Now ifψmis an eigenstate ofLz, thenLzψm=m ̄hψm, thus
Hψm=
μBB
̄h∗(m ̄hψm) = (mμBB)ψmHence the normalized eigenstates must be just those of the operatorLzitself, i.e., for the three
values ofm(eigenvalues ofLz), we have
ψm=+1=
1
0
0
ψm=0=
0
1
0
ψm=− 1 =
0
0
1
.
and the energy eigenvalues are just the values thatE=mμBBtakes on for the three values of m
i.e.,
Em=+1= +μBB Em=0= 0 Em=− 1 =−μBB.18.10.7A series of Stern-Gerlachs
Now that we have theshorthand notation for a Stern-Gerlachapparatus, we can put some
together and think about what happens. The following is a simple example in which three successive
apparati separate the atomic beam using a field gradient along the zdirection.
Oven(I 0 )→
+|
0
−
z(I 1 )→
+
0 |
−
z(I 2 )→
+|
0
−|
z(I 3 )→
If the intensity coming out of the oven isI 0 , what are the intensities at positions 1,
2, and 3?We assume an unpolarized beam coming out of the oven so that 1/3 ofthe atoms will
go into each initial beam in apparatus 1. This is essentially a classical calculation since we don’t
know the exact state of the particles coming from the oven. Now apparatus 1 removes them= 1
component of the beam, leaving a state with a mixture ofm= 0 andm=−1.
I 1 =
2
3
I 0
We still don’t know the relative phase of those two components and,in fact, different atoms in the
beam will have different phases.
The beam will split into only two parts in the second apparatus since there is nom= 1 component
left. Apparatus 2 blocks them= 0 part, now leaving us with a state that we can write.
I 2 =
1
3
I 0
All the particles in the beam are in the same state.
ψ=ψ(−z)