Dirac Equation and Hydrogen Atom

Transcription

1 Technische Universität Darmstadt October 19, 2006

2 Outline The Dirac Equation Dirac Equation with a Central Potential Separation of the Radial and Angular Components Energy Levels of the Hydrogen Atom

3 Schrödinger Equation and Lorentz Transformation i [ ] t ψ = 2 2m 2 + V ψ The Schrödinger equation is based on a non-relativistic ansatz. The Schrödinger equation is linear in time, but not in space. The Schrödinger equation is invariant under Galilei transformation, but not under Lorentz transformation.

4 Formalism (1) The Dirac Equation Coordinates A point in space and time is represented by a 4-vector x µ (x 0, x k ) (x 0, x 1, x 2, x 3 ) where x 0 = ct and x k = r Covariant and contravariant vectors We have to distinguish between covariant and contravariant vectors. They are connected by a µ = ν g µν a ν = g µν a ν where g µν is a metric tensor with g µν =

5 Formalism (2) The Dirac Equation Lorentz Transformation The Lorentz transformation along the x-axis is given by x = γ(x vt) t = γ(t v c 2 x) where γ = 1/ 1 v 2. By introducing the matrix Λ µ c 2 ν the transformation becomes x µ = Λ µ ν x ν.

6 Formalism (3) The Dirac Equation Scalar product of vectors The scalar product of two vectors a µ and b µ is defined as follows: a µ b µ = a µ b µ = a 0 b 0 a.b Gradient We keep the differential operator and define a covariant and contravariant vector µ ( / ct, ) µ g µν ν ( / ct, )

7 Formalism (4) The Dirac Equation Electromagnetic fields The electrical potential ϕ(r, t) and the vector potential A(r, t) are combined to a 4-vector A µ (ϕ, A) Units Without getting any problems we can set = c = 1. x µ = (ct, r) (t, r) p µ = (E/c, p) (E, p)

8 Relativistic Wave Equation In order to obtain a relativistic wave equation we start with the relativistic relation between the energy, the momentum and the mass of a particle: E 2 = p 2 + m 2 From analytical mechanics we get the Hamiltonian for the motion of a charged particle in an electromagnetic field: H = 1 2m (p ea)2 + eϕ Comparing this to the Hamiltonian for a free particle (H 0 = p2 2m ) we obtain a substitution for H and p: H H eϕ and p p ea

9 The Dirac Equation Making the substitutions for E and p leads us to (E eϕ) 2 = (p ea) 2 + m 2. Now we can perform the substitutions given by Schrödinger (E i t, p i ) to get a quantum mechanical equation, the : [ (i t eϕ)2 ( 1 i ea)2] ψ = m 2 ψ

10 (2) Now we can rewrite the to obtain a more compact form. [ (i t eϕ)2 + ( i ea) 2 + m 2 ] ψ = 0 [( t + ieϕ)2 + ( iea)( + iea) + m 2 ] ψ = 0 By introducing the 4-vector operators D µ µ + iea µ ( t + ieϕ, iea) and D µ µ + iea µ ( t + ieϕ, + iea) we finally get [ Dµ D µ + m 2] ψ = 0.

11 Problems of the Klein-Gordon Equation Energies can be negative Probability density is negative for negative energies The time dependence is not linear The equation describes particles with spin 0 Solution We have to linearize the equation, i.e. to find an equation that is linear in p µ and leads to the.

12 Ansatz In order to find a wave equation that describes a free electron Dirac made the ansatz H D = α.p + βm. α (α x, α y, α z ) and β are hermitian operators only working on the spin variables of the system. The wave equation becomes Eψ = H D ψ = [α.p + βm] ψ or [E α.p βm] ψ = 0.

13 Properties of α and β The Dirac Equation Furthermore we claim that solutions to [E α.p βm] ψ = 0 also solve the. To obtain an equation with second order in time we multiply with [E + α.p + βm] from the left side and compare the result to the [ E 2 p 2 m 2] ψ = 0. This leads to the following constaints for α and β: (α k ) 2 = 1 (β) 2 = 1 α k α l + α l α k = 0 (k l) α k β + βα k = 0

14 with electromagnetic fields Starting from the for a free electron [E α.p βm] ψ = 0 we perform the substitutions E E eϕ and p p ea and get [(E eϕ) α.(p ea) βm] ψ = 0 [ or (i t eϕ) α.( i ea) βm] ψ = 0. The Hamiltonian becomes H D = eϕ + α.(p ea) + βm.

15 Covariant To get a covariant form of the we multiply [(i t eϕ) α.( i ea) βm ] ψ = 0 from the left side with β and get [β(i t eϕ) βα.( i ea) ββm ] ψ = 0. Now we define some new operators: γ µ (γ 0, γ 1, γ 2, γ 3 ) (γ 0, γ) γ 0 β γ βα

16 Covariant (2) Finally we put it all together and get (β 2 = 1, D µ = µ + iea µ ): [γ 0 (i t eϕ) γ.( i ea) m ] ψ = 0. [iγ µ ( µ + iea µ ) m] ψ = 0 [iγ µ D µ m] ψ = 0

17 Representation of α and β We found earlier: (α k ) 2 = 1 (β) 2 = 1 α k α l + α l α k = 0 (k l) α k β + βα k = 0 The relations above can not be satisfied with real or complex numbers α and β could be matrices. The Hamiltonian H D = eϕ + α.(p ea) + βm is hermitian, so must be α and β. (α k ) 2 = I and (β) 2 = I The Eigenvalues are ±1.

18 Representation of α and β (2) The trace of α and β is zero: Tr(α k ) = Tr(ββα k ) = Tr(βα k β) = Tr(α k ββ) = Tr(α k ) The dimension N of α and β is even. For N = 2 there are only three linear independant hermitian matrices with a trace of zero (Pauli matrices): σ 1 = ( ) σ 2 = ( 0 i i 0 ) σ 3 = The smallest possible dimension is N = 4 and one representation is ( ) ( ) 0 σk I 0 α k = β =. σ k 0 0 I ( )

19 Representation of γ µ And finally we can write down γ µ (β, βα): γ 0 = γ 2 = i 0 0 i 0 0 i 0 0 i γ1 = γ3 =

20 Solutions with negative Energies For an electron in rest the becomes i ( ) ( ) ( ) φ 1 0 φ = m. t χ 0 1 χ The solutions are φ = e iω 0t and χ = e +iω 0t. The energies become E φ = + ω 0 = +mc 2 and E χ = ω 0 = mc 2. Interpretation Dirac All energy levels below zero are occupied (Dirac sea). Stückelberg The solutions with negative energies are propagating backwards in time. So these solutions describe antiparticles propagating forward in time with a positive ernergy.

21 g factor of the electron The Dirac Equation i t ( ) [( φ 0 σ = χ σ 0 ) ( 1 0 (p ea) ) ] ( ) φ m + eϕ χ Using the approximation χ 1 2m (σ.p)φ and P = p ea we obtain a differential equation for φ: i φ t = 1 (σ.p)(σ.p)φ + eϕφ 2m Rewriting the equation by using B = A and e = e (electron charge) leads to a Schrödinger like equation: i φ [ 1 t = 2m ( i + e A)2 + e ] σ.b e ϕ φ 2m

22 g factor of the electron (2) i φ [ 1 t = 2m ( i + e A)2 + e ] σ.b e ϕ φ 2m Now we compare this equation to the Pauli equation (Schrödinger equation for a particle with spin): i ψ [ ] s 1 = t 2m ( i + e A)2 + µ.b e ϕ ψ s Comparing with µ = g e 2m s, where s = 1 2σ, leads to g = 2.

23 Dirac Equation with a Central Potential Separation of the Radial and Angular Components Energy Levels of the Hydrogen Atom We aim to find the energy levels of the hydrogen atom. Therefore we have to solve the with a central potential V (r) = Ze2 r. The Hamiltonian for such a problem is H D α.p + βm + V (r). As the problem is spherical symmetric, the Hamiltonian obeys the relations [H D, J] = 0 and [H D, P] = 0 where J and P are the operator of the total angular momentum and the parity operator respectively.

24 Dirac Equation with a Central Potential Separation of the Radial and Angular Components Energy Levels of the Hydrogen Atom Quantum Numbers We assume that the wave function ( ) φ ψ χ is an eigenfunction of J, J z and P with the quantum numbers J, M and { +1 for states with parity ( ) J+ 1 2 κ = 1 for states with parity ( ) J 1 2. ( ) ( ) ( ) ( ) φ φ φ φ J 2 = J(J + 1) J z = M χ χ χ χ ( ) ( ) φ P (0) = ( ) J+ 1 φ 2 κ χ χ

25 Dirac Equation with a Central Potential Separation of the Radial and Angular Components Energy Levels of the Hydrogen Atom Eigenfunctions of the Angular Momentum The function YLJ M may be the eigenfunction of the state (JM). The functions parity is ( ) L. L can be obtained in two different ways: L = l = J κ L = l = J 1 2 κ Considering P (0)( ) φ χ = ( ) J+ 1 κ( φ 2 χ) we can define the wave equation ψκj M = 1 ( F (r) Y M ) lj r ig(r) Yl M J where F and G are arbitrary functions. We separated the radial and angular components.

26 Dirac Equation with a Central Potential Separation of the Radial and Angular Components Energy Levels of the Hydrogen Atom Separation of the Angular and Radial Components of H D We have to solve the problem H D ψ M κj = EψM κj. Therefore we introduce p r = i 1 r r r and α r = (α.ˆr). In order to obtain another expression for α.p we calculate (α.r)(α.p) = r.p + iσ.l = rp r + i(1 + σ.l). Now we multiply this equation with α r /r from the left side and use the identity αr 2 = 1: ( α.p = α r p r + i ) r (1 + σ.l)

27 Dirac Equation with a Central Potential Separation of the Radial and Angular Components Energy Levels of the Hydrogen Atom Derivation of F and G With the expression for α.p we write down the problem again: [α r (p r + ir ) ] (1 + σ.l) + mβ + V (r) ψκj M = EψM κj From this equation we get two coupled differential equations: [ d dr + κ(j ) ] G = (E m V )F r [ d dr + κ(j ) ] F = (E + m V )G r The differential equations can be solved be assuming that F and G can be represented by a power series. This leads to an expression for the energy levels of the bound states of the hydrogen atom.

28 Dirac Equation with a Central Potential Separation of the Radial and Angular Components Energy Levels of the Hydrogen Atom Energy Levels Finally we can write down an expression for the energy levels of the hydrogen atom: E nj = m [1 + Z 2 e 4 ] 1 2 (n ɛ J ) 2 ɛ J = J (J )2 Z 2 e 4 E nj = m [ or 1 Z 2 e 4 2n 2 (Z 2 e 4 ) 2 2n 4 ( n J ) ]

29 Dirac Equation with a Central Potential Separation of the Radial and Angular Components Energy Levels of the Hydrogen Atom Energy Levels for several Quantum Numbers 1S 1 2 2S 1 2 2P 1 2 2P 3 2 n l J E nj m 1 Z 2 e Z m 2 e Z m 2 e m Z 2 e 4 The states of the principle quantum number n are not degenerated completely any more. The states 2S 1 and 2P 1 are still degenerated. They have the 2 2 same quantum numbers n and J, but a different parity. Experiments show that 2S 1 has a higher energy than 2P Lamb shift.

30 Bibliography Bjorken/Drell, Relativistische Quantenmechanik A. Messiah, Quantenmechanik 1 & 2 P. Schmüser, Feynman-Graphen und Eichtheorien für Experimentalphysiker