Quantum Theory 2015/16 4. Relativistic Quantum Theory 4.1 Quick Review of Special Relativity Four-Vector Notation: The coordinates of an object or `event' in four-dimensional space- time, Minkowski space, form a contravariant four-vector whose components have `upper' indices: xµ ≡ (x0; x1; x2; x3) ≡ (ct; x) Similarly, we define a covariant four-vector whose components have `lower' indices: xµ ≡ (x0; x1; x2; x3) ≡ (ct; −x) A general four-vector aµ is defined in the same way: aµ ≡ (a0; a1; a2; a3) ≡ (a0; a) 0 aµ ≡ (a0; a1; a2; a3) ≡ (a ; −a) 0 i so that a = a0 and a = −ai; i = 1; 2; 3. Upper and lower indices are related by the metric tensor gµν: µ µν ν a = g aν aµ = gµν a where 0 1 0 0 0 1 B 0 −1 0 0 C µν B C g = gµν = B C @ 0 0 −1 0 A 0 0 0 −1 and we use the Einstein summation convention where there is an implicit sum over the repeated index: ν = 0; 1; 2; 3. The scalar product in Minkowski space is defined, for general 4-vectors aµ and bµ by µ µ µν µ ν a · b ≡ a bµ = aµb = aµbνg = a b gµν = a0b0 − a · b where a and b are ordinary 3-vectors. NB we do not underline 4-vectors; every pair of repeated indices is implicitly summed over and each pair consists of one upper & one lower index. An expression with two identical upper (or lower) indices (eg aµbµ ) is simply wrong! Lorentz transformations: Lorentz transformations are linear transformations on the components of 4-vectors which leave invariant this scalar product: 0 µ µ ν 0 µ µ ν a = Λ ν a eg x = Λ ν x Strictly, these are homogeneous Lorentz transformations { translations are not included. 1 The `standard' Lorentz transformation is a `boost' along the x direction 0 cosh ! − sinh ! 0 0 1 B − sinh ! cosh ! 0 0 C µ B C Λ ν = B C @ 0 0 1 0 A 0 0 0 1 where the `rapidity' ! satisfies tanh ! ≡ β ≡ v=c cosh ! ≡ γ = (1 − β2)−1=2 = (1 − (v=c)2)−1=2 sinh ! = γ β Hence ct0 = γ (ct − (v=c)x) and x0 = γ (x − vt) as usual, relating the time and space co- ordinates of a given event in two inertial frames in relative motion: S S' v Boost x x' Differential operators @ 1 @ ! @ ≡ = ; r µ @xµ c @t @ 1 @ ! @µ ≡ = ; −∇ @xµ c @t 1 @2 d'Alembertian: @ @µ = @µ @ = @2 = − r2 ( = ) µ µ c2 @t2 (NB sometimes is called 2, so we will almost always use @2.) Momentum and energy: The conserved 4-momentum is denoted by: E pµ ≡ ; p c E2 p2 = − p · p = m2c2 for a free particle c2 or E2 = jpj2c2 + m2c4 where m is the mass of the particle. We shall follow the historical development of relativistic quantum mechanics, beginning with the standard simple-minded approach to making the Schr¨odinger equation consistent with special relativity. 2 4.2 The Klein-Gordon equation Recall that the Schr¨odingerequation for a free particle @ ( h¯2 ) ih¯ (r; t) = − r2 (r; t) @t 2m can be obtained from the (non-relativistic) classical total energy jpj2 E = H = 2m by means of the operator substitution prescriptions @ E ! ih¯ and p ! −ih¯r @t The relativistic expression for the total energy of a free particle is E2 = jpj2c2 + m2c4 Schr¨odinger(& Klein, Gordon, & Fock) suggested this as a starting point, thus obtaining @2 −h¯2 φ(r; t) = − h¯2 c2r2φ(r; t) + m2c4φ(r; t) (1) @t2 which is the Klein Gordon (KG) equation for the wavefunction φ(r; t) of a free relativistic particle.1 We can write this in a manifestly covariant form as m2c2 ! @2 + φ(x) = 0 h¯2 where x is the four-vector (ct; x1; x2; x3), so the operator prescription in covariant form is (1 @ ) @ pµ ! p^µ = ih¯ ; −∇ = ih¯ = ih@¯ µ c @t @xµ For a massless particle, m = 0, the KG equation reduces to the classical wave equation @2φ = 0. Free particle solutions: By substitution into the KG equation (1) we see it has plane- wave solutions φ(r; t) = expfik · r − i! tg provided that !, k & m are related by h¯2!2 =h ¯2c2jkj2 + m2c4 n o1=2 Taking the square-root, we get:h! ¯ = ± h¯2c2jkj2 + m2c4 . Such solutions are readily seen to be eigenfunctions of the momentum and energy operators, with eigenvalues p ≡ hk¯ and E ≡ h!¯ , respectively. 1The KG equation was first written down by Schr¨odingerbut, due to the problems we will discover below, he discarded it in favour of the non-relativistic equation that bears his name. 3 Thus, if we interpreth! ¯ as an allowed total energy of the free particle solution, there is an ambiguity in the sign of the total energy: there are both +ve and −ve energy solutions, and these have energy q E = ± jpj2c2 + m2c4 The positive-energy eigenvalues are in agreement with the classical relativistic relation be- tween energy, mass, and momentum, but what should we make of particles with negative total energy? ! If we define the four-vector kµ ≡ ( ; k) we can write the solution in covariant form c µ µ φ(x) ≡ exp(−ik · x) ≡ exp(−ik xµ) ≡ exp(−ip xµ=h¯) and thus interpret the four-momentum as pµ =hk ¯ µ. Continuity equation and probability interpretation Denote the Schr¨odingerequation by (SE) and its complex-conjugate by (SE)∗. Considering ∗ (SE) − (SE)∗ ; @ gives a continuity equation ρ + r · j = 0 @t ih¯ where ρ = ∗ and j = − ( ∗r − r ∗) 2m are the probability density and probability current density, respectively. (Integrate over any volume, and use the divergence theorem to see why { tutorial.) If we repeat this for the Klein Gordon equation, we obtain the quantities ih¯ @φ @φ∗ ! ih¯ ρ = φ∗ − φ and j = − (φ∗rφ − φrφ∗) 2mc2 @t @t 2m Note: 1. j is identical in form to the non-relativistic Schr¨odinger current density (we have chosen to normalise j so that this is the case.). 2. ρ can be shown to reduce to φ∗φ in the non-relativistic limit. 3. The candidate for the probability density, ρ(x), is no longer positive definite { negative energy solutions have ρ < 0 (exercise). Therefore there is no obvious probability- density interpretation. Summary: The Klein Gordon (KG) equation is the simplest relativistically-covariant gen- eralisation of the Schr¨odingerequation. Its solutions have the usual desirable properties for the description of a relativistic quantum particle, but they also describe particles of negative total energy, together with negative probabilities for finding them! Considering the positive energy solutions only, the KG equation with a Coulomb potential can be solved exactly for the energy levels of the hydrogen atom. The non-relativistic expansion reproduces exactly the relativistic kinetic energy correction ∆EKE obtained in time-independent perturbation theory, but it doesn't account for the spin-orbit correction or the Darwin term, so something else is required. 4 4.3 The Dirac Equation Dirac tried to avoid the twin difficulties of negative energy and negative probability by @ proposing a relativistic wave equation which, like the Schr¨odingerequation, is linear in @t , hoping to avoid the sign ambiguity in the square-root of E2, and also the presence of time derivatives in the `probability density'. Relativity then dictates that the equation should also be linear in the spatial derivatives in order to treat space and time on an equal footing. Following Dirac, we start with a Hamiltonian equation of the form @ ih¯ (r; t) = H^ (r; t) @t and write @ ( @ @ @ ) ih¯ (r; t) = −ihc¯ α1 + α2 + α3 (r; t) + β mc2 (r; t) (2) @t @x1 @x2 @x3 n o = c α · p^ + β mc2 (r; t) = H^ (r; t) (3) 3 ! i i i @ i i X i i where α · p^ ≡ α p^ = − ihα¯ i with α p^ ≡ α p^ etc. @x i=1 Initially, we attempt to construct an equation for a free particle, so no terms in the Hamilto- nian H^ should depend on r or t as these would describe forces. By assumption, the quantities αi and β are independent of derivatives, therefore αi and β commute with r, t,p ^ and E but not necessarily with each other. Since relativistic invariance must be maintained, ie E2 = jpj2c2 + m2c4, Dirac demanded that H^ 2 (r; t) = c2 jp^j2 + m2c4 (r; t) (4) From equation (3) we have n o n o H^ 2 (r; t) = c α · p^ + β mc2 c α · p^ + β mc2 (r; t) Expand the RHS of this equation, being careful about the ordering of the, as yet undeter- mined, quantities αi and β H^ 2 (r; t) h i = c2 (α1)2 (^p1)2 + (α2)2 (^p2)2 + (α3)2 (^p3)2 + m2c4 β2 (r; t) + c2 α1α2 + α2α1 p^1p^2 + α2α3 + α3α2 p^2p^3 + α3α1 + α1α3 p^1p^3 (r; t) + mc3 α1β + βα1 p^1 + α2β + βα2 p^2 + α3β + βα3 p^3 (r; t) Condition (4) is satisfied if (α1)2 = (α2)2 = (α3)2 = β2 = 1 αi αj + αj αi = 0 (i 6= j) αi β + β αi = 0 5 or, more compactly, n o αi; αj = 2 δij the anticommutator of αi and αj n o αi; β = 0 ; β2 = 1 1. From these relations it's clear that the αi and β cannot be ordinary numbers. If we assume they're matrices, then since H^ is hermitian, αi and β must also be hermitian (and therefore square) n × n matrices.