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ContentsKlein-Gordon equation

Dirac equationFree-electron solution of Dirac equation

Electron magnetic momentLiterature

Relativistic quantum mechanicsQuantum mechanics 2 - Lecture 11

Igor Lukacevic

UJJS, Dept. of Physics, Osijek

January 15, 2013

Igor Lukacevic Relativistic quantum mechanics




1 Klein-Gordon equation

2 Dirac equation

3 Free-electron solution of Dirac equation

4 Electron magnetic moment

5 Literature





Contents


2 Dirac equation



5 Literature





Non-relativistic physics

E =p2

2m99K

E 7→ E = i~ ∂∂t

~p 7→ p = −i~∇

=⇒ i~∂ψ∂t

= − ~2

2m∆ψ

↓free-particle S.E.





Non-relativistic physics

E =p2

2m99K

E 7→ E = i~ ∂∂t

~p 7→ p = −i~∇

=⇒ i~∂ψ∂t

= − ~2

2m∆ψ

↓free-particle S.E.

Statistical interpretation of ψ(r, t):

ρ(r, t) = |ψ(r, t)|2





Relativistic physics

E 2 = m2c4 + c2p2 99K

E 7→ E = i~ ∂∂t

~p 7→ p = −i~∇

=⇒

(i~ ∂∂t

)2

ψ =[m2c4 + c2 (−i~∇)2

]ψ

↓relativistic S.E. (Fock’s equation)





Relativistic physics

E 2 = m2c4 + c2p2 99K

E 7→ E = i~ ∂∂t

~p 7→ p = −i~∇

=⇒

(i~ ∂∂t

)2

ψ =[m2c4 + c2 (−i~∇)2

]ψ

=⇒(�− κ2

)ψ = 0 Klein-Gordon equation

� = ∆− 1

c2∂2

∂t2

κ =mc

~,

1

κ=

~mc reduced Compton wavelength [3]





A question

What is Compton wavelength for an electron?





Statistical interpretation of ψ in Schrodinger theory:

S.E. ⇒ equation of continuity

∂ρ

∂t+ divj = 0

ρ = ψ∗ψ probability density

j = − ~2im

[ψ∗∇ψ − (∇ψ∗)ψ]





Statistical interpretation of ψ in Klein-Gordon theory:

K-G equation ⇒ equation of continuity

∂ρ

∂t+ divj = 0

ρ = ψ∗ψ − ψ∗ψj = −c2 [ψ∗∇ψ − (∇ψ∗)ψ]





Statistical interpretation of ψ in Klein-Gordon theory:

K-G equation ⇒ equation of continuity

∂ρ

∂t+ divj = 0

ρ = ψ∗ψ − ψ∗ψ R 0 problem!

j = −c2 [ψ∗∇ψ − (∇ψ∗)ψ]

Problem

ρ depends on the initial conditions: ψ(0) and ψ(0)

ρ cannot be interpreted as the probability density

Good side

K-G equation describes well the spinless bosons, like π-mesons.





Contents


2 Dirac equation



5 Literature





Requirements for the relativistic wave equation

1 keep the statistical interpretation of ψ

2 must be relativistically invariant

3 must be of the 1st order in time variable

4 agrees with the K-G equation in the limit of large quantum numbers





Requirements for the relativistic wave equation

1 keep the statistical interpretation of ψ

2 must be relativistically invariant

3 must be of the 1st order in time variable

4 agrees with the K-G equation in the limit of large quantum numbers

(2) ⇒ symmetrical in spatialand time derivatives

(3) in analogy with S.E.

⇒ must be linear in spatial derivatives:

H = cα · p + βmc2





Factorisation of K-G equation gives [5](E − cα · p− βmc2

)(E + cα · p + βmc2

)ψ = 0

Comparisson with K-G equation imposes

β2 = 1 , αkβ + βαk = 0 , (1)α2x = 1 , αxαy + αyαx = 0 , (2)α2y = 1 , αyαz + αzαy = 0 , (3)α2z = 1 , αzαx + αxαz = 0 (4)





A task

1 Write down conditions (1)-(4) using anticommutators. (Hint: consult Ref.[1].)

2 Pauli matrices

σx =

[0 11 0

], σy =

[0 −ii 0

], σx =

[1 00 −1

],

satisfy conditions (2)-(4). Please, verify if they satisfy condition (1).





Dirac’s matrices

αi =

[0 σi

σi 0

], β =

[I 00 −I

]

αx =

0 0 0 10 0 1 00 1 0 01 0 0 0

, αy =

0 0 0 −i0 0 i 00 −i 0 0i 0 0 0

,

αz =

0 0 1 00 0 0 −11 0 0 00 −1 0 0

, β =

1 0 0 00 1 0 00 0 −1 00 0 0 −1





A task

Please, verify if Dirac’s matrices satisfy condition (1).





Dirac’s equation (i~ ∂∂t− cα · p− βmc2

)ψ = 0

Solution is a four-component column matrix (spinor)

ψ(r, t) =

ψ1(r, t)ψ2(r, t)ψ3(r, t)ψ4(r, t)





Interpretation of ψ in Dirac equation

Dirac equation ⇒ equation of continuity

∂ρ

∂t+ divj = 0

ρ = ψ†ψ = |ψ1|2 + |ψ2|2 + |ψ3|2 + |ψ4|2 ≥ 0 probability density

j = −cψ†αψ probability density

current

requirement (1) is satisfied





Contents


2 Dirac equation



5 Literature





Dirac equation (E − cα · p− βmc2

)ψ = 0

Suppose a plane wave solution

ψ(r, t) = uei~ (p·r−Et) , E =

p2

2m





Dirac equation (E − cα · p− βmc2

)ψ = 0

Suppose a plane wave solution

ψ(r, t) = uei~ (p·r−Et) , E =

p2

2m

=⇒(Eu − cα · pu − βmc2u

)= 0

E −mc2 0 0 00 E −mc2 0 00 0 E +mc2 00 0 0 E +mc2

u1

u2u3u4

−c

0 0 pz px − ipy0 0 px + ipy −pzpz px − ipy 0 0

px + ipy −pz 0 0

u1

u2u3u4

= 0





∣∣∣∣∣∣∣∣E −mc2 0 −c(pz) −c(px − ipy )

0 E −mc2 −c(px + ipy ) c(pz)−c(pz) −c(px − ipy ) E + mc2 0

−c(px + ipy ) c(pz) 0 E + mc2

∣∣∣∣∣∣∣∣ = 0

E 2 = c2p2 + m2c4

E = ±√

c2p2 + m2c4





1 E+ = +√

c2p2 + m2c4

u(+)↑ = N

10cpz

E+ + mc2c(px + ipy )

E+ + mc2

, u(+)↓ = N

01

c(px − ipy )

E+ + mc2−cpz

E+ + mc2

2 E− = −

√c2p2 + m2c4

u(−)↑ = N

cpz

E− −mc2c(px + ipy )

E− −mc2

10

, u(−)↓ = N

c(px − ipy )

E− −mc2−cpz

E− −mc2

01





HW

Calculate the normalization constant N. (Solution can be found, for example, inRef. [1])





Why are there ↑ and ↓ arrows in the subscripts?

u(+) relate to nonrelativistic limit

cpzE+ + mc2

c(px + ipy )

E+ + mc2

∼v

cv�c−−−→ 0

u(+)↑ (r, t) ∼

[10

]e

i~ (p·r−Et) ,

u(+)↓ (r, t) ∼

[01

]e

i~ (p·r−Et)

free spin 1/2 particles





Interpretation of E+ and E−































Positron - experimentaldiscovery:

D. Skobelstyn (1929).

C.-Y. Chao (1929).

C. D. Anderson (1932). -Nobel Prize (1936).





Contents


2 Dirac equation



5 Literature





Consider an electron in an emg field:

H = cα ·(

p− q

cA)

+ βmc2 + qφ(r)

D.E. ⇒ [cα ·

(p− q

cA)

+ βmc2 + qφ(r)]ψ = Eψ





Consider an electron in an emg field:

H = cα ·(

p− q

cA)

+ βmc2 + qφ(r)

D.E. ⇒ [cα ·

(p− q

cA)

+ βmc2 + qφ(r)]ψ = Eψ

α symmetry ⇒[c(

p− q

cA)· σ]

+(mc2 + qφ(r)

)W = EW[

c(

p− q

cA)· σ]−(mc2 − qφ(r)

)V = EV

where

W =

[ψ(1)

ψ(2)

], V =

[ψ(3)

ψ(4)

]





V =

[c (p− (q/c)A) · σE − qφ+ mc2

]W

c2[(

p− q

cA)· σ(E − qφ+ mc2

)−1

(p− (q/c)A) · σ]W

+(mc2 + qφ(r)

)W = EW





V =

[c (p− (q/c)A) · σE − qφ+ mc2

]W

c2[(

p− q

cA)· σ(E − qφ+ mc2

)−1

(p− (q/c)A) · σ]W

+(mc2 + qφ(r)

)W = EW

v/c → 0⇒

‖V ‖‖W ‖ → 0

E ′ = E −mc2 , ‖qφ‖ � mc2

⇒(E ′ − qφ+ 2mc2

)−1

=1

2mc2

(1− E ′ − qφ

2mc2+ · · ·

)





D.E. in nonrelativistic limit[1

2m

(p− q

cA)2− q

mcS · B + qφ

]W = E ′W

where

S =~2σ





D.E. in nonrelativistic limit[1

2m

(p− q

cA)2− q

mcS · B︸︷︷︸

µ · B

+qφ

]W = E ′W

where

S =~2σ

⇒ µ =q

mcS magnetic moment of an electron





Contents


2 Dirac equation



5 Literature





Literature

1 R. L. Liboff, Introductory Quantum Mechanics, Addison Wesley, SanFrancisco, 2003.

2 I. Supek, Teorijska fizika i struktura materije, II. dio, Skolska knjiga,Zagreb, 1989.

3 Compton wavelength

4 P.A.M. Dirac - life & interesting facts

5 P. A. M. Dirac, ”The Quantum Theory of the Electron”, Proceedings ofthe Royal Society A: Mathematical, Physical and Engineering Sciences 117(778): 610 (1928).

6 C. D. Anderson - Nobel lecture about the dicovery of positron


http://math.ucr.edu/home/baez/lengths.html

http://www.dirac.ch/PaulDirac.html

http://www.nobelprize.org/nobel_prizes/physics/laureates/1936/anderson-lecture.html