Relativistic quantum mechanics

Transcription

1 Free-electron solution of Quantum mechanics 2 - Lecture 11 UJJS, Dept. of Physics, Osijek January 15, 2013

2 Free-electron solution of Free-electron solution of 4 5

3 Contents Free-electron solution of Free-electron solution of 4 5

4 Free-electron solution of Non-relativistic physics E = p2 2m E Ê = i t p ˆp = i = i ψ t = 2 2m ψ free-particle S.E.

5 Free-electron solution of Non-relativistic physics E = p2 2m E Ê = i t p ˆp = i = i ψ t = 2 2m ψ free-particle S.E. Statistical interpretation of ψ(r, t): ρ(r, t) = ψ(r, t) 2

6 Free-electron solution of Relativistic physics E 2 = m 2 c 4 + c 2 p 2 E Ê = i t p ˆp = i = ( i ) 2 [ ψ = m 2 c 4 + c 2 ( i ) 2] ψ t relativistic S.E. (Fock s equation)

7 Free-electron solution of Relativistic physics = E 2 = m 2 c 4 + c 2 p 2 E Ê = i t ( i t ) 2 ψ = [ m 2 c 4 + c 2 ( i ) 2] ψ p ˆp = i = ( κ 2) ψ = 0 = 1 c 2 2 t 2 κ = mc, 1 κ = mc reduced Compton wavelength [3]

8 Free-electron solution of A question What is Compton wavelength for an electron?

9 Free-electron solution of Statistical interpretation of ψ in Schrödinger theory: S.E. equation of continuity ρ t + divj = 0 ρ = ψ ψ probability density j = 2im [ψ ψ ( ψ ) ψ]

10 Free-electron solution of Statistical interpretation of ψ in Klein-Gordon theory: K-G equation equation of continuity ρ t + divj = 0 ρ = ψ ψ ψ ψ j = c 2 [ψ ψ ( ψ ) ψ]

11 Free-electron solution of Statistical interpretation of ψ in Klein-Gordon theory: K-G equation equation of continuity ρ t + divj = 0 ρ = ψ ψ ψ ψ 0 problem! j = c 2 [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.

12 Contents Free-electron solution of Free-electron solution of 4 5

13 Free-electron solution of 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

14 Free-electron solution of 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 spatial and time derivatives (3) in analogy with S.E. must be linear in spatial derivatives: Ĥ = cα p + βmc 2

15 Free-electron solution of Factorisation of K-G equation gives [5] (Ê cα p βmc 2 ) ( Ê + cα p + βmc 2) ψ = 0 Comparisson with K-G equation imposes β 2 = 1, α k β + βα k = 0, (1) αx 2 = 1, α xα y + α y α x = 0, (2) αy 2 = 1, α y α z + α zα y = 0, (3) αz 2 = 1, α zα x + α xα z = 0 (4)

16 Free-electron solution of A task 1 Write down conditions (1)-(4) using anticommutators. (Hint: consult Ref. [1].) 2 Pauli matrices σ x = [ ], σ y = [ 0 i i 0 ], σ x = [ ], satisfy conditions (2)-(4). Please, verify if they satisfy condition (1).

17 Free-electron solution of Dirac s matrices α x = α z = α i = [ 0 σi σ i ], β =, αy =, β = [ I 0 0 I ] i 0 0 i 0 0 i 0 0 i 0 0 0,

18 Free-electron solution of A task Please, verify if Dirac s matrices satisfy condition (1).

19 Free-electron solution of Dirac s equation (i t cα p βmc2 ) ψ = 0 Solution is a four-component column matrix (spinor) ψ 1(r, t) ψ(r, t) = ψ 2(r, t) ψ 3(r, t) ψ 4(r, t)

20 Free-electron solution of Interpretation of ψ in equation of continuity ρ t + divj = 0 ρ = ψ ψ = ψ ψ ψ ψ probability density j = cψ αψ probability density current requirement (1) is satisfied

21 Contents Free-electron solution of Free-electron solution of 4 5

22 Free-electron solution of ( E cα p βmc 2) ψ = 0 Suppose a plane wave solution ψ(r, t) = ue i (p r Et), E = p2 2m

23 Free-electron solution of ( E cα p βmc 2) ψ = 0 Suppose a plane wave solution ψ(r, t) = ue i (p r Et), E = p2 2m c = ( ) Eu cα pu βmc 2 u = 0 E mc E mc E + mc E + mc p z p x ip y 0 0 p x + ip y p z p z p x ip y 0 0 p x + ip y p z 0 0 u 1 u 2 u 3 u 4 u 1 u 2 u 3 u 4 = 0

24 Free-electron solution of E mc 2 0 c(p z) c(p x ip y ) 0 E mc 2 c(p x + ip y ) c(p z) c(p z) c(p x ip y ) E + mc 2 0 c(p x + ip y ) c(p z) 0 E + mc 2 = 0 E 2 = c 2 p 2 + m 2 c 4 E = ± c 2 p 2 + m 2 c 4

25 Free-electron solution of 1 E + = + c 2 p 2 + m 2 c 4 u (+) = N 1 0 cp z E + + mc 2 c(p x + ip y ) E + + mc 2, u (+) = N 0 1 c(p x ip y ) E + + mc 2 cp z E + + mc 2 2 E = c 2 p 2 + m 2 c 4 u ( ) = N cp z E mc 2 c(p x + ip y ) E mc 2 1 0, u ( ) = N c(p x ip y ) E mc 2 cp z E mc 2 0 1

26 Free-electron solution of HW Calculate the normalization constant N. (Solution can be found, for example, in Ref. [1])

27 Free-electron solution of Why are there and arrows in the subscripts? u (+) relate to nonrelativistic limit cp z E + + mc 2 c(p x + ip y ) E + + mc 2 v c v c 0 u (+) (r, t) [ 1 0 u (+) (r, t) [ 0 1 ] e i (p r Et), ] e i (p r Et) free spin 1/2 particles

28 Free-electron solution of Interpretation of E + and E

29 Free-electron solution of Interpretation of E + and E

30 Free-electron solution of Interpretation of E + and E

31 Free-electron solution of Interpretation of E + and E

32 Free-electron solution of Interpretation of E + and E

33 Free-electron solution of Interpretation of E + and E

34 Free-electron solution of Interpretation of E + and E Positron - experimental discovery: D. Skobelstyn (1929). C.-Y. Chao (1929). C. D. Anderson (1932). - Nobel Prize (1936).

35 Contents Free-electron solution of Free-electron solution of 4 5

36 Free-electron solution of Consider an electron in an emg field: H = cα (p q ) c A + βmc 2 + qφ(r) D.E. [ cα (p q ) ] c A + βmc 2 + qφ(r) ψ = Eψ

37 Free-electron solution of Consider an electron in an emg field: H = cα (p q ) c A + βmc 2 + qφ(r) D.E. [ cα (p q ) ] c A + βmc 2 + qφ(r) ψ = Eψ α symmetry [ c (p q ) c A [ c ] σ ] (p q c A ) σ ( ) + mc 2 + qφ(r) W = EW ( ) mc 2 qφ(r) V = EV where [ ψ (1) W = ψ (2) ] [ ψ (3), V = ψ (4) ]

38 Free-electron solution of [ ] c (p (q/c)a) σ V = W E qφ + mc 2 [ ( c 2 p q ) c A σ (E qφ + mc 2) ] 1 (p (q/c)a) σ W ( ) + mc 2 + qφ(r) W = EW

39 Free-electron solution of [ ] c (p (q/c)a) σ V = W E qφ + mc 2 [ ( c 2 p q ) c A σ (E qφ + mc 2) ] 1 (p (q/c)a) σ W ( ) + mc 2 + qφ(r) W = EW v/c 0 V W 0 E = E mc 2, qφ mc 2 ( E qφ + 2mc 2) 1 1 = 2mc 2 ( 1 E qφ 2mc 2 + )

40 Free-electron solution of D.E. in nonrelativistic limit [ 1 2m ( p q c A ) 2 ] q mc S B + qφ W = E W where S = 2 σ

41 Free-electron solution of D.E. in nonrelativistic limit [ 1 ( p q ) ] 2 2m c A q mc S B +qφ W = E W }{{} µ B where S = 2 σ µ = q S magnetic moment of an electron mc

42 Contents Free-electron solution of Free-electron solution of 4 5

43 Contents Free-electron solution of 1 R. L. Liboff, Introductory Quantum Mechanics, Addison Wesley, San Francisco, I. Supek, Teorijska fizika i struktura materije, II. dio, Školska knjiga, Zagreb, Compton wavelength 4 P.A.M. Dirac - life & interesting facts 5 P. A. M. Dirac, The Quantum Theory of the Electron, Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences 117 (778): 610 (1928). 6 C. D. Anderson - Nobel lecture about the dicovery of positron