Let's deal with spin $1/2$ systems and fix a value (e.g. $+1/2$) in a given direction (e.g. $z$). Following a measurement along an orthogonal direction (e.g. $x$), we obtain 50% probability for $+1/2$ and 50% for $-1/2$. Fine.

Let's do the same with spin-one systems and fix the value $+1$ in the $z$ direction. When measuring the $x$ component, the probabilities turn out to be $1/4,1/2,1/4$ and not $1/3,1/3,1/3$ as one would naively extrapolated from spin one-half case.

Besides doing the correct calculation, is there any heuristic/intuitive argument to explain the difference?


1 Answer 1


I guess you can simply see this, if we use a classical comparison and treat the spin, as it would rotate on sphere. Now quantum mechanics tells the spin, that only certain orientations are allowed. In a spin-one system these are the spins I illustrated in the picture below.

spins on sphere for the spin-one system

If you now want to measure the spin component in the $z$ direction you can see, that there is one configuration with $S_z = 1$ and one configuration with $S_z = -1$, but two possibilities for $S_z = 0$. So we get a $\frac{1}{4}$, $\frac{1}{4}$, $\frac{1}{2}$ distribution as you can calculate. As mentioned, this is a classical analogon and explains in no way the correct quantum mechanics, but it can give us a picture, why this is happening.


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