SME
Quantum Mechanics · Problem type
Measuring spin along a tilted axis
Use this for “An electron leaves a Stern–Gerlach magnet spin up along z and enters a second magnet tilted 120° from z. Find the chances of up and down and the average reading.”
Chart
The bars add to 1.
⟨S⟩ = −1 × 0.25 + 1 × 0.75 = 0.5
This is an example. Type your own values to start a new problem.
P(up)
=
cos²(
θ
°
÷
2)
=
P(down)
Chance of down
calculated
⟨S⟩
Average reading
calculated · ħ/2
Formulas
P(up)
=
(cos(
θ
2
))²
0.75 = (cos(60° ÷ 2))²
P(down)
=
(sin(
θ
2
))²
0.25 = (sin(60° ÷ 2))²
⟨S⟩
=
cos(
θ
)
0.5 = cos(60°)
Assumptions
•
A spin-½ particle prepared up along one axis, then measured along another axis θ away from it.
•
Each measurement gives only +ħ/2 (up) or −ħ/2 (down); the chances are cos²(θ/2) and sin²(θ/2).
•
⟨S⟩ is the average over many particles, in units of ħ/2: (+1)P(up) + (−1)P(down) = cos θ.
Step-by-step
WE KNOW
θ = 60°
FIND
chance of up (P(up)), chance of down (P(down)), average reading (⟨S⟩)
Step 1 · Find chance of up (P(up))
Use P(up) = (cos(θ ÷ 2))²
The
up
state
along
the
new
axis
overlaps
the
old
one
by
cos(
θ
2
);
square
it
for
the
chance.
P(up) = (cos(60° ÷ 2))²
P(up) = (cos(30°))²
P(up)
=
0.866
2
P(up) = 0.75
Step 2 · Find chance of down (P(down))
Use P(down) = (sin(θ ÷ 2))²
The
down
state
along
the
new
axis
overlaps
the
old
up
state
by
sin(
θ
2
).
P(down) = (sin(60° ÷ 2))²
P(down) = (sin(30°))²
P(down)
=
0.5
2
P(down) = 0.25
Step 3 · Find average reading (⟨S⟩)
Use ⟨S⟩ = cos(θ)
Average
the
readings:
P(up)
−
P(down)
=
cos²(
θ
2
)
−
sin²(
θ
2
)
=
cos
θ
.
⟨S⟩ = cos(60°)
⟨S⟩ = 0.5 ħ/2
Check
Put the numbers back in. Both sides should match.
0.75 = (cos(60° ÷ 2))² ✓
0.25 = (sin(60° ÷ 2))² ✓
0.5 = cos(60°) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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