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I'm trying to get an idea of what spin in quantum mechanics means. I have the following questions regarding spin:

The eigenvalues of the operator corresponding to the z component of spin is $\hat{S}_{z}$ with eigenvalues $\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$ respectively. We then have two corresponding eigenvectors $|\frac{1}{2}(\frac{1}{2}) \rangle$ and $|\frac{1}{2}(-\frac{1}{2}) \rangle$. As I understand from the postulates of QM, any quantum state can be expressed as a linear combination of these eigenvectors. What I don't understand is how we know that these eigenvectors can be expressed as $\chi_+ = \left( \begin{array}{c} 1\\ 0\\ \end{array} \right) \text{ and } \chi_- = \left( \begin{array}{c} 0\\ 1\\ \end{array} \right)$?

What exactly does it mean for a state to be in z spin up state $\chi_+ = \left( \begin{array}{c} 1\\ 0\\ \end{array} \right)$ where it has eigenvalue $\frac{\hbar}{2}$? Is the following the correct interpretation: Does it simply mean that the spin around the z axis can take only two values $\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$ which corresponds to counter clockwise spin around the $\hat{z}$ axis (spin up) so that the vector points up the z axis and respectively counter clockwise spin around the $- \hat{z}$ axis (spin down) so the vector points down the z axis?

If this is the correct interpretation then how does the form of the eigenvectors $\left( \begin{array}{c} 1\\ 0\\ \end{array} \right) \text{ and } \left( \begin{array}{c} 0\\ 1\\ \end{array} \right)$ correspond to a vector up the z axis and one down the z axis?

Thanks for any assistance.

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  • $\begingroup$ The $z$-axis isn't the spatial $z$-direction that you may think of: it is one of the three components of the spin and has no connection at all with the spatial dimensions. There is no (counter)-clockwise rotation anywhere and the state being a composition of the spin eigenstates work like any other state being the composition of eigenstates of any other operator (under the assumption that the operators spectra span the entire state space). $\endgroup$ – gented Nov 22 '16 at 19:04
  • $\begingroup$ @GennaroTedesco Okay thanks, if you have a chance please see the question in my post. $\endgroup$ – Alex Nov 22 '16 at 19:09
  • $\begingroup$ @GennaroTedesco What do these components of spin of a particle refer to? $\endgroup$ – Alex Nov 22 '16 at 19:10
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    $\begingroup$ Spin is a set of three observables that any particle possesses that happen to have the same commutation relations as the orbital angular momentum; that is why in the literature their components are referred to as $(x,y,z)$. In general you can consider it as any set of observables $A_i$ that you use to describe a quantum system. $\endgroup$ – gented Nov 22 '16 at 19:15
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Good question---that's awesome you are studying quantum mechanics---keep it up!

I think it may help to understand what $|\frac{1}{2} \frac{1}{2}\rangle$ and $|\frac{1}{2} \frac{-1}{2}\rangle$ mean. For example, I have a state $|\Psi \rangle = |\frac{1}{2} \frac{1}{2}\rangle$ . What is the inner product of $\Psi$ with respect to $|\frac{1}{2}\frac{1}{2}\rangle$ (i.e., $\langle\frac{1}{2}\frac{1}{2}|\Psi\rangle$? What is the inner product of $\Psi$ with respect to $|\frac{1}{2} \frac{-1}{2}\rangle$ (i.e., $\langle\frac{1}{2}\frac{-1}{2}|\Psi\rangle$)? (If you cannot answer these questions, let me know and I will alter my teaching method.)

Now, think about a putatively different example. Suppose I am given a vector $\vec{r} = \hat{x} + \hat{y}$. How would you represent this vector in Cartesian coordinate? How about the vector $\vec{r} = \hat{x}$?

Now try to find a relationship between the two different examples above. Did you answer your own question?

Once you understand this mathematical representation of quantum states, you can begin to answer your second question: The quantum state, $\Psi = | \frac{1}{2}\frac{1}{2}\rangle$ is an eigenstate of the angular momentum, spin-z component with eigenvalue equal to $\frac{\hbar}{2}$. Thus, in this quantum state, when the spin-z angular momentum component is measured, you will obtain the value $\frac{\hbar}{2}$.

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  • $\begingroup$ I think I understand your point thanks. The $(1 , 0)$ just corresponds to the component in a given basis which is not explicitly shown in the notation. In this case the basis would be $| \frac{1}{2}(\frac{1}{2}) \rangle$. $\endgroup$ – Alex Nov 22 '16 at 19:49
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It's the convention.

The spin for the electron was detected experimentally - the math came later.

A non-homogeneous magnetic field (or a magnetic filter) picks out the z direction - which has 2 eigenvalues, namely, $+\hbar/2$ and $-\hbar/2$.

Simply having one eigenvector pointing up and the other one pointing down along the same axis is odd.

But if you haven't tried to predict the outcomes in the Stern-Gerlach experiment by blocking one of the beams or rotating the magnetic, things get weirder.

See the Stern-Gerlach experiments in the Feynman lecture series chapter 5 and chapter 6.

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