An electron has two spins $1/2$ and $-1/2$
I wanted to represent them using bra-ket notation. I used the following…
\begin{align} 1/2 &\to |0\rangle \\ -1/2 &\to |1\rangle \end{align}
is this notation correct? Or did I do it backwards?
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$\begingroup$ Afaik, spin is not a straightforward concept, as there is no straight component with a simply computed value of 1/2 or -1/2 being the spin. Quantum mechanics using Schrodinger equation has no inherent spin solutions. Spin arose when using Dirac's equation with the wavefunction now each being 4-component quantities $\Psi = \begin{bmatrix}\Psi_1 \\ \Psi_2 \\ \Psi_3 \\ \Psi_4 \end{bmatrix}$. Then solutions of certain similarities occur when solving for the Dirac equation using a given constraint, which possess many similarities and only small differences that can be interpreted as "spin". $\endgroup$– JamesMay 30 at 20:06
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$\begingroup$ I think you might be mixing your concepts. Spin arises as a fairly straightforward result of angular momentum quantization, and is reasonably easy to visualize by shooting a proton or electron beam through a magnetic field (i.e. a Stern-Gerlach experiment). I believe the difference between fermionic/half-integer vs bosonic/integer spins only arise as solutions to the Dirac equation, and are a bit more complicated to motivate $\endgroup$– Chris EMay 30 at 22:37
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$\begingroup$ @ChrisE thank you. I am learning too and would welcome discussions on this. I was thinking about when we solve for the hydrogen atom (for instance chem.libretexts.org/Courses/University_of_California_Davis/…) and get solutions like $\Psi_{100} = ke^{-r}$, is the spin incorporated within these wavefunctions? Which of the parameters n, m, l represent the spin in these hydrogen solutions? $\endgroup$– JamesMay 31 at 0:51
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$\begingroup$ It's a little bit further down the page, but it's the $m_s$ quantum number discussed in "Electron Spin: The Fourth Quantum Number". I might have gotten my elementary QM mixed up (my bad) because indeed spin quantization is just asserted in that source since like a number of things in quantum mechanics, it's something we observed and that's given as its motivation. Apparently you do need the Dirac equation to actually motivate it mathematically. Derp, sorry about that $\endgroup$– Chris EMay 31 at 2:45
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Yup, you got it, although it is really only a matter of convention. Typically when talking about spins (at least spin-1/2's), we think about them physically in a basis of eigenvectors of $\hat{S_z}$: $|m_s=\frac{1}{2}\rangle, |m_s=-\frac{1}{2}\rangle$ with eigenvalues of $\frac{1}{2}, -\frac{1}{2}$ respectively (really $\frac{\hbar}{2}, -\frac{\hbar}{2}$ since spin values should have units of angular momentum, but we like to set $\hbar=1$ for convenience). In this way, $\hat{S_z} = \frac{\hbar}{2} \hat{\sigma_z}$, and we generally define computational basis states as the eigenvectors $|0\rangle, |1\rangle$ of $\hat{\sigma_z}$ with eigenvalues +1,-1 respectively.
