~ ) e · (5.6) (2⇡)3 2E s Z p~ X At this stage we use the outer product formulae (4.128)and(4.129)whichtellus us(p~ )¯us(p~ )= p/+ m and vs(p~ )¯vs(p~ )= p/ m,sothat s s 3 P d p 1P 0 ip~ (~x ~y ) 0 ip~ (~x ~y ) [ (~x ), †(~y )] = ( p/+ m) e · +(p/ m) e · (2⇡)3 2E Z p~ 3 d p 1 0 i 0 0 i 0 +ip~ (~x ~y ) = (p + p + m) +(p p m) e · (2⇡)3 2E 0 i 0 i Z p~ where, in the second term, we’ve changed p~ p~ under the integration sign. Now, ! using p0 = Ep~ we have

3 d p +ip~ (~x ~y ) (3) [ (~x ), †(~y )] = e · = (~x ~y )(5.7) (2⇡)3 Z as promised. Notice that it’s a little tricky in the middle there, making sure that i the pi terms cancel. This was the reason we needed the minus sign in the [c, c†] terms in (5.5). ⇤

–107– 5.1.1 The Hamiltonian To proceed, let’s construct the Hamiltonian for the theory. Using the momentum

⇡ = i †,wehave

= ⇡ ˙ = ¯( ii@ + m) (5.8) H L i which means that H = d3x agrees with the conserved energy computed using H Noether’s theorem (4.92). We now wish to turn the Hamiltonian into an . R Let’s firstly look at

3 i d p 1 s i s +ip~ ~x s i s ip~ ~x · † · ( i @i + m) = 3 bp~ ( pi + m)u (p~ ) e + cp~ ( pi + m)v (p~ ) e (2⇡) 2Ep~ Z h i where, for once we’ve left thep sum over s =1, 2 implicit. There’s a small subtlety with the minus signs in deriving this equation that arises from the use of the Minkowski metric in contracting indices, so that p~ ~x xipi = xip . Now we use the defining · ⌘ i i equations for the spinors us(p~ )andvs(p~ )givenin(4.105)and(4.111), to replace P ( ip + m)us(p~ )=0p us(p~ )and(ip + m)vs(p~ )= 0p vs(p~ )(5.9) i 0 i 0 so we can write

3 i d p Ep~ 0 s s +ip~ ~x s s ip~ ~x ( i @ + m) = b u (p~ ) e · c † v (p~ ) e · (5.10) i (2⇡)3 2 p~ p~ Z r h i We now use this to write the operator Hamiltonian

3 0 i H = d x † ( i @ + m) i Z 3 3 3 d xd pd q Ep~ r r iq~ ~x r r +iq~ ~x † † · † · = 6 bq~ u (~q ) e + cq~ v (~q ) e (2⇡) s4Eq~ · Z h i s s +ip~ ~x s s ip~ ~x b u (p~ )e · c † v (p~ ) e · p~ p~ 3 h i d p 1 r s r s r s r s = b †b [u (p~ )† u (p~ )] c c †[v (p~ )† v (p~ )] (2⇡)3 2 p~ p~ · p~ p~ · Z h r s r s r s r s bp~ †c †p~ [u (p~ )† v ( p~ )] + cp~ b p~ [v (p~ )† u ( p~ )] · · i where, in the last two terms we have relabelled p~ p~ .Wenowuseourinnerproduct ! formulae (4.122), (4.124)and(4.127)whichread

r s r s rs r s r s u (p~ )† u (p~ )=v (p~ )† v (p~ )=2p and u (p~ )† v ( p~ )=v (p~ )† u ( p~ )=0 · · 0 · ·

–108– giving us

3 d p s s s s H = E b †b c c † (5.11) (2⇡)3 p~ p~ p~ p~ p~ Z 3 ⇣ ⌘ d p s s s s 3 (3) = E b †b c †c +(2⇡) (0) (5.12) (2⇡)3 p~ p~ p~ p~ p~ Z ⇣ ⌘ (3) The term is familiar and easily dealt with by ordering. The b†b term is familiar and we can check that b† create positive energy states as expected,

s s [H, bp~ †]=Ep~ bp~ †

The minus sign in front of the c†c term should make us nervous. If we think of c† as creation operators then there’s no problem since, using the commutation relation (5.5), we still find that c† creates positive energy states,

s s [H, cp~ †]=Ep~ cp~ †

However, as we noted after (5.5), these states have negative norm. To have a sensible Hilbert , we need to interpret c as the creation operator. But then the Hamiltonian is not bounded below because

[H, cs]= E cs p~ p~ p~ This is a disaster. Taken seriously it would tell us that we could tumble to states of lower and lower energy by continually producing c particles. As the English would say, it’s all gone a bit Pete Tong. (No relation).

Since the above calculation was a little tricky, you might think that it’s possible to rescue the theory to get the minus signs to work out right. You can play around with di↵erent things, but you’ll always find this minus sign cropping up somewhere. And, in fact, it’s telling us something important that we missed.

5.2 Fermionic The key piece of physics that we missed is that spin 1/2particlesarefermions,meaning that they obey Fermi-Dirac statistics with the picking up a minus sign upon the interchange of any two particles. This fact is embedded into the structure of relativistic quantum field theory: the spin-statistics theorem says that spin fields must be quantized as , while half-integer spin fields must be quantized as . Any attempt to do otherwise will lead to an inconsistency, such as the unbounded Hamiltonian we saw in (5.12).

–109– So how do we go about quantizing a field as a ? Recall that when we quantized the scalar field, the resulting particles obeyed bosonic statistics because the creation and annihilation operators satisfied the commutation relations,

[a†,a†]=0 a†a† 0 p,~ ~q = ~q,p~ (5.13) p~ q~ ) p~ q~ | i⌘| i | i To have states obeying fermionic statistics, we need anti-commutation relations, A, B { }⌘ AB + BA.Ratherthan(5.3), we will ask that the fields satisfy

(~x ), (~y ) = † (~x ), † (~y ) =0 { ↵ } { ↵ } (3) (~x ), † (~y ) = (~x ~y )(5.14) { ↵ } ↵

We still have the expansion (5.4)of and † in terms of b, b†,c and c†.Butnowthe same proof that led us to (5.5)tellsusthat

r s 3 rs (3) b ,b † =(2⇡) (p~ ~q ) { p~ q~ } r s 3 rs (3) c ,c † =(2⇡) (p~ ~q )(5.15) { p~ q~ } with all other anti-commutators vanishing,

r s r s r s r s b ,b = c ,c = b ,c † = b ,c = ...=0 (5.16) { p~ q~ } { p~ q~ } { p~ q~ } { p~ q~ } The calculation of the Hamiltonian proceeds as before, all the way through to the penultimate line (5.11). At that stage, we get

3 d p s s s s H = E b †b c c † (2⇡)3 p~ p~ p~ p~ p~ Z 3 h i d p s s s s 3 (3) = E b †b + c †c (2⇡) (0) (5.17) (2⇡)3 p~ p~ p~ p~ p~ Z h i The anti-commutators have saved us from the indignity of an unbounded Hamiltonian. Note that when normal ordering the Hamiltonian we now throw away a negative contri- bution (2⇡)3(3)(0). In principle, this could partially cancel the positive contribution from bosonic fields. problem anyone?!

5.2.1 Fermi-Dirac Statistics Just as in the bosonic case, we define the 0 to satisfy, | i bs 0 = cs 0 =0 (5.18) p~ | i p~ | i

–110– Although b and c obey anti-commutation relations, the Hamiltonian (5.17)hasnice commutation relations with them. You can check that

r r r r [H, b ]= E b and [H, b †]=E b † p~ p~ p~ p~ p~ p~ r r r r [H, c ]= E c and [H, c †]=E c † (5.19) p~ p~ p~ p~ p~ p~ This means that we can again construct a tower of energy eigenstates by acting on the r r, vacuum by bp~ † and cp~ † to create particles and , just as in the bosonic case. For example, we have the one- states

r p~ , r = b † 0 (5.20) | i p~ | i The two particle states now satisfy

r1 r2 p~ 1,r1; p~ 2,r2 b † b † 0 = p~ 2,r2; p~ 1,r1 (5.21) | i⌘ p~ 1 p~ 2 | i | i confirming that the particles do indeed obey Fermi-Dirac statistics. In particular, we have the Pauli-Exclusion principle p~ , r ; p~ , r = 0. Finally, if we wanted to be sure | i about the spin of the particle, we could act with the operator (4.96)toconfirmthatastationaryparticle p~ =0,r does indeed carry intrinsic angular | i momentum 1/2 as expected.

5.3 Dirac’s Hole Interpretation “In this attempt, the success seems to have been on the side of Dirac rather than logic” Pauli on Dirac

Let’s pause our discussion to make a small historical detour. Dirac originally viewed his equation as a relativistic version of the Schr¨odinger equation, with interpreted as the wavefunction for a single particle with spin. To reinforce this interpretation, he wrote (i @/ m) =0as @ i = i~↵ ~ + m Hˆ (5.22) @t · r ⌘ where ~↵ = 0~ and = 0. Here the operator Hˆ is interpreted as the one-particle Hamiltonian. This is a very di↵erent viewpoint from the one we now have, where is a classical field that should be quantized. In Dirac’s view, the Hamiltonian of the system is Hˆ defined above, while for us the Hamiltonian is the field operator (5.17). Let’s see where Dirac’s viewpoint leads.

–111– With the interpretation of as a single-particle wavefunction, the plane- solu- tions (4.104) and (4.110)totheDiracequationarethoughtofasenergyeigenstates, with

ip x @ = u(p~ ) e · i = E ) @t p~ +ip x @ = v(p~ ) e · i = E (5.23) ) @t p~ which look like positive and solutions. The spectrum is once again unbounded below; there are states v(p~ )witharbitrarilylowenergy E .Atfirst p~ glance this is disastrous, just like the unbounded field theory Hamiltonian (5.12). Dirac postulated an ingenious solution to this problem: since the are fermions (a fact which is put in by hand to Dirac’s theory) they obey the Pauli-exclusion principle. So we could simply stipulate that in the true vacuum of the , all the negative energy states are filled. Only the positive energy states are accessible. These filled negative energy states are referred to as the . Although you might worry about the infinite negative charge of the vacuum, Dirac argued that only charge di↵erences would be (a trick reminiscent of the normal ordering prescription we used for field operators).

Having avoided disaster by floating on an infinite sea comprised of occupied negative energy states, Dirac realized that his theory made a shocking prediction. Suppose that anegativeenergystateisexcitedtoapositiveenergystate,leavingbehindahole. The hole would have all the properties of the , except it would carry positive charge. After flirting with the idea that it may be the proton, Dirac finally concluded that the hole is a new particle: the . Moreover, when a positron comes across an electron, the two can annihilate. Dirac had predicted anti-matter, one of the greatest achievements of . It took only a couple of years before the positron was discovered experimentally in 1932.

Although Dirac’s physical insight led him to the right answer, we now understand that the interpretation of the as a single-particle wavefunction is not really correct. For example, Dirac’s argument for anti-matter relies crucially on the particles being fermions while, as we have seen already in this course, anti-particles exist for both fermions and bosons. What we really learn from Dirac’s analysis is that there is no consistent way to interpret the as describing a single particle. It is instead to be thought of as a classical field which has only positive energy solutions because the Hamiltonian (4.92)ispositivedefinite.Quantizationofthisfieldthengives rise to both particle and anti-particle excitations.

–112– This from :

“Until now, everyone thought that the Dirac equation referred directly to physical particles. Now, in field theory, we recognize that the equations refer to a sublevel. Experimentally we are concerned with particles, yet the old equations describe fields.... When you begin with field equations, you operate on a level where the particles are not there from the start. It is when you solve the field equations that you see the emergence of particles.”

5.4 Let’s now move to the . We define the spinors (~x , t )ateverypoint in such that they satisfy the operator equation @ = i[H, ](5.24) @t We solve this by the expansion

2 3 d p 1 s s ip x s s +ip x (x)= b u (p~ )e · + c †v (p~ )e · (2⇡)3 p~ p~ s=1 2Ep~ X Z h i 2 3 d p p1 s s +ip x s s ip x †(x)= b †u (p~ )†e · + c v (p~ )†e · (5.25) (2⇡)3 p~ p~ s=1 2Ep~ X Z h i Let’s now look at the anti-commutatorsp of these fields. We define the fermionic prop- agator to be

iS = (x), ¯ (y) (5.26) ↵ { ↵ } In what follows we will often drop the indices and simply write iS(x y)= (x), ¯(y) , { } but you should remember that S(x y)isa4 4 . Inserting the expansion (5.25), ⇥ we have

3 3 d pd q 1 s r s r i(p x q y) † · · iS(x y)= 6 bp~ ,bq~ u (p~ )¯u (~q )e (2⇡) 4Ep~ Eq~ { } Z h s r s r +i(p x q y) p + c †,c v (p~ )¯v (~q )e · · { p~ q~ } 3 i d p 1 s s ip (x y) s s +ip (x y) = u (p~ )¯u (p~ )e · + v (p~ )¯v (p~ )e · (2⇡)3 2E Z p~ 3 ⇥ ⇤ d p 1 ip (x y) +ip (x y) = (p/+ m)e · +(p/ m)e · (5.27) (2⇡)3 2E Z p~ ⇥ ⇤

–113– where to reach the final line we have used the outer product formulae (4.128) and (4.129). We can then write iS(x y)=(i@/ + m)(D(x y) D(y x)) (5.28) x in terms of the for a real scalar field D(x y)which,recall,canbewritten as (2.90)

3 d p 1 ip (x y) D(x y)= e · (5.29) (2⇡)3 2E Z p~ Some comments: For spacelike separated points (x y)2 < 0, we have already seen that D(x y) • D(y x)=0.Inthebosonictheory,wemadeabigdealofthissinceitensured that [(x),(y)] = 0 (x y)2 < 0(5.30) outside the lightcone, which we trumpeted as proof that our theory was causal. However, for fermions we now have (x), (y) =0 (x y)2 < 0(5.31) { ↵ } outside the lightcone. What happened to our precious ? The best that we can say is that all our are bilinear in fermions, for example the Hamiltonian (5.17). These still commute outside the lightcone. The theory re- mains causal as long as fermionic operators are not observable. If you think this is alittleweak,rememberthatnoonehaseverseenaphysicalmeasuringapparatus come back to minus itself when you rotate by 360 degrees!

At least away from singularities, the propagator satisfies • (i@/ m)S(x y)=0 (5.32) x which follows from the fact that (@/2 + m2)D(x y)=0usingthemassshell x condition p2 = m2.

5.5 The Feynman Propagator By a similar calculation to that above, we can determine the , 3 d p 1 ip (x y) ¯ / · 0 ↵(x) (y) 0 = 3 (p + m)↵ e h | | i (2⇡) 2Ep~ Z 3 d p 1 +ip (x y) 0 ¯ (y) (x) 0 = (p/ m) e · (5.33) h | ↵ | i (2⇡)3 2E ↵ Z p~

–114– We now define the Feynman propagator S (x y), which is again a 4 4matrix,as F ⇥ the time ordered product, 0 (x) ¯(y) 0 x0 >y0 SF (x y)= 0 T (x) ¯(y) 0 h | | i (5.34) h | | i⌘( 0 ¯(y) (x) 0 y0 >x0 h | | i Notice the minus sign! It is necessary for Lorentz invariance. When (x y)2 < 0, there is no invariant way to determine whether x0 >y0 or y0 >x0.Inthiscasetheminussignis necessary to make the two definitions agree since (x), ¯(y) =0outsidethelightcone. { } We have the 4-momentum representation for the Feynman propagator, 4 d p ip (x y) p + m S (x y)=i e · · (5.35) F (2⇡)4 p2 m2 + i✏ Z which satisfies (i@/ m)S (x y)=i(4)(x y), so that S is a Green’s for x F F the .

The minus sign that we see in (5.34)alsooccursforanystringofoperatorsinside atimeorderedproductT (...). While bosonic operators commute inside T ,fermionic operators anti-commute. We have this same behaviour for normal ordered products as well, with fermionic operators obeying : := : :. With the understanding 1 2 2 1 that all fermionic operators anti-commute inside T and ::, Wick’s theorem proceeds just as in the bosonic case. We define the contraction

(x) ¯(y)=T ( (x) ¯(y)) : (x) ¯(y):=S (x y)(5.36) F 5.6 Yukawa Theoryz }| { The interaction between a of m and a real scalar field of mass µ is governed by the Yukawa theory, = 1 @ @µ 1 µ22 + ¯(iµ@ m) ¯ (5.37) L 2 µ 2 µ which is the proper version of the baby scalar Yukawa theory we looked at in Section 3. Couplings of this type appear in the , between fermions and the Higgs . In that context, the fermions can be (such as the electron) or .

Yukawa originally proposed an interaction of this type as an e↵ective theory of nuclear . With an eye to this, we will again refer to the particles as , and the particles as . Except, this time, the nucleons have spin. (This is still not a particularly realistic theory of interactions, not least because we’re omitting isospin. Moreover, in Nature the relevant mesons are which are pseudoscalars, so acouplingoftheform ¯5 would be more appropriate. We’ll turn to this briefly in Section 5.7.3).

–115– Note the of the various fields. We still have []=1,butthekinetic terms require that [ ]=3/2. Thus, unlike in the case with only scalars, the coupling is dimensionless: []=0.

We’ll proceed as we did in Section 3, firstly computing the amplitude of a particular process then, with that calculation as a guide, writing down the Feynman rules for the theory. We start with:

5.6.1 An Example: Putting Spin on Nucleon Scattering Let’s study scattering. This is the same calculation we performed in Section ! (3.3.3)exceptnowthefermionshavespin.Ourinitialandfinalstatesare

s r i = 4E E b † b † 0 p~ , s ; ~q , r | i p~ q~ p~ q~ | i⌘| i s0 r0 f = p4Ep~ Eq~ b † b † 0 p~ 0,s0; ~q 0,r0 (5.38) | i 0 0 p~ 0 q~ 0 | i⌘| i p We need to be a little cautious about minus signs, because the b†’s now anti-commute. In particular, we should be careful when we take the adjoint. We have

r0 s0 f = 4Ep~ Eq~ 0 b b (5.39) h | 0 0 h | q~ 0 p~ 0 p We want to calculate the order 2 terms from the S-matrix element f S 1 i . h | | i ( i)2 d4x d4x T ¯(x ) (x )(x ) ¯(x ) (x )(x ) (5.40) 2 1 2 1 1 1 2 2 2 Z where, as usual, all fields are in the . Just as in the bosonic calcula- tion, the contribution to nucleon scattering comes from the contraction

: ¯(x1) (x1) ¯(x2) (x2): (x1)(x2)(5.41) z }| { We just have to be careful about how the spinor indices are contracted. Let’s start by looking at how the fermionic operators act on i .Weexpandoutthe fields, | i leaving the ¯ fields alone for now. We may ignore the c† pieces in since they give no contribution at order 2.Wehave

3 3 s r d k1 d k2 m n : ¯(x ) (x ) ¯(x ) (x ): b † b † 0 = [ ¯(x ) u (~k )] [ ¯(x ) u (~k )] 1 1 2 2 p~ q~ | i (2⇡)6 1 · 1 2 · 2 Z ik1 x1 ik2 x2 e · · m n s r b~ b~ b †b † 0 (5.42) 4E E k1 k2 p~ q~ | i ~k1 ~k2 p

–116– where we’ve used square brackets [ ] to show how the spinor indices are contracted. The · minus sign that sits out front came from moving (x1)past ¯(x2). Now anti-commuting the b’s past the b†’s, we get

1 r s ip x2 iq x1 = [ ¯(x1) u (~q )] [ ¯(x2) u (p~ )]e · · 2 Ep~ Eq~ · · s r ip x1 iq x2 [ ¯(x ) u (p~ )] [ ¯(x ) u (~q )]e · · 0 (5.43) p 1 · 2 · | i Note, in particular, the relative minus sign that appears between these two terms. Now let’s see what happens when we hit this with f .Welookat h | e+ip0 x1+iq0 x2 r0 s0 r s · · s0 r r0 s 0 b b [ ¯(x1) u (~q )] [ ¯(x2) u (p~ )] 0 = [¯u (p~ 0) u (~q )] [¯u (~q 0) u (p~ )] h | q~ 0 p~ 0 · · | i · · 2 Ep~ 0 Eq~ 0

+ip0 x2+iq0 x1 e · · r r s s p [¯u 0 (~q 0) u (~q )] [¯u 0 (p~ 0) u (p~ )] · · 2 Ep~ 0 Eq~ 0 p The [ ¯(x ) us(p~ )] [ ¯(x ) ur(~q )] term in (5.43)doublesupwiththis,cancellingthe 1 · 2 · factor of 1/2infrontof(5.40). Meanwhile, the 1/pE terms cancel the relativistic state normalization. Putting everything together, we have the following expression for f S 1 i h | | i

4 4 4 ik (x1 x2) 2 d x1d x2d k ie · s s r r +ix1 (q q)+ix2 (p p) ( i) [¯u 0(p~ 0) u (p~ )] [¯u 0(~q 0) u (~q )]e · 0 · 0 (2⇡)4 k2 µ2 + i✏ · · Z ⇣ s r r s ix1 (p q)+ix2 (q p) [¯u 0(p~ 0) u (~q )] [¯u 0 (~q 0) u (p~ )]e · 0 · 0 · · ⌘ where we’ve put the propagator back in. Performing the over x1 and x2, this becomes,

4 2 4 (2⇡) i( i) s s r r (4) (4) d k [¯u 0(p~ 0) u (p~ )] [¯u 0 (~q 0) u (~q )] (q0 q + k) (p0 p k) k2 µ2 + i✏ · · Z ⇣ s r r s (4) (4) [¯u 0(p~ 0) u (~q )] [¯u 0 (~q 0) u (p~ )] (p0 q + k) (q0 p k) · · ⌘ And we’re almost there! Finally, writing the S-matrix element in terms of the amplitude 4 (4) in the usual way, f S 1 i = i (2⇡) (p + q p0 q0), we have h | | i A s s r r s r r s [¯u 0(p~ 0) u (p~ )] [¯u 0 (~q 0) u (~q )] [¯u 0(p~ 0) u (~q )] [¯u 0 (~q 0) u (p~ )] =( i)2 · · · · A (p p)2 µ2 + i✏ (q p)2 µ2 + i✏ ✓ 0 0 ◆ which is our final answer for the amplitude.

–117– 5.7 Feynman Rules for Fermions It’s important to bear in mind that the calculation we just did kind of blows. Thankfully the Feynman rules will once again encapsulate the combinatoric complexities and make life easier for us. The rules to compute amplitudes are the following To each incoming fermion with momentum p and spin r,weassociateaspinor • ur(p~ ). For outgoing fermions we associateu ¯r(p~ ).

p p ur (p) u r(p)

Figure 21: An incoming fermion Figure 22: An outgoing fermion

To each incoming anti-fermion with momentum p and spin r,weassociateaspinor • v¯r(p~ ). For outgoing anti-fermions we associate vr(p~ ).

p p vr (p) v r(p)

Figure 23: An incoming anti-fermion Figure 24: An outgoing anti-fermion

Each vertex gets a factor of i. • Each internal line gets a factor of the relevant propagator. • i p for scalars p2 µ2 + i✏ p i( p/+ m) for fermions (5.44) p2 m2 + i✏ The arrows on the fermion lines must flow consistently through the diagram (this ensures fermion number conservation). Note that the fermionic propagator is a 4 4 matrix. The matrix indices are contracted at each vertex, either with further ⇥ propagators, or with external spinors u,¯u, v orv ¯.

Impose momentum conservation at each vertex, and integrate over undetermined • loop momenta.

Add extra minus signs for statistics. Some examples will be given below. •

–118– 5.7.1 Examples Let’s run through the same examples we did for the scalar Yukawa theory. Firstly, we have

Nucleon Scattering For the example we worked out previously, the two lowest order Feynman diagrams are shown in Figure 25. We’ve drawn the second with the legs crossed

p,s p,s p,s/ / p,s/ / + q,r/ / q,r/ / q,r q,r

Figure 25: The two Feynman diagrams for nucleon scattering to emphasize the fact that it picks up a minus sign due to statistics. (Note that the way the legs in the Feynman diagram doesn’t tell us the direction in which the particles leave the scattering event: the momentum label does that. The two diagrams above are di↵erent because the incoming legs are attached to di↵erent outgoing legs). Using the Feynman rules we can read o↵the amplitude. s s r r s r r s [¯u 0(p~ 0) u (p~ )] [¯u 0 (~q 0) u (~q )] [¯u 0(p~ 0) u (~q )] [¯u 0 (~q 0) u (p~ )] =( i)2 · · · · (5.45) A (p p )2 µ2 (p q )2 µ2 ✓ 0 0 ◆ The denominators in each term are due to the propagator, with the momen- tum determined by conservation at each vertex. This agrees with the amplitude we computed earlier using Wick’s theorem.

Nucleon to Meson Scattering Let’s now look at ¯ .ThetwolowestorderFeynmandiagramsareshownin ! Figure 26. Applying the Feynman rules, we have r µ s r µ s v¯ (~q )[ (pµ p0 )+m]u (p~ ) v¯ (~q )[ (pµ q0 )+m]u (p~ ) =( i)2 µ + µ A (p p )2 m2 (p q )2 m2 ✓ 0 0 ◆ Since the internal line is now a fermion, the propagator contains (p p0 )+m factors. µ µ µ This is a 4 4 matrix which sits on the top, sandwiched between the two external ⇥ spinors. Now the exchange statistics applies to the final meson states. These are bosons and, correspondingly, there is no relative minus sign between the two diagrams.

–119– p,s p,s p,s/ / p,s/ / + q,r/ / q,r/ / q,r q,r

Figure 26: The two Feynman diagrams for nucleon to meson scattering

Nucleon-Anti-Nucleon Scattering For ¯ ¯, the two lowest order Feynman diagrams are of two distinct types, just ! like in the bosonic case. They are shown in Figure 27. The corresponding amplitude is given by,

s s r r r s s r [¯u 0(p~ 0) u (p~ )] [¯v (~q ) v 0 (~q 0)] [¯v (~q ) u (p~ )] [¯u 0 (p~ 0) v 0 (~q 0)] =( i)2 · · + · · (5.46) A (p p )2 µ2 (p + q)2 µ2 + i✏ ✓ 0 ◆ As in the bosonic diagrams, there is again the di↵erence in the momentum dependence in the denominator. But now the di↵erence in the diagrams is also reflected in the spinor contractions in the numerator.

More subtle are the minus signs. The fermionic statistics mean that the first diagram has an extra minus sign relative to the scattering of Figure 25.Sincethisminussign will be important when we come to figure out whether the Yukawa is attractive or repulsive, let’s go back to basics and see where it comes from. The initial and final states for this scattering process are

s r i = 4E E b † c † 0 p~ , s ; ~q , r | i p~ q~ p~ q~ | i⌘| i s0 r0 f = p4Ep~ Eq~ b † c † 0 p~ 0,s0; ~q 0,r0 (5.47) | i 0 0 p~ 0 q~ 0 | i⌘| i

The ordering of b† and c† in thesep states is crucial and reflects the scattering ¯ ¯, ! as opposed to ¯ ¯ which would di↵er by a minus sign. The first diagram in ! Figure 27 comes from the term in the perturbative expansion, ¯ ¯ s r m ~ ¯ n ~ m n s r f : (x1) (x1) (x2) (x2): b † c † 0 f [¯v (k1) (x1)] [ (x2) u (k2)]c~ b~ b †c † 0 h | p~ q~ | i⇠h | · · k1 k2 p~ q~ | i 4 where we’ve neglected a bunch of objects in this equation like d ki and exponential factors because we only want to keep track of the minus signs. Moving the annihilation R operators past the creation operators, we have

+ f [¯vr(~q ) (x )] [ ¯(x ) us(p~ )] 0 (5.48) h | · 1 2 · | i

–120– p,s p,s / / p,s/ / p,s + q,r/ / q,r/ / q,r q,r

Figure 27: The two Feynman diagrams for nucleon-anti-nucleon scattering

Repeating the process by expanding out the (x1)and ¯(x2)fieldsandmovingthem to the left to annihilate f ,wehave h | r0 s0 m n r m ~ n ~ s r r0 s0 s 0 cq~ bp~ c~ †b~ † [¯v (~q ) v (l1)] [¯u (l2) u (p~ )] 0 [¯v (~q ) v (~q 0)] [¯u (p~ 0) u (p~ )] h | 0 0 l1 l2 · · | i⇠ · · m s where the minus sign has appeared from anti-commuting c † past b 0 . This is the ~l1 p~ 0 overall minus sign found in (5.46). One can also follow similar contractions to compute the second diagram in Figure 27.

Meson Scattering Finally, we can also compute the scattering of p ! 1 p / which, as in the bosonic case, picks up its leading contribution 1 at one-loop. The amplitude for the diagram shown in the figure

/ p2 is p2 4 4 d k k/ + m k/ + p/10 + m i = ( i) 4 Tr 2 2 2 2 A (2⇡) (k m + i✏) ((k + p0 ) m + i✏) Figure 28: Z 1 k/ + p/10 p/1 + m k/ p/20 + m 2 2 2 2 ⇥((k + p 0 p ) m + i✏) ((k p 0) m + i✏) 1 1 2 Notice that the high momentum limit of the integral is d4k/k4,whichisnolonger finite, but diverges logarithmically. You will have to wait until next term to make sense R of this integral.

There’s an overall minus sign sitting in front of this amplitude. This is a generic feature of diagrams with fermions running in loops: each fermionic loop in a diagram gives rise to an extra minus sign. We can see this rather simply in the diagram

–121– which involves the expression

¯↵(x) ↵(x) ¯(y) (y)= (y) ¯↵(x) ↵(x) ¯(y) z }| { z }| { = Trz (S}|(y {xz)S (}|x y{)) F F After passing the fermion fields through each other, a minus sign appears, sitting in front of the two propagators.

5.7.2 The Revisited We saw in Section 3.5.2,thattheexchangeofarealscalarparticlegivesrisetoa universally attractive Yukawa potential between two spin zero particles. Does the same hold for the spin 1/2particles?

Recall that the strategy to compute the potential is to take the non-relativistic limit of the scattering amplitude, and compare with the analogous result from quantum . Our new amplitude now also includes the spinor degrees of freedom u(p~ ) and v(p~ ). In the non-relativistic limit, p (m,p ~ ), and ! pp ⇠ ⇠ u(p~ )= · pm pp ⇠¯ ! ! ⇠ ! · pp ⇠ ⇠ v(p~ )= · pm (5.49) pp ⇠¯ ! ! ⇠ ! · In this limit, the spinor contractions in the amplitude for scattering (5.45) ! becomeu ¯s0 us =2mss0 and the amplitude is ·

p,s p,s/ / s0s r0r s0r r0s 2 = i( i) (2m) 2 2 (5.50) q,r/ / (p~ p~ 0)+µ (p~ ~q 0)+µ q,r ✓ ◆

The symbols tell us that spin is conserved in the non-relativistic limit, while the momentum dependence is the same as in the bosonic case, telling us that once again the particles feel an attractive Yukawa potential,

2e µr U(~r )= (5.51) 4⇡r Repeating the calculation for ¯ ¯,therearetwominussignswhichcanceleach ! other. The first is the extra overall minus sign in the scattering amplitude (5.46),

–122– due to the fermionic nature of the particles. The second minus sign comes from the non-relativistic limit of the spinor contraction for anti-particles in (5.46), which is v¯s0 vs = 2mss0 .Thesetwosignscancel,givingusonceagainanattractiveYukawa · potential (5.51).

5.7.3 Pseudo-Scalar Coupling Rather than the standard Yukawa coupling, we could instead consider

= ¯5 (5.52) LYuk This still preserves if is a pseudoscalar, i.e.

P : (~x , t ) ( ~x , t )(5.53) ! We can compute in this theory very simply: the Feynman rule for the interaction vertex is now changed to a factor of i5.Forexample,theFeynmandiagramsfor ! scattering are again given by Figure 25,withtheamplitudenow

[¯us0(p~ )5us(p~ )] [¯ur0 (~q )5ur(~q )] [¯us0(p~ )5ur(~q )] [¯ur0 (~q )5us(p~ )] =( i)2 0 0 0 0 A (p p )2 µ2 (p q )2 µ2 ✓ 0 0 ◆ We could again try to take the non-relativistic limit for this amplitude. But this time, things work a little di↵erently. Using the expressions for the spinors (5.49), we haveu ¯s0 5us 0inthenon-relativisticlimit.Tofindthenon-relativisticamplitude, ! s 5 s we must go to next to leading order. One can easily check thatu ¯ 0 (p~ 0) u (p~ ) s T s ! m⇠ 0 (p~ p~ 0) ~ ⇠ . So, in the non-relativistic limit, the leading order amplitude arising · from pseudoscalar exchange is given by a spin-spin coupling,

p,s p,s/ / s0 T s r0 T r 2 [⇠ (p~ p~ 0) ~ ⇠ ][⇠ (p~ p~ 0) ~ ⇠ ] +im( i) · 2 2 · (5.54) q,r/ / ! (p~ p~ 0) + µ q,r

–123–