The Dirac Equation May 18, 2015


 Colin Nelson
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1 The Dirac Equation May 8, 25
2 . Introduction In nonrelativistic quantum mechanics, wave functions are descibed by the timedependent Schrodinger equation : 2m 2 ψ + V ψ = i () This is really just energy conservation ( kinetic energy ( p2 ) plus potential energy (V) equals total 2m energy (E)) with the momentum and energy terms replaced by their operator equivalents p i ; E i (2) In relativistic quantum theory, the energymomentum conservation equation is E 2 p 2 = m 2 (note that we are working in the standard particle physics units where h = c = ). Proceeding with the same replacements, we can derive the KleinGordon equation : In covariant notation this is E 2 p 2 m 2 2 ψ ψ m 2 = (3) µ µ ψ m 2 ψ = (4) Suppose ψ is solution to the KleinGordon equation. Multiplying it by iψ we get iψ 2 ψ 2 iψ 2 ψ + iψ m 2 = (5) Taking the complex conjugate of the KleinGordon equation and multiplying by iψ we get iψ 2 ψ 2 iψ 2 ψ + iψm 2 = (6) If we subtract the second from the first we obtain [ ( i ψ )] ψ + [ i (ψ ψ ψ ψ )] = (7) This has the form of an equation of continuity with a probability density defined by ρ + j = (8) ( ρ = i ψ ) ψ (9) and a probability density current defined by j = i (ψ ψ ψ ψ ) () Now, suppose a solution to the KleinGordon equation is a free particle with energy E and momentum p ψ = Ne ipµxµ ()
3 Substitution of this solution into the equation for the probability density yields ρ = 2E N 2 (2) The probability density is proportional to the energy of the particle. Now, why is this a problem? If you substitute the free particle solution into the KleinGordon equation you get, unsurpisingly, the relation E 2 p 2 = m 2 (3) so the energy of the particle could be E = ± p 2 + m 2 (4) The fact that you have negative energy solutions is not that much of a problem. What is a problem is that the probability density is proportional to E. This implies a possibility for negative probability densities...which don t exist. This (and some others) problem drove Dirac to think about another equation of motion. His starting point was to try to factorise the energy momentum relation. In covariant formalism where p µ is the 4momentum : (E, p x, p y, p z ). Dirac tried to write E 2 p 2 m 2 p µ p µ m 2 (5) p µ p µ m 2 = (β κ p κ + m)(γ λ p λ m) (6) where κ and λ range from to 3. This notation looks a bit confusing. It may help if we write both sides out longhand. On the lefthand side we have On the righthand side we have p µ p µ m 2 = p 2 p p = p 2 p 2 p 2 2 p 2 3 m 2 (7) (β κ p κ + m)(γ λ p λ m) = (β p β p β 2 p 2 β 3 p 3 + m)(γ p γ p γ 2 p 2 γ 3 p 3 m) (8) Expanding the righthand side (β κ p κ + m)(γ λ p λ m) = β κ γ λ p κ p λ m 2 + mγ λ p λ mβ κ p κ (9) This should be equal to p µ p µ m 2, so we need to get rid of the terms that are linear in p. We can do this just be choosing β κ = γ κ. Then p µ p µ m 2 = γ κ γ λ p κ p λ m 2 (2) Now, since κ, λ runs from to 3, the right hand side can be fully expressed by γ κ γ λ p κ p λ m 2 = (γ ) 2 p 2 + (γ ) 2 p 2 + (γ 2 ) 2 p (γ 3 ) 2 p 2 3 (2) + (γ γ + γ γ )p p + (γ γ 2 + γ 2 γ )p p 2 (22) + (γ γ 3 + γ 3 γ )p p 3 + (γ γ 2 + γ 2 γ )p p 2 (23) + (γ γ 3 + γ 3 γ )p p 3 + (γ 2 γ 3 + γ 3 γ 2 )p 2 p 3 m 2 (24) 2
4 This must be equal to p µ p µ m 2 = p 2 p 2 p 2 2 p 2 3 m 2 (25) One could make the squared terms equal by choosing (γ ) 2 = and (γ ) 2 = (γ 2 ) 2 = (γ 3 ) 2 =, but this would not eliminate the crossterms. Dirac realised that what you needed was something which anticommuted : i.e. γ µ γ ν + γ ν γ µ = if µ ν. He realised that the γ factors must be 4x4 matrices, not just numbers, which satisfied the anticommutation relation where {γ µ, γ ν } = 2g µν (26) g µν = (27) We also define, for later use, another γ matrix called γ 5 = iγ γ γ 2 γ 3. This has the properties (you should check for yourself) that (γ 5 ) 2 =, {γ 5, γ µ } = (28) We will want to take the Hermitian conjugate of these matrices at various times. The Hermitian conjugate of each matrix is γ = γ γ 5 = γ 5 γ µ = γ γ µ γ = γ for µ (29) If the matrices satisfy this algebra, then you can factor the energy momentum conservation equation p µ p µ m 2 = (γ κ p κ + m)(γ λ p λ m) = (3) The Dirac equation is one of the two factors, and is conventionally taken to be Making the standard substitution, p µ equation γ λ p λ m = (3) i µ we then have the usual covariant form of the Dirac (iγ µ µ m)ψ = (32) where µ = (,,, ), m is the particle mass and the γ matrices are a set of 4dimensional x y z matrices. Since these are matrices, ψ is a 4element column matrix called a bispinor. This bispinor is not a 4vector and doesn t transform like one..2 The BjorkenDrell convention Any set of four 4x4 matrices that obey the algebra above will work. The one we normally use includes the Pauli spin matrices. Recall that each component of the spin operator S for spin /2 particles is defined by its own Pauli spin matrix : S x = 2 σ = 2 ( ) S y = 2 σ 2 = 2 3 ( ) i i S z = 2 σ 3 = 2 ( ) (33)
5 In terms of the Pauli spin matrices, the γ matrices have the form ( ) ( ) σ γ = γ µ µ = σ µ γ 5 = ( ) Each element of the matrices in Equations 34 are 2x2 matrices. denotes the 2 x 2 unit matrix, and denotes the 2 x 2 null matrix..3 The Probability Density and Current In order to understand the probability density and probability flow we will want to derive an equation of continuity for the probability. The first step is to write the Dirac equation out longhand : iγ + iγ x + iγ2 y We want to take the Hermitian conjugate of this : [iγ + iγ x + iγ2 y + iγ3 z (34) mψ = (35) + iγ3 z mψ] (36) Now, we must remember that the γ are matrices and that ψ is a 4component column vector. Thus [γ ] = [ 2 3 ] (37) 4 2 = 3 (38) 4 ( ) = (39) = γ (4) Using the Hermitian conjugation properties of the γ matrices defined in the previous section we can write Equation 36 as i γ i x γ i y γ2 i z γ3 mψ (4) which, as γ µ = γ µ for µ means we can write this as i γ i x ( γ ) i y ( γ2 ) i z ( γ3 ) mψ (42) 4
6 We have a problem now  this is no longer covariant. That is, I would like to write this in the form ψ ( i µ γ µ m) where the derivative µ now operates to the left. I can t because the minus signs on the spatial components coming from the γ µ spoils the scalar product. To deal with this we have to multiply the equation on the right by γ, since we can flip the sign of the γ matrices using the relationship γ µ γ = γ γ µ. Thus or i γ γ i x ( γ γ ) i y ( γ2 γ ) i z ( γ3 γ ) mψ γ (43) i γ γ i x (γ γ ) i y (γ γ 2 ) i z (γ γ 3 ) mψ γ (44) Defining the adjoint spinor ψ = ψ γ, we can rewrite this equation as and finally we get the adjoint Dirac equation i γ i x (γ ) i y (γ2 ) i z (γ3 ) mψ (45) Now, we multiply the Dirac equation on the left by ψ : and the adjoint Dirac equation on the right by ψ : and we add them together, which eliminates the two mass terms or, more succinctly,. If we now identify the bit in the brackets as the 4current : ψ(i µ γ µ + m) = (46) ψ(iγ µ µ m)ψ = (47) ψ(i µ γ µ + m)ψ = (48) ψ(γ µ µ ψ) + (ψ µ γ µ )ψ = (49) µ (ψγ µ ψ) = (5) j µ = (ρ, j) (5) where ρ is the probability density and j is the probability current, we can write the conservation equation as µ j µ = with j µ = ψγ µ ψ (52) which is the covariant form for an equation of continuity. The probability density is then just ρ = ψγ ψ = ψ γ γ ψ = ψ ψ and the probability 3current is j = ψ γ γ µ ψ. This current is the same one which appears in the Feynman diagrams. It is called a Vector current, and is the current responsible for the electromagnetic interaction. 5
7 ψ i e ψ f e q γ e e φ i φ f For the interaction in Figure.3, with two electromagnetic interactions, the matrix element is then.4 Examples M = e2 4π [ψ fγ µ ψ i ] q 2 [φ fγ ν φ i ] (53) Now let s look at some solutions to the Dirac Equation. The first one we will look at is for a particle at rest..4. Particle at rest Suppose the fermion wavefunction is a plane wave. We can write this as ψ(p) = e ixµpµ u(p) (54) where p µ = (E, p x, p y, p z ) and x µ = (t, x, y, z) and so ix µ p µ = i(et x p). This is just the phase for an oscillating plane wave. For a particle at rest, the momentum term disappears : ix µ p µ = i(et). Furthur, since the momentum is zero, the spatial derivatives must be zero : reads x,y,z =. The Dirac equation therefore or which gives us iγ mψ = (55) iγ ( ie)u(p) mu(p) = (56) Eγ u(p) = mu(p) (57) 6
8 Now, u(p) is a 4component bispinor, so properly expanding the gamma matrix in Equation 57, we get u E = m u 2 u 3 (58) u 4 This is an eigenvalue equation. There are four independent solutions. Two with energy E = m and two with E = m. The solutions are just u = u 2 = u 3 = u 4 = (59) with eigenvalues m, m, m and m respectively. The first two solutions can be interpreted as positive energy particle solutions with spin up and spin down. One can see this be checking if the function is an eigenfunction of the spin matrix : S z. S z u = 2 = 2 u (6) and similarly for u 2. Note that we have yet to normalise these solutions. We won t either for the purposes of this discussion. What about these negative energy solutions? We are required to keep them (we can t just say that they are unphysical and throw them away) since quantum mechanics requires a complete set of states. What happens if we allow the particle to have momentum?.4.2 Particle in Motion Again, let s look for a plane wave solution ψ(p) = e ixµpµ u(p) (6) Plugging this into the Dirac equation means that we can replace the µ by p µ and remove the oscillatory term to obtain the momentumspace version of the Dirac Equation (γ µ p µ m)u(p) = (62) Notice that this is now purely algebraic and can be easily(!) solved γ µ p µ m = Eγ p x γ p y γ 2 p z γ 3 m (63) ( ) ( ) ( ) σ = E p m (64) σ ( ) (E m) σ p = (65) σ p (E + m) 7
9 where each element in this 2x2 matrix is actually a 2x2 matrix itself. In this spirit, let s write the 4component bispinor solution as 2component vector ( ) ua u = (66) u B then the Dirac Equation implies that (γ µ p µ m)u(p) = = leading two coupled simultaneous equations ( ) ( ) (E m) σ p ua = σ p (E + m) u B ( ) (67) (σ p)u B = (E m)u A (68) (σ p)u A = (E + m)u B (69) Now, let s expand that (σ p) : (σ p) = = and right back to the Dirac Equation ( ) ( ) p x + p i y + ( ) pz p x ip y p x + ip y p z ( ) (7) (7) (σ p)u B = (E m)u A (72) (σ p)u A = (E + m)u B (73) gives u B = = (σ p) E + m u A (74) ( ) pz p x ip y u E + m p x + ip y p A (75) z Now, we just need to make choices for the form of u A. Let s make the obvious choice and remember that u A is a 2component spinor so we need to specify two solutions: ( ) ( ) u A = or u A = (76) These give u = N p z p x+ip y and u 2 = N 2 p x ip y p z (77) where N and N 2 are normalisation factors we won t go into. Note that, if p = these correspond to the E > solutions we found for a particle at rest, which is good. 8
10 Repeating this for u B = ( ) and u B = u 3 = N 3 p z p x+ip y ( ) which gives solutions u 3 and u 4 p x ip y and u p z 4 = N 4 (78) and collecting them together u = N p z p x+ip y u 2 = N 2 p x ip y p z u 3 = N 3 p z p x+ip y u 4 = N 4 p x ip y p z (79) All of which, if put back into the Dirac Equation, yields : E 2 = p 2 + m 2 as you might expect. Comparing with the p = case we can identify u and u 2 as positive energy solutions with energy E = p 2 + m 2, and u 3, u 4 as negative energy solutions with energy E = p 2 + m 2. How do we interpret these negative energy solutions? The conventional interpretation is called the FeynmanStuckelberg interpretation : A negative energy solution represents a negative energy particle travelling backwards in time, or equivalently, a positive energy antiparticle going forwards in time. With this interpretation we tend to rewrite the negative energy solutions to represent positive antiparticles. Starting from the E < solutions p z p x ip y p x+ip y u 3 = N 3 and u p z 4 = N 4 (8) Here we implicitly assume that E is negative. We define antiparticle states by just flipping the sign of E and p following the FeynmanStuckelberg convention. v (E, p)e i(et x p) = u 4 ( E, p)e i(et x p) (8) v 2 (E, p)e i(et x p) = u 3 ( E, p)e i(et x p) (82) where E = p 2 + m 2. Note that v is associated with u 4 and v 2 is associated with u 3. We do this because the spin matrix, S antiparticle, for antiparticles is equal to S particle i.e. the spin flips for antiparticles as well and the spin eigenvalue for v = u 4 is spinup and v 2 = u 3 is spindown. In general u, u 2, v and v 2 are not eigenstates of the spin operator (Check for yourself). In fact we should expect this since solutions to the Dirac Equation are, by definition, eigenstates of the Hamiltonian operator, Ĥ, and the S z does not commute with the Hamiltonian : [Ĥ, Ŝz], and hence we can t find solutions which are simultaneously solutions to S z and Ĥ. However if the zaxis is aligned with particle direction : p x = p y =, p z = ± p then we have the following Dirac states u = N ± p u 2 = N p v = N 9 p ± p v 2 = N (83)
11 These are eigenstates of S z S z u = + 2 u S z u 2 = 2 u 2 S anti z v = S z v = + 2 v (84) S anti z v 2 = S z v 2 = 2 v 2 (85) So we have found solutions of the Dirac Equation which are also spin eigenstates...but only if the particle is travelling along the zaxis..4.3 Charge Conjugation Classical electrodynamics is invariant under a change in the sign of the electric charge. In particle physics, this concept is represented by the charge conjugation operator that flips the signs of all the charges. It changes a particle into an antiparticle, and vice versa: Ĉ p >= p > (86). Application of Ĉ twice brings us back to the the orignal state : Ĉ2 = and so eigenstates of Ĉ are ±. In general most particles are not eigenstates of Ĉ. If p > were an eigenstate of Ĉ then Ĉ p >= ± p >= p > (87) implying that only those particles which are their own antiparticles are eigenstates of Ĉ. This leaves us with photons and the neutral mesons like π, η and ρ. Ĉ changes a particle spinor into an antiparticle spinor. The relevant operation is ψ ψ C = Ĉγ ψ (88) with Ĉ = iγ2 γ. We can apply this to, say, the u solution to the Dirac Equation. if ψ = u e i(et x p), the ψ C = iγ 2 ψ = iγ 2 u e i(et x p) i iγ 2 u = i i i N i p z p x+ip y = N p x ip y p z = v (89) or, in full, Ĉ(u e i(et x p) ) = v e i(et x p). Hence charge conjugation changes the particle eigenspinors into their respective antiparticle spinors..4.4 Helicity The fact that we can find spin eigenvalues for states in which the particles are travelling along the spindirection indicates that the quantity we need is not spin but helicity. The helicity is defined as the projection of the spin along the direction of motion: ( ) σ ĥ = Σ ˆp = 2S ˆp = ˆp (9) σ
12 Figure : The two helicity states. On the left the spin vector is aligned in the direction of motions of the particle. This is a righthanded helicity state and has helicity +. On the right the spin vector is antiparallel to the particle momentum. This is a lefthanded helicity state with helicity . and has eigenvalues equal to + (called righthanded where the spin vector is aligned in the same direction as the momentum vector) or  (called lefthanded where the spin vector is aligned in the opposite direction as the momentum vector), corresponding to the diagrams in Figure. It can be shown that the helicity does commute with the Hamiltonian and so one can find eigenstates that are simultaneously states of helicity and the Hamiltonian. The problem, and it is a big problem, is that helicity is not Lorentz invariant in the case of a massive particle. If the particle is massive it is possible to find an inertial reference frame in which the particle is going in the opposite direction. This does not change the direction of the spin vector, so the helicity can change sign. The helicity is Lorentz invariant only in the case of massless particles..4.5 Chirality We d rather have operators which are Lorentz invariant, than commute with the Hamiltonian. In general wave functions in the Standard Model are eigenstates of a Lorentz invariant quantity called the chirality. The chirality operator is γ 5 and it does not commute with the Hamiltonian. Due to this, it is somewhat difficult to visualise. The best picture I can get comes from the definition : something is chiral if it s mirror image does not superimpose on itself. Think of your left hand. It s mirror image (from the point of view of someone in that mirror looking back at you) is actually a right hand. It and it s mirror image cannot be superimposed andit is therefore an intrinsically chiral object. In the limit that E >> m, or that the particle is massless, the chirality is identical to the helicity. For a massive particle this is no longer true. In general the eigenstates of the chirality operator are γ 5 u R = +u R γ 5 u L = u L γ 5 v R = v R γ 5 v L = +v L (9), where we define u R and u L are right and lefthanded chiral states. We can define the projection operators P L = 2 ( γ5 ) P R = 2 ( + γ5 ) (92) such that P L projects outs the lefthanded chiral particle states and righthanded chiral antiparticle states. P R projects out the righthanded chiral particle states and lefthanded chiral antiparticle states. The projection operators have the following properties : P 2 L,R = P L,R ; P R P L = P L P R = ; P R + P L = (93)
13 Any spinor can be written in terms of it s left and righthanded chiral states: ψ = (P R + P L )ψ = P R ψ + P L ψ = ψ R + ψ L (94) Chirality holds an important place in the standard model. Let s take a standard model probability current ψγ µ φ (95) We can decompose this into it s chiral states ψγ µ φ = (ψ L + ψ R )γ µ (φ L + φ R ) = ψ L γ µ φ L + ψ L γ µ φ R + ψ R γ µ φ L + ψ R γ µ φ R Now, ψ L = ψ L γ = ψ 2 ( γ5 )γ = ψ γ 2 ( + γ5 ) = ψ 2 ( + γ5 ) = ψp R where I have used the fact that γ 5 = γ 5 and γ 5 γ = γ γ 5. Similarly ψ R = ψp L. Using this, it is easy to show that the terms ψ L γ µ φ R and ψ R γ µ φ L are equal to : ψ L γ µ φ R = ψp R γ µ P R φ (96) = ψ 2 ( + γ5 )γ µ 2 ( + γ5 )φ (97) = ψ 4 ( + γ5 )(γ µ + γ µ γ 5 )φ (98) = ψ 4 ( + γ5 )(γ µ γ 5 γ µ )φ (99) = ψ 4 ( + γ5 )( γ 5 )γ µ φ () = ψ 4 ( + γ5 γ 5 (γ 5 ) 2 )γ µ φ () = (2) since (γ 5 ) 2 = and γ µ γ 5 = γ 5 γ µ. Similarly for the other cross term, ψ R γ µ φ L. Hence, ψγ µ φ = ψ L γ µ φ L + ψ R γ µ φ R (3) So lefthanded chiral particles couple only to lefthanded chiral fields, and righthanded chiral fields couple to righthanded chiral fields. One must be very careful with how one interprets this statement. What it does not say is that there are lefthanded chiral electrons and righthanded chiral electrons which are distinct particles. Useful though it is when describing particle interactions, chirality is not conserved in the propagation of a free particle. In fact the chiral states φ L and φ R do not even satisfy the Dirac equation. Since chirality is not a good quantum number it can evolve with time. A massive particle starting off as a completely left handed chiral state will soon evolve a righthanded chiral component. By contrast, helicity is a conserved quantity during free particle propagation. Only in the case of massless particles, for which helicity and chirality are identical and are conserved in freeparticle propagation, can leftand righthanded particles be considered distinct. For neutrinos this mostly holds. 2
14 .5 What you should know Understand the covariant form of the Dirac equation, and know how to manipulate the γ matrices. Know where the definition of a current comes from. Understand the difference between helicity and chirality. Be able to manipulate gamma matrices..6 Futhur reading Griffiths Chapter 7, Sections 7., 7.2 and 7.3 Griffiths Chapter, Section.7 3
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