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5 the Dirac Equation and Spinors 5 The Dirac Equation and Spinors In this section we develop the appropriate wavefunctions for fundamental fermions and bosons. 5.1 Notation Review The three dimension differential operator is : ∂ ∂ ∂ = , , (5.1) ∂x ∂y ∂z We can generalise this to four dimensions ∂µ: 1 ∂ ∂ ∂ ∂ ∂ = , , , (5.2) µ c ∂t ∂x ∂y ∂z 5.2 The Schr¨odinger Equation First consider a classical non-relativistic particle of mass m in a potential U. The energy-momentum relationship is: p2 E = + U (5.3) 2m we can substitute the differential operators: ∂ Eˆ i pˆ i (5.4) → ∂t →− to obtain the non-relativistic Schr¨odinger Equation (with = 1): ∂ψ 1 i = 2 + U ψ (5.5) ∂t −2m For U = 0, the free particle solutions are: iEt ψ(x, t) e− ψ(x) (5.6) ∝ and the probability density ρ and current j are given by: 2 i ρ = ψ(x) j = ψ∗ ψ ψ ψ∗ (5.7) | | −2m − with conservation of probability giving the continuity equation: ∂ρ + j =0, (5.8) ∂t · Or in Covariant notation: µ µ ∂µj = 0 with j =(ρ,j) (5.9) The Schr¨odinger equation is 1st order in ∂/∂t but second order in ∂/∂x. However, as we are going to be dealing with relativistic particles, space and time should be treated equally. 25 5.3 The Klein-Gordon Equation For a relativistic particle the energy-momentum relationship is: p p = p pµ = E2 p 2 = m2 (5.10) · µ − | | Substituting the equation (5.4), leads to the relativistic Klein-Gordon equation: ∂2 + 2 ψ = m2ψ (5.11) −∂t2 The free particle solutions are plane waves: ip x i(Et p x) ψ e− · = e− − · (5.12) ∝ The Klein-Gordon equation successfully describes spin 0 particles in relativistic quan- tum field theory. There are problems with the interpretation of the positive and negative energy solutions of the Klein-Gordon equation, E = p2 + m2, since the negative energy solutions have negative probability densities ρ.± 5.4 The Dirac Equation The problems with the Klein-Gordon equation led Dirac to search for an alternative relativistic wave equation in 1928, in which the time and space derivatives are first order. The Dirac equation can be thought of in terms of a “square root” of the Klein-Gordon equation. In covariant form it is written: ∂ iγ0 + iγ m ψ =0 (iγµ∂ m) ψ = 0 (5.13) ∂t · − µ − where we have introduced the coefficients γµ =(γ0,γ )=(γ0, γ1, γ2, γ3), which have to be determined. As we will see in equation (5.20), the Dirac equation is simply four coupled differential equations, describing a wavefunction ψ with four components. 5.5 The Gamma Matrices To find what the γµ, µ =0, 1, 2, 3 objects are, we first multiply the Dirac equation by its conjugate equation: 0 ∂ 0 ∂ ψ† iγ iγ m iγ + iγ m ψ = 0 (5.14) − ∂t − · − ∂t · − and demand that this be consistent with the Klein-Gordon equation, (5.11). This leads to the following conditions on the γµ: (γ0)2 =1, (γi)2 = 1 γµγν + γνγµ =0forµ = ν − with i =1, 2, 3, µ, ν =0, 1, 2, 3 (5.15) 26 Equivalently in terms of anticommutation relations and the metric tensor (equation (3.3)): γµ, γν = γµ, γν + γν, γµ =2gµν µ, ν =0, 1, 2, 3 (5.16) { } The simplest solution for the γµ, that satisfies these anticommutation relations, are 4 4 unitary matrices. We will use the following representation for the γ matrices: × I0 0 σi γ0 = γi = (5.17) 0 I σi 0 − − where I denotes a 2 2 identity matrix, 0 denotes a 2 2 null matrix, and the σi are the Pauli spin matrices× : × 01 0 i 10 σ = σ = σ = (5.18) x 10 y i −0 z 0 1 − Let’s write out the gamma matrices in full: 10 0 0 0001 01 0 0 0010 γ0 = γ1 = 00 10 0 100 − − 10 0 1 1000 − − 000 i 0010 00i −0 000 1 γ2 = γ3 = (5.19) 0 i 00 100− 0 − i 00 0 0100 − Please note, despite the µ superscript, the γµ are not four vectors. However they do remain constant under Lorentz transforms. Finally let’s write out the Dirac Equation in full: i ∂ m 0 i ∂ i ∂ + ∂ 1 ∂t − ∂z ∂x ∂y ψ 0 0 i ∂ mi∂ ∂ i ∂ 2 ∂t ∂x ∂y ∂z ψ 0 ∂ ∂ − ∂ ∂ − − 3 = (5.20) i ∂z i ∂x ∂y i ∂t m 0 ψ 0 −∂ ∂ − ∂− − − ∂ 4 i + i 0 i m ψ 0 − ∂x ∂y ∂z − ∂t − 5.6 Spinors The Dirac equation describes the behaviour of spin-1/2 fermions in relativistic quantum field theory. For a free fermion the wavefunction is the product of a plane wave and a Dirac spinor, u(pµ): µ µ ip x ψ(x )=u(p )e− · (5.21) Substituting the fermion wavefunction, ψ, into the Dirac equation: (γµp m)u(p) = 0 (5.22) µ − 27 For a particle at rest, p = 0, we find the following equations: ∂ mI 0 iγ0 m ψ = γ0E m ψ =0 Euˆ = u (5.23) ∂t − − 0 mI − The solutions are four eigenspinors: 1 0 0 0 0 1 0 0 u1 = u2 = u3 = u4 = (5.24) 0 0 1 0 0 0 0 1 and the associated wavefunctions of the fermion is: 1 imt 1 2 imt 2 3 +imt 3 4 +imt 4 ψ = e− u ψ = e− u ψ = e u ψ = e u (5.25) Note that the spinors are 1 4 column matrices, and that there are four possible states. The spinors are, however,× not four-vectors: the four components do not represent t, x, y, z. The four components are a suprise: we would expect only two spin states for a spin-1/2 fermion! Note also the change of sign in the exponents of the plane waves in the states ψ3 and ψ4. The four solutions in equations (5.24) and (5.25) describe two different spin states ( and ) with E = m, and two spin states with E = m. ↑ ↓ − 5.7 Negative Energy Solutions & Antimatter To describe the negative energy states, Dirac postulated that an electron in a positive energy state is produced from the vacuum accompanied by a hole with negative energy. The hole corresponds to a physical antiparticle, the positron, with charge +e. Another interpretation (Feynman-St¨uckelberg) is that the E = m solutions can either describe a negative energy particle which propagates backwards− in time, or a positive energy antiparticle propagating forward in time: i[( E)( t) ( p) ( x)] i[Et p x] e− − − − − · − = e− − · (5.26) 5.8 Spinors for Moving Particles For a moving particle, p = 0 the Dirac equation becomes (using (5.13) and (5.17)): µ E m σ p uA (γ pµ m) uA uB = − − · = 0 (5.27) − σ p E m uB · − − where u and u denote the 1 2 upper and lower components of u respectively. The A B × equations for uA and uB are coupled: σ p σ p u = · u u = · u (5.28) A E m B B E + m A − 28 1 0 The solutions are obtained by successively setting: u = , u = , u = A 0 A 1 B 1 0 , and u = , to give: 0 B 1 1 0 0 1 u1 = u2 = p /(E + m) (p ip )/(E + m) z x − y (p + ip )/(E + m) p /(E + m) x y − z pz/( E + m) ( px + ipy)/( E + m) ( p− ip−)/( E + m) − p /( E +−m) u3 = x y u4 = z (5.29) − − 1 − −0 0 1 The u1 and u2 solutions describe an electron of energy E =+ m2 + p 2, and momen- tum p: 1 µ ip x 2 µ ip x ψ = u (p )e− · ψ = u (p )e− · (5.30) The u3 and u4 of equation (5.29) describe a positron of energy E = m2 + p 2, and momentum p. − It is usual to change to the spinors v2(p) u3( p) and v1(p) u4( p) to describe these positive energy antiparticle states, E =+≡ −m2 + p 2 ≡ − pz/(E + m) 2 µ 3 µ (px + ipy)/(E + m) 2 µ ip x 3 µ i( p) x v (p ) u ( p )= ψ = v (p )e− · = u ( p )e − · ≡ − 1 − 0 (p ip )/(E + m) x − y 1 µ 4 µ pz/(E + m) 1 µ ip x 4 µ i( p) x v (p ) u ( p )= − ψ = v (p )e− · = u ( p )e − · ≡ − 0 − 1 (5.31) The u and v are the solutions of: (iγµp m)u =0 (iγµp + m)v = 0 (5.32) µ − µ 5.9 Spin and Helicity The two different solutions for each of the fermions and antifermions corresponds to two 1 µ ip x possible spin states. For a fermion with momentum p along the z-axis, ψ = u (p )e− · 2 µ ip x describes a spin-up fermion and ψ = u (p )e− · describes a spin-down fermion. For 1 µ ip x an antifermion with momentum p along the z-axis, ψ = v (p )e− · describes a spin-up 2 µ ip x antifermion and ψ = v (p )e− · describes a spin-down antifermion. 29 Figure 5.1: Helicity eigenstates for a particle or antiparticle travelling along the +z axis. 1 2 1 2 The u ,u ,v ,v spinors are only eigenstates of Sˆz for momentum p along the z-axis. There’s nothing special about projecting out the component of spin along the z-axis, that’s just the conventional choice. For our purposes it makes more sense to project the spin along the particle’s direction of flight, this defines the helicity, h of the particle.
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