# The physical meaning of Pauli Spin Matrices

1. Jun 7, 2008

### gn0m0n

Not really a specific homework problem, more of a conceptual problem that is going to come up again and again in problems:

OK, I understand the physical interpretation of spin and magnetic quantum number, so much as we can give one. That is, m_s = the component of s (or is it s^2? oh, dear....) in the arbitrarily chosen z direction.

Specifically, I am looking at Griffiths, p. 155-157. The eigenstates for spin 1/2 systems are easy enough to understand, although the derivations of them as actually presented in the context of magnetical orbital angular momentum through the lowering and raising operators was a bit obscure to me. I suppose I am even there struggling to recognize the physical significance of the operators (this is p. 147); if anyone can elucidate them at all I would much appreciate it.

Anyway, then we get to spin and Griffiths states that the "algebraic theory of spin is a carbon copy of the theory of orbital angular momentum," which I generally get. I also understand that we can then represent any spin 1/2 particle's state as a linear combination of the two vectors |1/2 1/2> and |1/2 -1/2>, ie, |s m> = ax(+) + bx(-) where x(+) is the spin-up eigenvector and x(-) is the spin-down eigenvector (right so far?). This can also be represented as the spinor (a b)^T (sorry, I'm making up how to convey the notation as I go along - this is the column vector with entries a, b) - now this essentially gives the coefficient of each basis vector? Ie, here, x(+) and x(-), right?

Fair enough, but then he says "the spin operators become 2x2 matrices"... can anyone expand on that? Is it that these operators are not conveying any action performed on the system but simply the observable, e.g. S_x=spin component in the x-direction? What do these operators act on? What about S(+) and S(-), the spin raising and lowering operators - these do not represent observables so what do they represent? I know they move us up and down the "ladder" of states but... how?

Again, it is mainly the Pauli spin matrices I want to understand better... For one thing, I don't understand how or why they are different, given that we choose the z-axis arbitrarily (although I do suspect that the indeterminacy principle has to do with it - perhaps once an axis is chosen we assume measurements are made first on that axis, which we call the z, then any subsequent spin measurements along either other axis will require these different matrices/operators?).

For x(+) we have simply (1 0)^T but for x(+)^(x) we have (1/sqrt(2) 1/sqrt(2)) ... why? What is physical meaning of this latter one? I know the physical meaning of first is that it is entirely "along the |1/2 1/2> vector", right? Now the latter one means we'd have equal probability of finding the spin in the x direction to be 1/2 or -1/2... but don't we also have equal probability (before we measure) to find spin in z direction to be 1/2 or -1/2?

As you can see, I'm going a little crazy here. Someone please save my sanity!

Sorry for the long post - I'm trying to be crystal clear.

We also first said that the generic spinor x= (a b)^T = ax(+) + bx(-) then also that x=((a+b)/sqrt(2))x(+)^(x) + ((a-b)/sqrt(2))x(-)^(x) So here we are representing the same generic spinor in terms of the same variable coefficients but with reference to different basis vectors, ie, eigenvectors for the different operator S_x? Actually, now that I think about it, were those first basis vectors eigenvectors for S_z?

Finally, Griffiths p. 157 says "if you measure S_x, the probability of getting h/2 is (1/2)|a+b|^2 and the probability of getting -h/2 is (1/2)|a-b|^2. You should check for yourself that these probabilities add up to 1." If anyone can elucidate this that would be great. I gather that the eigenvalues for the operators are possible values for the corresponding observables. But how do we use the quoted equations to obtain the probability for each?

I'm sorry if I posted in the wrong section - please help, and I'll get the hang of the forum - I'm new here! And any moderator can feel free to relabel the thread if desired. I posted also in general quantum forum - not sure which is better for this...

Thanks...!

2. Jun 11, 2008

### Mindscrape

Phew, you made a looong post. Don't worry, it's fine. Griffith's doesn't do a great job with spin, in my opinion, because he avoids Dirac notation and makes a mucky notation himself. I'll give it a shot though, anyone else feel free to correct.

m_s is a quantum number that takes the values of -1/2 and 1/2 for spin.

Operators don't have any physical significance in themselves. They raise and lower states to the next or last state, which is physically significant, but the operators are merely a mathematical tool.

Electron spin angular momentum is merely a special case of general angular momentum. Take l=1/2 and give it a special case name of s=1/2 and then take J and give it a special case name of S, so if s=1/2 then m_s=-s,-s+1,...,s-1,s=-1/2,1/2. You can always represent a general solution, a ket, as a linear combination of special states. Yes, you can represent any ket as a column vector, as Dirac showed in his state notation that the Schr. wave mechanics and Heis. matrix mechanics are two equivalent representations. So, it only makes sense that you can represent the state vector as a "spinor."

Think about what an operator does. It takes a state, and spits out another state. What happens when you multiply a 2x2 matrix by a 2x1 column vector? It's the operator doing its thing on a state, and returning another state. You now ask a lot of questions that would take a while to expand upon, so try and work with it and see if you can get anywhere.

Yes, you got it right in the later part. The z-axis does not an any way mean, for example, "up." It's an axis that is chosen arbitrarily, and thus defines the other axis to be relative to it. The S_z operator takes your spin angular momentum and projects it onto the z-axis. The z-axis is not a spatial axis!

I'm lost in the notation, sorry. As far as probability goes, just use the postulate.

Finally, yes, this is the right section.

3. Jun 11, 2008

### gn0m0n

Thanks for the reply! I actually posted here https://www.physicsforums.com/showthread.php?p=1760719 as well but it helps to hear a multitude of explanations. I know the L+ or S+ and S_ operators in some way move us between states rather than give a value for a physical observable, but as Griffiths gives it there is also the Sz operator, which returns the spin component in the z direction. The variety of uses for the operators confused me, but I'm getting the hang of it.

Thanks, everybody.