A. Proof.
B:
A.12. Proof.
3.11. 0.16
3.38.
(a). Eigenvalues of H: hbar w,
2 hbar w,
2 hbar w.
Eigenvalues of A: 2 l,
l,
- l.
Eigenvalues of B: 2 m,
m,
- m.
Eigenvectors of H:
(1) (0) (0)
(0) & (1) & (0)
(0) (0) (1)
[the last two can actually be any orthonormal linear combination of
the ones shown]
Eigenvectors of A:
(0) (1) ( 1)
(0) & [1/Ö2](1) & [1/Ö2](-1)
(1) (0) ( 0)
Eigenvectors of B:
(1) (0) ( 0)
(0) & [1/Ö2](1) & [1/Ö2]( 1)
(0) (1) (-1)
(b).<H> = hbar w ( |c1|2 + 2 |c2|2 + 2 |c3|2)
<A> = l. ( c1* c2 + c2* c1 + 2 |c3|2). [How do you know that this is real?]
<B> = m ( 2 |c1|2 + c2* c3 + c3* c2).
(c).
( c1 exp(iwt) )
|S(t)> = exp(-2wt) ( c2 )
( c3 )
For each operator, the possible values of a measurement are the values of the eigenvalues of the operator. The probabilities:
H: |c1|2 and |c2|2 + |c3|2
A: |c3|2, (1/2) [ |c1|2 + |c2|2 + 2 Re (c1* c2 e -iwt )], (1/2) [ |c1|2 + |c2|2 - 2 Re (c1* c2 e -iwt )].
B: |c1|2, (1/2) [ |c2|2 + |c3|2 + 2 Re ( c2* c3)], (1/2) [ |c2|2 + |c3|2 - 2 Re ( c2* c3)],
[The middle probability for B does not agree with the answer in the solutions manual, but I am rather sure that it is correct. The solutions-manual result is definitely a misprint since the manual's probabilities do not sum to 1.]
| Last revised 2005/04/06 |