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This is a conceptual question about a problem in Sakurai. I understand how to solve the problem, but there's something about it that irks me, and it feels like I'm missing something.

In the problem, we seek to find the eigenvalues and vectors of matrix S-dot-n in the basis of the z-axis. Where n is defined as an arbitary vector described by $\beta$ and $\alpha$, where $\beta$ is the azumuthal angle and $\alpha$ is radial.

The goal is to find this: $$ \left |\text{s.n;+}\right\rangle = \cos{\frac{\beta}{2}}\left |\text{+}\right\rangle + e^{i \alpha}\sin{\frac{\beta}{2}}\left |\text{-}\right\rangle $$ We start out by constructing this vector, n, in a basis: $$ n = (\sin{\beta}\cos{\alpha}) \hat x + (\sin{\beta}\sin{\alpha}) \hat y + cos{\beta} \hat z $$

And now here's the step that irks me. We represent our each of our bases as their cooresponding Pauli spin matrices and construct our s.n matrix by this. $$ s.n = (\sin{\beta}\cos{\alpha}) \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) + (\sin{\beta}\sin{\alpha}) \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) + cos{\beta} \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) $$

So I understand the general idea here: We can decompose our n.s into an x, y and z component. And that for each component there is an associated eigenstate basis which can be re-represented in the direction of the z-axis's spin eigenstate basis (as the pauli matrices).

It's hard for me to explain, but something about this step (generating our new operator that finds the spin from our n-vector) really bothers me. Am I missing a step in my logic? Is it possible we can go from n to s.n a litte bit more formally (maybe by appealing to some particular axiom)?

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The operator $\vec s\cdot \vec n$ is just the inner product $$ \vec s\cdot \vec n = s_x n_x + s_y n_y + s_z n_z $$ Here, you substitute Pauli matrices for $s_x,s_y,s_z$ (they should be called $\sigma$, not $s$, but I adopted your conventions) and $$ (n_x,n_y,n_z) = (\sin\beta\cos\alpha,\sin\beta\sin\alpha,\cos\beta)$$ for the unit vector in the desired direction. The equation above is just the unit vector translated from polar to Cartesian coordinates.

The inner product gives you what you want if you substitute $\vec n=(1,0,0)$ or $(0,1,0)$ or $(0,0,1)$, namely one of the Pauli matrices, right? It works for the most general direction, too. The point is that the inner product with $\vec n$ simply gives you the projection of the vector to the axis. This is true for number-valued vectors as well as vectors of operators.

One may easily see that $\vec n\cdot \vec s$ has the eigenvalues $+1$ and $-1$, for example. If you don't see it, check it explicitly. Even though one of the vectors in the inner product is a "vector of operators", the inner product still inherits all the nice features from inner products, e.g. you may imagine that it is equal to $|s|\cdot |n| \cdot \cos \phi$ as long as you appreciate that $\cos\phi$ is an operator, too.

It means that the operator (two times two matrix) $\vec s\cdot \vec n$ is conjugate ("qualitatively the same, up to rotations") as $s_x$ or $s_y$ or $s_z$ themselves, and you may verify that what you wrote above is indeed one of the eigenvectors if you know how to calculate eigenvectors of a $2\times 2$ matrix.

There's something illogical about the way you present it. You seem to start with the result that you obtained somewhere and then ask about the first step of the derivation as if it were harder than the result itself. The right order is that you define the operator $\vec s\cdot \vec n$ which is the "counterpart" of $s_z$, but for a different axis, and then find its eigenvalues and eigenstates.

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  • $\begingroup$ +1 for this very nice and clear answer! From the answers I've seen from you, I think you explain QM very clearly. It would be nice e.g. to have your take on this discussion, where the current answers are so confusing and complicated... Best. $\endgroup$ – user929304 Dec 12 '14 at 17:08

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