572
FALL PHYSICS 2007
Solution 8
1.(a) It is easy to see that B = z/2 z/2 + z = 0 , and B = (y/2 + y/2) + (x/2 x/2) = 0 .1 x y The magnitude of the magnetic eld is B= (x zx/2)2 + (y zy/2)2 + (B0 + (2z 2 x2 y 2 )/4)2 .
Assuming that B0 > 0, we expand B near the origin of R3 to the lowest order B B0 + Hence B = B0 + 1 2 2 B0 2 1 (x2 + y 2 ) + z 2 + higher order terms. 2 1 (2 x2 + 2 y 2 + 2B0 (2z 2 x2 y 2 )/4) . 2B0
Let 2 /B0 /2 = , then up to quadratic order in the coordinates 1 B = B0 + (x2 + y 2 + z 2 ) . 2 (b) S is unitary because SS = = = =
1 Actually,
1 2 1 2 1 2 1 2
1+bzi
bz 1+bz
1+bz+i
bz 1+bz
bz bz 1+bz+ 1+bz 1+bz (b z) (b z) 1+bz+ 1+bz 2 1 (b z) 1+bz+ 1+bz =1.
= 0 for any smooth function .
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And since Hmag = B, SHmag = 2 Then, SHmag S = ((1 + b z) B + B (b z) + 2 B (b z) (b z) B (b z) ) 1+bz ((1 + b z) B + 2B((b z)b z) 2 (b z) (b z) + B ) 1+bz ((1 + b z) B + 2B((b z)b z) + (1 b z) B) 2 (2(b z)( B) + 2(b z)( B) 2Bz ) 2 Bz . B (b z) 1+bz B+ 1+bz .
=
= = =
(c) We know that S
2
S =
2
+ 2S( S )
+ S(
2
S ),
Let A = S( S ), then ( + A) ( + A) = =
2 2
+ 2A + A + A A + 2S( S ) + (S( S )) + S( S ) S( S ) .
By the identity S( S ) = ( S)S , we can rewrite it as + 2S( S ) = + 2S( S ) 2 = + 2S( ) S = S 2 S .
2 2
+ (S( S )) ( S)S S( S ) + ( S) ( S ) + S( 2 S ) ( S) ( S ) + S( 2 S )
(d) Apply the unitary transformation S on H, combining (b) and (c), we get, SHS =
2
2M
(
+ A) (
+ A) B z .
Ignoring the vector potential (A), the Hamiltonian becomes SHS =
2 2
2M
B z .
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For a spin-down particle, SHS
2
=
2
2M
2 2
+ B 1 + B0 + (x2 + y 2 + z 2 ) 2
2M
It is a three-dimensional harmonic oscillator, so the ground state is M 3/8 M r2 ), (x, y, z) = ( 2 2 ) exp( 2 with the eigenvalue E0 = 2. Dene O1 = We need to prove | Oi = Oi | . For the rst operator, O1 | = = = With integral by part, O1 | = = dx d1 d2 + dx 2 dx 1 2 0 d dx
d dx
3 2
d dx
+ B0 . M , O2 = 0
d i dx d i dx 0
0 d dx
0
dx dx dx
0 d dx
d 2 dx
d dx
0
1 2 1 2
1 2
d 1 dx
d d 2 1 1 2 dx dx
dx 1
0
1 2
= | O1 . By the denition, O1 is Hermitian. Alternatively, set = , we get | O1 = O1 | = which implies the expected value, O1 = O1
| O1
,
.
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PHYSICS 572
FALL 2007
It is straightforward to repeat this process to prove O2 is Hermitian, O2 | = = = = = dx 0
d i dx
d i dx 0
1 2
d 1 dx
1 2
dx i d2 dx dx dx i
i
1 2
d d 2 1 i 1 2 dx dx d2 d1 +i i dx 2 dx 1 2 0
d i dx d i dx 0
dx 1
1 2
= | O2 . However, alternatively, since O2 = U O1 U where U= ei/4 0 0 ei/4
is a unitary matrix and O1 is Hermitian, O2 is also Hermitian.
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