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5. Quantizing the Dirac Field

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5. Quantizing the Dirac Field 5. Quantizing the Dirac Field 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 scalar 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 Spin-Statistics Theorem We start in the usual way and define the momentum, @ 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 ), β(~y )] = [ ↵† (~x ), β† (~y )] = 0 (3) [ (~x ), † (~y )] = δ δ (~x ~y )(5.3) ↵ β ↵ − 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 waves, we may write the quantum 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 particles associated to the spinors 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 annihilation and creation operators –106– Claim: The field commutation relations (5.3)areequivalentto r s 3 rs (3) [b ,b †]=(2⇡) δ δ (p~ ~q ) p~ q~ − r s 3 rs (3) [c ,c †]= (2⇡) δ δ (p~ ~q )(5.5) p~ q~ − − with all other commutators vanishing. Note the strange minus sign in the [c, c†]term.Thismeansthatwecan’tdefinethe r ground state 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~ E~q 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~ )γ e · − + v (p~ )¯v (p~ )γ 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)γ e · − +(p/ m)γ e− · − (2⇡)3 2E − Z p~ 3 d p 1 0 i 0 0 i 0 +ip~ (~x ~y ) = (p γ + p γ + m)γ +(p γ p γ m)γ 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 · − = δ (~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 piγ terms cancel. This was the reason we needed the minus sign in the [c, c†] commutator 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 = ⇡ ˙ = ¯( iγi@ + 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 operator. R Let’s firstly look at 3 i d p 1 s i s +ip~ ~x s i s ip~ ~x · † − · ( iγ @i + m) = 3 bp~ ( γ pi + m)u (p~ ) e + cp~ (γ 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 ( γip + m)us(p~ )=γ0p us(p~ )and(γip + m)vs(p~ )= γ0p 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 ( iγ @ + m) = γ 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 †γ ( iγ @ + m) − i Z 3 3 3 d xd pd q Ep~ r r iq~ ~x r r +i~q ~x † † − · † · = 6 b~q u (~q ) e + cq~ v (~q ) e (2⇡) s4E~q · 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 δ 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⇡) δ (0) (5.12) (2⇡)3 p~ p~ p~ − p~ p~ Z ⇣ ⌘ (3) The δ term is familiar and easily dealt with by normal 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 space, 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 Quantization The key piece of physics that we missed is that spin 1/2particlesarefermions,meaning that they obey Fermi-Dirac statistics with the quantum state 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 integer spin fields must be quantized as bosons, while half-integer spin fields must be quantized as fermions. 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 fermion? 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 spinor fields satisfy (~x ), (~y ) = † (~x ), † (~y ) =0 { ↵ β } { ↵ β } (3) (~x ), † (~y ) = δ δ (~x ~y )(5.14) { ↵ β } ↵ − 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⇡) δ δ (p~ ~q ) { p~ q~ } − r s 3 rs (3) c ,c † =(2⇡) δ δ (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).
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