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Relativistic Mechanics

The Klein -Gordon equation Interpretation of the Klein -Gordon equation The Dirac representation for the matrices α and β Covariant form of the Dirac equation

Why Relativistic ?

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 zero The Klein -Gordon equation

So, how to come up with such an equation? For the relativistic particle with rest m and 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 on right-hand side. If expanded in power series it lead to 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 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 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 : ∂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 ÿ . 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 . 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 ÿ 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 ÿ …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 Ψ = ∆ 2 ÷ is a four-component ∆@ ÷ ∆ ÷ 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 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 ∂ ÿ 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 ◊

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