# Spin-½ and beyond: Measuring spin components other than ± ħ / 2: How to formulate the probability function?

It is my understanding that in quantum mechanics (for 1/2 spin particles) the

probability function that describes the direction of a particle's spin state is

proportional to the overlap of the theoretical hemispheres formed by the motion of

the particle (in either spin state). However, with this theory, it is also

imperative to assume that the only eigenvalues to describe a particle's spin state

are ± ħ / 2. If we do assume that the spin state of a particle is well defined,

even if it is not observed,how would we formulate a function to determine the

expected value of a spin state for particles with more than two spin states (such

as the 3/2 delta baryons)? Are the results in any way linked to an overlapping of

the hemispheres in the various spin states? In addition, if it were ever

necessary, could we formulate such a function for particles with an even higher

number of spin states?

Here we see a visual of this theory. The spin components of particles (a) and (b) have an angle between their vectors, $\phi$. Using this, we can see that the probability of finding both particles in a certain spin state is proportional to the overlap of their hemispheres.

$\textbf{These calculations are mine, but I thought that it made sense intuitively }$

$\textbf{}$

This is the basic Idea behind the Hemisphere mathematics I described to you:

The expected value will be denoted as $\langle\sigma^{(ab)}\rangle$ and:

$\langle\sigma^{(ab)}\rangle$ = $P(u,u) + P(d,d) − P(u,d) − P(d,u)$, where $(d)$ donotes down and $(u)$ donotes up.

This spin values are relative to a set field direction that corresponds to a Stern-Gorlach apparatus.

Consequently, we can assume that if we observe the value $+\cfrac\hslash2$, the spin of the particle will lie in the $"up"$ hemisphere, and contrarily, if we observe a value of $-\cfrac\hslash2$, our spin will reside in the $"down"$ hemisphere.

Now, if we denote angle between the spin states of 2 particles as $(\phi)$, the probability of observing both of the particles in a down state with be equivalent to $(P(u,u)=(P(d,d))=(\cfrac\phi\pi)$

Therefore, the proportionality coefficient has to equal (1) when $\phi=\pi$ and the probability of finding the particles in two different spin states will be $P(u,d) = x\phi + y$

This probability is equal to (1) when the two hemispheres coincide ($\phi$ = 0) and consequently $y = 1$ [note that this disappears if $\phi = \pi$]. Thus, $x = −\cfrac1π$ and $P(u,d) = P(d,u) = 1 − \phi$

Just like $j=1/2$ (spin-half) particles have two-complex-component wave functions, spinors, particles with spin $j$ have $(2j+1)$-dimensional wave functions describing the spin degrees of freedom. The dimension is what it is because $j_z$ always goes from $-j$ to $+j$ with the spacing equal to one.

All the transformation rules under rotations may be extracted from the representation theory of Lie groups, in this case $SU(2)\sim Spin(3)$, essentially $SO(3)$. The $(2j+1)$-dimensional representation for spin $j$ is the symmetric tensor products of $2j$ copies of the $j=1/2$ representation.

For $j=3/2$, we get a particular four-dimensional complex representation, i.e. four complex probability amplitudes describing the spin with respect to the $z$-axis (which may however be translated to amplitudes to any other axis, using well-defined rules).

For integer values of $j$, the complex amplitudes form a representation of $SU(2)$ that is also a representation of $SO(3)$. In particular, the 3-dimensional $j=1$ representation is completely equivalent to a (complexified) 3-component vector of $SO(3)$ while the 5-dimensional representation for $j=2$ is isomorphic the symmetric traceless tensor. Note that the symmetric $3\times 3$ table has 6 independent entries and one of them is removed by requiring the trace to vanish.

There are lots of relationships between different representations of different (or isomorphic) groups.

• Thank you for answering my question(which, essentially, you did quite well). What I am still curious about, however, is the overlapping of the theoretical hemispheres (made by the particle in various spin states), and its proportionality to the functions which you described. Aug 5 '14 at 6:35
• Maybe I don't really know the piece of maths you are asking about. Where did you hear about "theoretical hemispheres" as being relevant for these amplitudes? Aug 6 '14 at 18:07
• I edited the maths into the question. Please get back when you are not too busy. Aug 7 '14 at 0:16
• Sorry for adding it to the answer, but I wanted to make certain you saw it Aug 7 '14 at 0:17
• Apologize, I don't think that any of the mathematics you added is correct. There is no discontinuity of the actual probabilities happening on the boundary of the hemispheres, and probabilities never have any terms linearly proportional to $\phi$. Is that your own research or did you see it somewhere? Aug 7 '14 at 3:27