Solutions to Problems 4.2 and 4.3
4.2 [10] The commutation relations of the angular momentum operator Lz with the
momentum operator P were to be worked out in chapter 3, problem 3.
(a) [3] Using these commutation relations, derive two non-trivial uncertainty
relationships.
The commutation relations were
 Lz , Px   iPy ,
 Lz , Py   iPx ,
 Lz , Pz   0
According to the generalized uncertainty relation, we therefore have
Lz Px  12  Py , Lz Py  12  Px , Lz Pz  0
Since the left side of each of these expressions is the product of two positive numbers,
the last inequality doesn’t give us any information, but the other two inequalities suffice.
(b) [3] Show that if you are in an eigenstate of any observable, the uncertainty in
that observable is zero.
Let A be any observable, and a a normalized eigenstate with A a  a a .
Then
 A
2
 A2  A  a A2 a  a A a
2
2
 a a A a  a a a

2
 a2 a a  a2  0
(c) [4] Show that if you are in an eigenstate of Lz, then you must have
Px  Py  0 .
According to our inequalities we found above, if we are in an eigenstate of Lz,
then we have Lz  0 , and therefore
0  12  Py , 0  12  Px .
Since absolute values are never negative, it follows that Py  Px  0 , which implies
Px  Py  0 .
3. [10] We will eventually discover that particles have spin, which is described by
three operators S   S x , S y , S z  with commutation relations
 S x , S y   iS z ,  S y , S z   iS x ,
 S z , S x   iS y
A particle in a magnetic field of magnitude B pointing in the z-direction will
have Hamiltonian H    BS z where  and B are constants.
(a) [5] Derive formulas for the time derivative of all three components of S
We use the standard formula for the time evolution of any operator, namely
d
i
i B
i B
Sx   H , Sx   
 S z , S x    i S y   B S y ,
dt



d
i
i B
i B
 S z , S y   
S y   H , S y   
 i  S x    B S x ,
dt



d
i
i B
i B
Sz   H , Sz   
 S z , S z    0  0.
dt



(b) [5] At time t = 0, the expectation values of S are given by
Sx
t 0
 a,
Sy
t 0
Determine the expectation value S
t
 0,
Sz
t 0
b
at later times.
Since the time derivative of S z vanishes, this will just remain constant. The
expectation value of the other two operators, however, are related by
d S x dt   B S y , d S y
dt    B S x .
One way to proceed is to take another derivative of the first equation, which yields
d 2 Sx
dt 2   B d S y
dt    2 B 2 S x .
This suggests solutions along the lines of
S x   cos   Bt    sin   Bt 
Taking the derivative we can find
S y   sin   Bt    cos   Bt 
Our boundary conditions at t = 0 then tell us that   a and   0 . In summary, we
have
Sx
t
 a cos   Bt  ,
Sy
t
  a sin   Bt  ,
Sz
t
b