Relativistic Quantum Mechanics The Klein -Gordon equation Interpretation of the Klein -Gordon equation The Dirac equation Dirac representation for the matrices α and β Covariant form of the Dirac equation Why Relativistic Quantum Mechanics? The Schrödinger equation: correctly describes the phenomena only if particle velocities are v 0 c . It is not invariant under a Lorentz change of the reference frame (required by the principle of relativity). Need: a relativistic generalization! 5 Free particle of spin zero The Klein -Gordon equation So, how to come up with such an equation? For the relativistic particle with rest mass m and momentum p, E= mc24 + p 22 c . Using the correspondence rule ∂ EEi→=Z; p →=−∇ p i Z op ∂t op ∂ 1/ 2 one can write: i ZΨ=() mc24 − Z 222 c ∇ Ψ . ∂t Problems: 1. It is not clear how to interpret the operator on right-hand side. If expanded in power series it lead to differential operator of infinite order. 2. The time and space coordinates do not appear in a symmetric way (no relativistic invariance ?). Free particle of spin zero The Klein -Gordon equation So we remove the square root and try again (there will be consequences of removing this square root!) E2= mc 24 + p 22 c ∂ EEi→=Z; p →=−∇ p i Z op ∂t op ∂2 Ψ 2 24 222 The Klein -Gordon −Z =()m c −∇Ψ Z c . ∂t 2 equation Notes: it is second-order differential equation with respect to the time unlike the Schrödinger equation. 6 Probabilistic interpretation of non- relativistic quantum mechanics 2 Pt(,)r=Ψ (,) r t =Ψ* (,)(,) rr tt Ψ Position probability density ∂ Probability is conserved: P(,)r t d r = 0 ∂t — ∂Ψ (r ,t ) Z 2 Using the Schrödinger equation iZ =− ∇+2 Vtt(,)r Ψ (,) r ∂t 2 m ∂P(r , t ) we obtain + ∇ j (,) r t = 0 , where j can be interpreted as ∂t probability current density Z j(,) r t =» Ψ*() ∇Ψ−∇Ψ() * Ψ ÿ . 2mi ⁄ Interpretation of the Klein -Gordon equation: Problem 1 We try to construct a position probability density P( r,t) and probability current density j(r,t) which satisfy the continuity equation: ∂P(r , t ) + ∇j( r ,t ) = 0. ∂t ∂2 Ψ *» 2 24222 ÿ multiply by Ψ− …Z =()m c −∇Ψ Z c Ÿ ∂t 2 ⁄ ∂2 Ψ * »2 24 222* ÿ multiply by Ψ− …Z =()m c −∇Ψ Z c Ÿ ∂t 2 ⁄ ≈∂Ψ2 ∂Ψ 2 * ’ Z2*∆Ψ −Ψ ÷ = Z 22*2c () Ψ∇Ψ−Ψ∇Ψ 2* «∂t2 ∂ t 2 ◊ 7 Interpretation of the Klein -Gordon equation: Problem 1 If we require that the expressions from j(r,t) and P( r,t) had correct non-relativistic limits we define Z j(,) r t = Ψ*() ∇Ψ−∇Ψ() * Ψ . 2mi ⁄ ∂P(r , t ) Then, we obtain the equation + ∇j( r ,t ) = 0. ∂t ∂Ψ ∂Ψ * iZ »* ÿ with P(,)r t =… Ψ −Ψ Ÿ . 2mc2 ∂ t ∂ t ⁄ P( r,t) is not positive-definite and can not be interpreted as position probability density. Interpretation of the Klein -Gordon equation: Problem 2 ∂2 Ψ Free particle Klein-Gordon equation: −Z2 =()m 24 c −∇Ψ Z 222 c . ∂t 2 −ω Plane wave solutions: Ψ(,)r t = Ae i(kr t ) . E = Zω Z22ω =mc 24 + Z 222 ck p= Z k We get additional “negative-energy” E= ± mc24 + cp 22 solutions and energy spectrum is not bound from below. Then, arbitrary large energy can be extracted from the mc 2 system if external perturbation leads to transition 2 between positive and negative energy states. 0 ∆E = 2 mc -mc 2 8 Interpretation of the Klein -Gordon equation In 1934 W. Pauli and V. Weisskopf reinterpreted Klein-Gordon equation as a field equation and quantized it using the formalism of quantum field theory. Klein-Gordon equation Relativistic wave equation for spinless particles in the framework of many-particle theory; negative energy states are interpreted in terms of antiparticles. Still, is it possible to define positive-definite position probability density within the framework of the relativistic theory? Dirac equation Note: we will still get negative-energy states… P.A. Dirac (1928) Dirac equation ∂ Ψ We start from the wave equation in the form iZ Ψ = H Ψ . 1 ∂t Ψ Ψ = 2 @ Spatial coordinates ( x1=x, x 2=y, x 3=z ) of a space-time point (event) Ψ N and the time coordinate ( x4=ict ) have to enter on the same footing. Therefore, the hamiltonian H must be linear in space derivatives as well. Free particle Simplest Hamiltonian: 1) Must be independent of r and t 2) Must be linear in p and m = ⋅ + β 2 = −Z ∇ H c pop mc pop i ααα, , and β 1) are independent of r, t, p, and E 1 2 3 2) do not have to commute with each other 9 How to determine α and β? ∂ ZΨ=− Z ⋅∇Ψ+β2 Ψ −⋅ − β 2 Ψ= i i c mc or Eop c p op mc ⁄ 0 ∂t The solution of the Dirac equation also must be a solution of the Klein-Gordon equation »E2− c 22p − mc 24 ÿ Ψ= 0. op op ⁄ We use it to determine the restrictions on the values of α and β by matching the coefficients in ¡ »−⋅−¡ β2 ÿ» −⋅− β 2 ÿ Ψ= Eop c p op mc ⁄ E op c p op mc ⁄ 0 and 2 22 24 »E− cp − mc ÿ Ψ= 0. Note: we drop the index op op ⁄ op in the derivation below. How to determine α and β? Some transformations ¢ ¢ 2 2 »E−⋅− c p β mc ⁄ ÿ» E −⋅− c p β mc ⁄ÿ Ψ= 0 ¢ ¢ ¢ ¢ 2¢ 2 2 2 »EEc−()()()()() ⋅−p Emccβ −⋅+p Ec ⋅p ⋅+⋅p cp β mc 2 3¢ 224 −βmc E + β mc() ⋅+p β m c ⁄ÿ Ψ= 0 ¢ ¢ ¢ ¢ » 2−¢ ⋅+β 2 +⋅⋅+ 2()()()() 3 ⋅+⋅+ βββ 2 2 4 ÿ Ψ= EEc2 ()p mccp p mc »p p ⁄ÿ mc ⁄ 0 ¢ ¢ ¢ 2 2¢ 3 2 2 4 »Ec−()()()() ⋅p ⋅−p mc » ⋅+⋅−p β βp ⁄ÿ β mc ⁄ÿ Ψ= 0 » 3 3 2 −2α 2 2 − 2 () αααα + −3() αββα +−Ψ= β 224 ÿ …E cƒii pc ƒ ijjiij ppmc ƒ i ii pmc ⁄ 0 i=1 i < j i = 1 10 How to determine α and β? Now we can match the coefficients of 3 3 2− 2α 2 2 − 2 () αααα + −3() αββα +−Ψ= β 224 …Ecƒii pc ƒ ijjiij ppmc ƒ i ii pmc ⁄ 0 i=1 i < j i = 1 α 2 = β 2 = 1 and i 1 αβ+ βα = 0, i = 1, 2, 3 2− 22 − 24 Ψ= i i »E cp mc ⁄ÿ 0. αα+ αα = ij ji 0 i≠ j Properties of α and β α 2 = 1 1 {α β } = {α α } = 1, 0 1, 2 0 α 2 = 1 α= α † 2 {}α β = {}α α = i i 2 , 0 2, 3 0 α 2 = 1 β= β † 3 {}α, β = 0 {}α, α = 0 β 2 = 1 3 1 3 α α α β Therefore, 1, 2, 3,and anticommute in pairs and their squares are equal to unity. Clearly, they must be matrices. The eigenvalues of α and β are ±1. βα= − αβ (α) =( β ) = i i Tri Tr 0 α=− βαβ = αββ = αβ 2 i ii i ()α=−= βαβ αβ2 = βαβ = Tri Tr( i )()( Tr i Tr i )0 11 What if the lowest rank of α and β? Dirac representation of α and β. What is the lowest rank of the representation for α and β? (α) =( β ) = Tri Tr 0 Therefore, rank N must be even. For 2x2 matrices we can not find a representation of more than 3 anticommuting matrices. Therefore, the lowest representation has N=4. Dirac representation: 0 σ I 0 α=i β = , i σ − i 0 0 I σ where i (i=1,2,3) are Pauli matrices 01 0−i 10 σ= σ = σ = x y z 10 i 0 01− Dirac equation Ψ 1 Ψ For N=4, the wave function Ψ = 2 is a four-component spinor @ Ψ N and describes spin ½ particles. We note that this result may be foreseen as in non-relativistic quantum mechanics spin ½ particles are described by 2-component spinors and each spin ½ particle has an antiparticle with the same mass and spin, which lead to 4-component wave function. Higher rank matrix representations correspond to particle with spin greater than ½. 12 Covariant form of the Dirac equation ∂ 2 iZΨ=− ic Z ⋅∇Ψ+β mc Ψ xµ ≡ (x , ict ) ∂t ∂Ψ3 ∂ β×− Z +i Z α Ψ− β mc Ψ= 0 …∂ƒ i ∂ Ÿ x4 i=1 x i ⁄ »3 ∂ ∂ mc ÿ −iβ α + β +Ψ= 0 …ƒ i ∂ ∂ Z Ÿ i=1 xi x 4 ⁄ »∂ mc ÿ …γ+Ψ= Ÿ 0 γβαγβ =−i = µ ∂ Z i i 4 …xµ ⁄Ÿ {γγµ, ν} = 2 δ µν , µν, = 1,2,3,4 −σ From the Dirac representation ≈0 i ’ ≈I 0 ’ γ=i γ = , for the α and β i ∆σ ÷ 4 ∆− ÷ «i 0 ◊ «0 I ◊ 13.