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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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    Just transform your arbitrary unit vector to spherical coordinates and use the formula you have. What's wrong with that? – Mauricio Apr 14 '23 at 16:44
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    Also, what do you take as the starting point and an acceptable method to derive the solution? Would you accept doing the "dot product" of the unit vector with the Pauli matrices and then diagonalising? Or would you only be happy (like me) with a derivation that starts from analysing Stern-Gerlach experiment with rotated measuring axes, and slowly deriving that the eigenfunctions behave in specific ways, thereby deriving the same results as simpler ways to do things? – naturallyInconsistent Apr 14 '23 at 17:42
  • @naturallyInconsistent I would gladly accept doing the "dot product" of the unit vector with the Pauli matrices and then diagonalizing as the start of a derivation! – m_1265 Apr 15 '23 at 23:12
  • @Mauricio Valid point; I guess that would work too. Oh, one more quick question: what if I was given a unit vector such as n= (sin,0,cos), would n(x)= sin, n(y)= 0, and n(z)=cos, or am I doing something wrong? The confusion I have is, since this is already in spherical coordinates, how would I format it into the generic unit vector that I had listed above (that's also in spherical coordinates); I really do apologize for asking another dumb question, but I've had this in the back of my mind for a while now. Many thanks! – m_1265 Apr 15 '23 at 23:20
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    Would I simply have to dot product it with the Pauli matrices [...] and then solve for the eigenvalues and eigenstates? - correct.

    Maybe it will help if I expand a little bit: the state of internal states is $2-$dimensional, since it is a spin-$\tfrac 1 2$ particle. When you say 'spin in $(x,y,z)$ direction', what you really mean is 'how do I decompose the particle state into a linear combination of $\pm$ eigenstates under a rotation in the axis $(x,y,z)$?' Well, as you suggested, you can construct the operator which does that rotation, then solve for its eigenvectors.

     – Myridium Apr 16 '23 at 22:04

2 Answers2

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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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  • Thank you for your answer! However, I am aware of the components that you listed above are the components of a unit vector in spherical coordinates. The main question that I had was, "what if I was given a unit vector in cartesian coordinates and didn't want to convert it to spherical coordinates: how would I find the spin states"? For example, what if I had =(1/sqrt(9))(1,2,2) or something of the sort? – m_1265 Apr 15 '23 at 23:15
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    first: $\sqrt 9 =3$, second write $v = (\sin\theta\cos\phi, \sin\theta\sin\theta, \cos\theta)$ with $\theta=0.841\ldots$ and $\phi=1.107\ldots$. – JEB Apr 16 '23 at 16:47
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    @m_1265 that is even easier. We expressed as spherical polars so that it can be easier to analyse. What is actually done is that the spherical polars were expressed as Cartesian and then the Pauli matrices were used. If you have $\hat{\vec n} = \frac13(1,2,2)$ then you just need to find the eigenvectors of $\frac13\begin{pmatrix}2&1-2i\1+2i&-2\end{pmatrix}$ – naturallyInconsistent Apr 17 '23 at 04:44
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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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