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Lecture #23

Plane solutions of the Dirac equation Spherical -like systems … again (Relativistic version) Dirac energy levels

Chapter 2, pages 48 -53, Lectures on Atomic Physics Chapter 15, pages 696 -716, Bransden & Joachain ,

Plane wave solutions of the Dirac equation

The Dirac equation for the with ½ is

∂

2 2 iZΨ=− ic Z ⋅∇Ψ+β mc Ψ or » Ec − ⋅−p β mc ⁄ÿ Ψ= 0 ∂t

We look for solutions in the form

( − ) Ψ()r,t = Aue ipr Et / Z 4-component constant

¡ 2 »c ⋅p +β mcuEu ⁄ÿ =

1 Plane wave solutions of the Dirac equation: p=0

First, we consider the case p=0. We label the solutions u(0). ≈mc 2 0 0 0 ’ ∆ ÷ ≈I 0 ’ 0mc 2 0 0 β = ∆ ÷ ∆ ÷ u(0)= Eu (0) «0 −I ◊ ∆0 0−mc 2 0 ÷ ∆ ÷ ∆2 ÷ «0 0 0 −mc ◊ 2 The eigenvalues are E+ = mc ( twice ) 2 E− = − mc( twice ).

≈’1 ≈’ 0 ≈’ 0 ≈’ 0 ∆÷ ∆÷ ∆÷ ∆÷ 0 1 0 0 The eigenvectors are u(1)(0)=∆÷ u (2) (0) = ∆÷ u (3) (0) = ∆÷ u (4) (0) = ∆÷ ∆÷0 ∆÷ 0 ∆÷ 1 ∆÷ 0 ∆÷ ∆÷ ∆÷ ∆÷ «◊0 «◊ 0 «◊ 0 «◊ 1

Plane wave solutions of the Dirac equation: general case σ ≈0 i ’ ≈I 0 ’ »¡ ⋅ +β 2 ÿ = α= β = c p mcuEu ⁄ i ∆σ ÷ ∆− ÷ «i 0 ◊ «0 I ◊ ≈’u ≈’ u ≈’ u =A =1 = 3 We use the following designations: u∆÷, uA ∆÷ u B ∆÷ . «◊uB «◊ u2 «◊ u 4

σσσ ⋅ 2 ≈0c p ’ ≈’uA≈mc I 0 ’ ≈’≈’ u A u A ∆ ÷ ∆÷+ ∆ ÷ ∆÷∆÷= E σσσ ⋅ − 2 «c p 0 ◊ «◊uB«0 mc I ◊ «◊«◊ u B u B ≈mc2 I c σσσ ⋅ p ’ ≈’u ≈’ u ∆ ÷ ∆÷A= E ∆÷ A σσσ ⋅ − 2 «c p mc I ◊ «◊uB «◊ u B c σσσ ⋅ p mcuc2 +⋅= σσσ p uEu → u = u A BA AE− mc 2 B c σσσ ⋅ p −⋅−c σσσ p u mc2 u = Eu →= u u A BB B E+ mc 2 A

2 Plane wave solutions of the Dirac equation: general case

c σ⋅p c σ ⋅ p Combining u = uu and = u AE− mc2 BB E + mc 2 A −2 + 22 =⋅⋅=σ σ 22 we obtain (EmcEmcu)( ) A cp p u A cu p A

using σ ⋅⋅ a σ b = ab+ ⋅ i σ [ ab × ] . 2− 24 = 22 (E mcu) A cp u A

Therefore, we get four eigenvalues:

24 22 E+ = + mc + cp ( twice )

24 22 E− = − mc + cp ( twice ).

Plane wave solutions of the Dirac equation

≈mc2 I c σσσ ⋅ p ’ ≈’u ≈’ u We can get the same result by expanding ∆ ÷ ∆÷A= E ∆÷ A σσσ ⋅ − 2 «c p mc I ◊ «◊uB «◊ u B ≈’01 ≈’ 0−i ≈’ 10 σ= σ = σ = x∆÷ y ∆÷ z ∆÷ «◊10 «◊i 0 «◊ 01− ≈p p− ip ’ σσσ ⋅ p = σp+ σ p + σ p = z x y xx yy zz ∆+ − ÷ «px ip y p z ◊

≈ 2 − − ’ mc E0 cpz c( p x ip y ) ∆ ÷ ≈u1 ’ ∆ 2 −() + − ÷ ∆ ÷ 0 mc E c px ip y cp z u ∆ ÷ ∆2 ÷ = 0 2 ∆ ÷ ∆ cp c() p− ip − mc − E 0 ÷ u3 ∆ z x y ÷ ∆ ÷ u ∆ ()+− −−2 ÷ «4 ◊ «c px ip y cp z 0 mc E ◊

3 Plane wave solutions of the Dirac equation

The corresponding is E 2 − mc 24 − cp 22 = 0 . > with(LinearAlgebra): > A := ([[m*c^2-E,0,c*pz,c*px-I*c*py], [0,m*c^2-E,c*px+I*c*py,-c*pz],[c*pz,c*px-I*c*py,-m*c^2-E,0], [c*px+I*c*py,-c*pz,0,-m*c^2-E]]); » 2 ÿ … m c − E 0 c pz c px − c py IŸ … Ÿ … Ÿ … 2 Ÿ … 0 m c − E c px + c py I −c pz Ÿ … Ÿ A := … Ÿ … 2 Ÿ … c pz c px − c py I −m c − E 0 Ÿ … Ÿ … 2 Ÿ c px + c py I −c pz 0 −m c − E ⁄ > B:=factor(Determinant(A)); 2 2 2 2 2 2 2 2 4 2 B := (−c px − c py − c pz − m c + E )

24 22 So we get the same results, as expected: E+ = + mc + cp ( twice )

24 22 E− = − mc + cp ( twice ).

Plane wave solutions of the Dirac equation

We note that if E 2 − mc 24 − cp 22 = 0 holds, then the determinant is of rank 2 (all 3x3 minors vanish). Therefore, there are two linearly

independent solutions corresponding to E +. ≈α(1) ’ ≈ α (2) ’ ≈’1 ≈’ 0 ∆ ÷ ∆ ÷ (1) (2) u(1) = Nσ⋅ u (2) = N σ ⋅ α=∆÷ , α = ∆÷ ∆c pα(1) ÷ ∆ c p α (2) ÷ «◊0 «◊ 1 ∆+ 2 ÷ ∆+ 2 ÷ «E+ mc ◊ « E+ mc ◊

(1) (2) The general solution is given by a linear combination u + = au + bu .

Solutions corresponding to E are ≈σ⋅ ’≈ σ ⋅ ’ − c pα(1) − c p α (2) (3) =∆−+2 ÷∆(4) = −+ 2 ÷ uN∆E− mc ÷∆ uN E− mc ÷ . ∆ (1) ÷ ∆(2) ÷ «α ◊« α ◊

4 Spherical spinors

ϕ= ϕ =⋅++ β 2 hED , in atomic units hcD p cVr ( )

Total angular is given by J=L+S.

J commutes with the Dirac Hamiltonian hD. Therefore, we may classify 2 the eigenstates of hD according to the eigenvalues of energy, J and Jz. 2 Ω (θ,φ) The eigenstates of J and Jz are spherical spinors κm .

2 We combine spherical harmonics, which are eigenstates of L and Lz 2 and spinors, which are eigenstates of S and Sz to form eigenstates of 2 Ω (θ,φ) J and Jz (refereed to as spherical spinors jlm ). l m-µ Ωθφ = θφχ jlm ( , )ƒ − Yl, m −µ ( , ) µ µ jm ½ µ

Spherical spinors

l m-µ Ωθφ = θφχ jlm ( , )ƒ − Yl, m −µ ( , ) µ µ jm µ ½ ≈+ + ’ l m 1/ 2 θ φ ∆ Yl, m − 1/2 (,) ÷ Ωθ φ = ∆2l + 1 ÷ = + l+1/ 2 lm ( , ) j l 1/ 2 ∆− + ÷ l m 1/ 2 θ φ ∆ Yl, m + 1/2 ( , ) ÷ «2l + 1 ◊ j= l ± 1/ 2

≈l− m + 1/ 2 ’ j= l − 1/ 2 − θ φ ∆ Yl, m − 1/2 (,) ÷ Ωθ φ = ∆2l + 1 ÷ l−1/ 2 lm ( , ) ∆+ + ÷ l m 1/ 2 θ φ ∆ Yl, m + 1/2 ( , ) ÷ «2l + 1 ◊

5 Spherical spinors

Spherical spinors are eigenfunctions of σ ·L and, therefore, of K =−1 − ⋅ L The eigenvalue for K is À −−l1 j =+ l 1/ 2 Ωθφ =Ω κ θφ κ = K jlm( , ) jlm ( , ) Ã Õ l j= l − 1/ 2

Note that the κ (which is referred to as relativistic ) uniquely defines the orbital with l and j Ω θ φ so we can use designations κ m ( , ) . s l= 0 j = 1/ 2 κ = −1 = =κ = p1/2 l 1 j 1/ 2 1 = =κ = − p3/ 2 l 1 j 3/ 2 2 = =κ = d3/ 2 l 2 j 3/ 2 2 = = κ = − d5/ 2 l 2 j 3/ 2 3

Dirac equation for a central potential

ϕ= ϕ =⋅++ β 2 hED , in atomic units hcD p cVr ( )

Total angular momentum is given by J=L+S.

J commutes with the Dirac Hamiltonian hD. Therefore, we may classify 2 the eigenstates of hD according to the eigenvalues of energy, J and Jz. 2 Ω (θ,φ) The eigenstates of J and Jz are spherical spinors κm .

1 ≈iPκ( r ) Ω κ (θ , φ ) ’ We seek solutions in a form ϕ ( r ) = ∆ m ÷ . k Ω ()θ φ r «Qκ( r ) − κ m , ◊ The resulting equations for the radial functions are »κ ÿ 2 +d − = »cVrPrc() ⁄ÿ κ () + Qr κ () EPr κ () …dr r ⁄Ÿ »κ ÿ −+d +−2 = c Prκ() » VrcQr () ⁄ÿ κ () EQr κ (). …dr r ⁄Ÿ

6 Hydrogen -like systems: Dirac energy levels »κ ÿ »2 + ÿ d − = cVrPrc() ⁄ κ () +… Ÿ Qr κ () EPr κ () Z dr r ⁄ V( r ) = − »κ ÿ r −+d +−2 = c… Ÿ Prκ() » VrcQr () ⁄ÿ κ () EQr κ () dr r ⁄ 2 =c γ = κ2 − α 2 2 Dirac energy levels: Enκ , Z α 2Z 2 1+ ()γ+n + κ

The non-relativistic energy levels depends only on the principal quantum number n. When relativistic effects are taken into accounts the non-relativistic will split into n different Dirac energy levels ( splitting). Note that the energy above depends only on the values of n and | κ|=j+1/2, therefore the levels with the same n and l but different j, for example levels 2p 1/2 and 2p 3/2 will have different energies. The energy difference between such levels is called the fine-structure interval.

Hydrogen -like systems: Dirac energy levels

2 =c γ = κ2 − α 2 2 Dirac energy levels: Enκ , Z α 2Z 2 1+ ()γ+n + κ

The levels the same n and j but different l will have the same energies,

for example levels 2s 1/2 and 2p 1/2 . The experimentally observed energy difference between these levels is called the and is explained by (QED) effects, known as radiative corrections. The expansion of the formula above in powers of αZ will give (in atomic units) 2α 2 4 » ÿ =−−2 Z Z 1 −+ 3 Enκ c 2 3 … Ÿ ... 2n 2 n κ 4 n ⁄ Non-relativistic ’s rest Coulomb- 2 Leading fine-structure correction energy (mc ) binding energy

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