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How the Universe Expands
Large-Scale Structure
 > Spin and Measurement
Quantum Teleportation

## Spin and Measurement

When I was writing about polarization (in the abstract sense), I said something slightly misleading.

Say you have a particle with spin ½ with its spin pointing in some horizontal direction. If you measure the component of spin in the vertical direction, you don't get zero, as you'd expect; you get one of the two values ±½, at random. If, however, you measure the same component again, you always get the same value as the first time. Strange but true!

The problem is, the spin is never really pointing entirely in the horizontal direction. So, what's strange isn't that a vertical component appears out of nowhere; what's strange, as usual, is quantum mechanics.

I should really just shut up now, but I've already written most of the rest of the essay, and I might as well finish it. Also, I thought I knew all about spin, but I found a way of thinking about it that was new to me, so perhaps it will be new to somebody else too. So here goes.

When talking about a particle with spin ½, it's customary to talk about its spin pointing in this direction or that, up or down or whatever, but that doesn't necessarily mean what you think it means. What it really means is that the particle is in the appropriate eigenstate of S · d, where S is the spin operator and d is the direction.

Just for definiteness, let's pick a horizontal direction, say, along the x axis, and imagine a particle in the appropriate eigenstate of Sx. Now, on the one hand, in that state the expectation value of the spin, <S>, does point along the x axis.

<Sx> = 1/2
<Sy> = 0
<Sz> = 0

On the other hand, the expectation values of the squared components are all nonzero; in fact, they're identical.

<Sx2> = 1/4
<Sy2> = 1/4
<Sz2> = 1/4

Knowing both sets of quantities, we can say something about the underlying probability distributions: Sx has mean 1/2 and standard deviation zero, while Sy and Sz have mean zero and standard deviation 1/2.

In fact, we know exactly what the probability distributions are, because they're what we'd get if we performed the corresponding measurements. Here's what they look like.

So, we can think of the spin as having magnitude 1/2 in all three directions. That's why it is not so surprising that a measurement in the vertical direction produces a nonzero result.

Actually, that's still not quite right. The components of the spin operator don't commute with each other, so, according to the uncertainty principle, we shouldn't even think of the spin as having more than one component at the same time. Still, it is true that the overall magnitude of the spin is (root 3)/2, not 1/2, and that extra spin has to go somewhere … right?