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I was wondering how does one go about solving for the spin (1/2) eigenstates in an arbitrary direction?

Let me specify my question. I had seen previously (such as Spin operator in an arbitrary direction) how to calculate such a problem when we are given the unit vector in spherical coordinates: ̂=(sincos,sinsin,cos) (which actually makes quite a lot of sense to me).

But, however improper my question may seem, I am quite unsure as to how would I apply this to a general case where I'm simply given any unit vector (not necessarily in spherical coordinates).

For example, let's say we are given a vector, for simplicity, $v= \tfrac1{\sqrt9}(1,2,2)$. What would be the spin in this direction?

Would I simply have to dot product it with the Pauli matrices (with an extra factor of $\hbar/2$) and then solve for the eigenvalues and eigenstates? How would I display this using the z-basis eigenstates (would I simply apply a change of basis?).

My main confusion seems to revolve around how regardless of where I've searched, all answers seem to be of the form

enter image description here

which, I know how to derive using the spherical unit vector, but am completely confused as to why it would apply to a general unit vector that might not necessarily be described using spherical coordinates.

I'd greatly appreciate any guidance or assistance.

m_1265
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2 Answers2

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As stressed in the comment, you first need to rectify your wrong (math) expression to the physics convention, ̂=(sincos, sinsin, cos) so $$ \hat n \cdot \vec \sigma /2 =\tfrac{1}{2}\begin{pmatrix} \cos \theta & \sin\theta e^{-i\phi} \\ \sin\theta e^{i\phi}&-\cos\theta \end{pmatrix} $$ with eigenvectors $$ \begin{pmatrix} \cos \theta/2 \\ \sin\theta/2 ~ e^{i\phi} \end{pmatrix}, \qquad \begin{pmatrix} \sin \theta/2 \\- \cos \theta/2~ e^{i\phi} \end{pmatrix} . $$ respectively, for eigenvalues $\pm 1/2$.

Likewise, for =0 in your comment, you have eigenvectors $$ \begin{pmatrix} \cos \theta/2 \\ \sin\theta/2 \end{pmatrix}, \qquad \begin{pmatrix} \sin \theta/2 \\- \cos \theta/2 \end{pmatrix} . $$

At no point do you have to use spherical coordinates, even though you could. For your unit vector $\hat n = (1,2,2)^T /3$ you have $$ \hat n \cdot \vec \sigma =\tfrac{1}{3}\begin{pmatrix} 2& 1-2i \\ 1+2i&- 2 \end{pmatrix} $$ with evident eigenvectors $$ {1\over \sqrt 6} \begin{pmatrix} 1 \\ -1-2i \end{pmatrix} , \qquad {1\over \sqrt 6}\begin{pmatrix} 1-2i \\ 1 \end{pmatrix} . $$ respectively, for eigenvalues $\mp 1$. You don't need to connect to the above, even though you could.

Cosmas Zachos
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$$n_x = \sin\theta\cos\phi $$ $$n_y = \sin\theta\sin\phi $$ $$n_z = \cos\theta $$

Invert those for polar and azimuth angles. A vector is just a vector, and doesn't matter which coordinated you use to write down its representation.

Note that the term "spherical vector" refers to a completely different (complex) basis:

$$(\hat e_{-1}, \hat e_0, \hat e_{+1})$$

where the $\hat e_m$ are eigenvalues of $z$-rotations with eigenvalue $e^{im\phi}$.

JEB
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