Solve for the exact levels of the th e Dirac particle in a uniform magnetic field B B0k A ˆ
1 2
B0 ( yi xj ) . ˆ
ˆ
Consult exercise 12.3.8, and write the equation for . Preliminaries: Dirac equation, in which the final step is made because { , } , definition of momentum in terms of canonical momentum momentum P , the property derived in the previous exercise, a formla from QM 08, p. 384, un-numbered, the matrix representation of { , } , and the Dirac equation in this representation,
[ ( p qc A ) 2 c 2 ( mc 2 )2 q ] i I t [ i 2 c2 ( mc2 )2 q ] [ c ( p cq A ) mc2 Iq ] ;
H
mv P c A i q
O I ; O O
q c
B ; ( A )( B ) ( A B ) I i ( A B ) ;
; ;
O
; ; I
We can expand out the Dirac equation into a 2x2 matrix representation, charge E mc2 E mc2 0 conjugation??? (c ) 2 0 E mc 0
[I.1]
( I E mc 2 ) c( )
( c ) [I.2] E mc2 0
One can “square” over electron space and positron space (note, the operator “rotates” between these spaces), 2 2 2 2 2 2 2 2 [( E mc )Ie ( E mc )Ie ] c ( ) ( ) I E (mc ) c ( ) [I.3]
We encounter the square of , which we handle with QM 08, p. 384, shown in [I.1], 2 2 q 2 q q ( ) ( )I i ( ) (P ( c A ) c {P, A})I i (i c B ) ; {P , A} P A A P
[I.4]
We then have, for the electron , the spinor-Hamiltonian. Note that for A 21 B0 ( yi xj ) we have ˆ
PA A P 1 c
2
E
2
B0 2
ˆ
([ P x y Py x ] [ yPx xPy ]) B0 Lz , which immediately makes life easy for us,
m2c4 (P2 ( qc A)2 i qc {P, A})I
q c
(B ) ( P x2 Py 2
q2 B02 4 c2
( x2 y2 )
B0 q ic
Lz )I
q c
B0 z [I.5]
317 - pr 12.3.8 - landau levels and the 2D SHO: we have seen this Hamiltonian before. A summary of the results from 12.3.8, we have, H ( P x , Py , X ,Y )
1 2
( Px
qB 2c
Y ) ( Py 2
qB 2c
X)
2
2
P
2
1
2Q 2 H SHO (Q , P ) H 12 0 , 21 0L z ; 2
[I.6]
E n, m H n, m ( H ( 12 , ) 12 L z ) k , m ((n 12 12 ) ( 12 ) ( 12 )(m )) n , m E (n 1 m ) 12
our math up until this point “thinks” our SHO is one-dimensional one-dimensional (except for the
1 2
12 brought out by the
H ( 12 , ) --we --we need to “tell” our math here that we have two ground-state degrees of freedom). However, recall
from problem 12.3.7, the treatment of the 2D isotropic SHO, the math “though” the same way, until we thought about the “angular degeneracy”, which told us n 2k m , where the m came from the angular solution. Making this replacement , we immediately conclude E (k 12 m 12 m 12 ) . This motivates rewriting [I.5] in a convenient form, 2
1 2 c
P x Py
2
E
2
m c 2 4
2
2
2
q B0 4c
2
2 2 (x y )
2
B0 q c
( z Lz )
H S z k 12 m 12 m 12
1 2
[I.7]
Including the rest-energy in the energy levels [I.7], we get, E 2 c2 k 12 m 12 m 12
1 2
m2c4
[I.8]
Solve for the exact levels of the Dirac particle in a uniform magnetic field B B0k A ˆ
1 2
B0 ( yi xj ) . ˆ
ˆ
Consult exercise 12.3.8, and write the equation for . Preliminaries: Dirac equation, in which the final step is made because { , } , definition of H
[ (p qc A)2 c2 (mc 2 )2 q ] i t [ i 2c 2 (mc 2 )2 q ] [c (p qc A ) mc 2 q ] ; q mv P c A i
q c
B;
(A )(B ) (A B )I i (A B ) ;
[I.9]
Ok, rewriting the Dirac equation as (i t mc 2 q ) [c (p cq A )] , in which , , we expand into 2x2 matrices of operators and spinors/antispinors,
i q mc2 cpi i qc2c ( y x x y ) O q c( p 2c ( yi xj )) [I.10] qc 2 cp ( y x ) O x y i q mc i i 2 c ˆ
ˆ
The two equations that result from this [I.10] are the coup led, 2 q i q mc ( 2 ( y x x y ) cpi i )
i q mc ( ( y x x y ) cpi i ) 2
q 2
[I.11]
