The beam in apparatus 3 all goes along the same path, the lower one.Apparatus 3 blocks that path.
I 3 = 0The following is a more complex example using a field gradients in the z andx directions (assuming
the beam is moving in y).
Oven(I 0 )→
+
0 |
−|
z(I 1 )→
+|
0 |
−
x(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?
Now we have a Quantum Mechanics problem. After the first apparatus, we have an intensity as
before
I 1 =1
3
I 0
and all the particles are in the state
ψ(+z)=
1
0
0
.
The second apparatus is oriented to separate the beam in the x direction. The beam separates into
3 parts. We can compute the intensity of each but lets concentrate on the bottom one because we
block the other two.
I 2 =∣
∣
∣〈ψ
(x)
−|ψ(z)
+〉∣
∣
∣
2
I 1We have written the probability that one particle, initially in the the stateψ(+z), goes into the state
ψ−(x)whenmeasuredin the x direction (times the intensity coming into the apparatus). Lets
compute that probability.
〈ψ(−x)|ψ(+z)〉=(
−^12 √^12 −^12
)
1
0
0
=−^1
2
So the probability is^14.
I 2 =1
4
I 1 =
1
12
I 0
The third apparatus goes back to a separation in z and blocks them= 1 component. The incoming
state is
ψ−(x)=
−^12
√^1
2
−^12
Remember that the components of this vector are just the amplitudes to be in the differentmstates
(using the z axis). The probability to get through this apparatus is just the probability to be in the
m= 0 beam plus the probability to be in them=−1 beam.
P=
∣
∣
∣
∣−
1
√
2
∣
∣
∣
∣
2
+∣
∣
∣
∣
1
2
∣
∣
∣
∣
2
=