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$\newcommand{\ket}[1]{\left| #1 \right>}$The problem is to write the ket vector for a particle with spin +1/2 along the x axis, in terms of the standard basis vectors $\ket{+1/2}$ and $\ket{-1/2}$ along the z axis.

This page gives the rotation matrix about y axis as:

\begin{pmatrix} \cos (\theta/2) & \sin(\theta/2)\\ -\sin (\theta/2) & \cos(\theta/2) \end{pmatrix} So I figure if I just rotate the vector $\left(\begin{smallmatrix} 1 \\0\end{smallmatrix} \right) $ 90 degrees around the y axis, that will produce the answer. (Since a z oriented vector rotated 90 degrees about the y axis produces an x oriented vector).

But multiplying the above matrix by the vector $\left(\begin{smallmatrix} 1 \\0\end{smallmatrix} \right) $ gives $\left(\begin{smallmatrix} \cos(90^\circ/2) \\-\sin(90^\circ/2) \end{smallmatrix} \right) $

which is $$\begin{pmatrix} \cos(45^\circ) \\ -\sin(45^\circ)\end{pmatrix} = \begin{pmatrix} 1/\sqrt2 \\ -1/\sqrt2\end{pmatrix} $$.

But the book says the answer is $\hbar/2 \cdot \ket{-1/2}$ which I believe is the same as $\hbar/2 \cdot \left(\begin{smallmatrix} 0 \\1\end{smallmatrix} \right) $ . What is wrong?

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Are you sure that's what the book is asking you to find?

$\hbar/2$ is the eigenvalue of the $S_{x}$ operator corresponding to spin up, but it is not part of the state vector. If the question is really asking you to express the $\mid S_{x};+\rangle$ ket in the $S_{z}$ basis, then you're nearly correct, just a minor sign error:

$$\mid S_{x};+\rangle = \frac{1}{\sqrt{2}}\mid+\rangle + \frac{1}{\sqrt{2}}\mid-\rangle$$

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  • $\begingroup$ I think you're right that I misread the question. Here is their solution: $\endgroup$
    – a00
    Commented Jun 16, 2015 at 18:24
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    $\begingroup$ Yes, that's right. $\mid \pm \rangle$ = $\mid (\pm1/2) \rangle$. $\hat S_{x}\mid (+1/2) \rangle$ corresponds to measuring the spin along the $x$-axis when the particle's spin is oriented along the positive $z$-axis. First they express the $\hat S_{x}$ operator in the ($\mid (\pm1/2) \rangle$) basis $$\hat S_{x} = \hbar/2 \mid (+1/2) \rangle \langle (-1/2) \mid + \hbar/2 \mid (-1/2) \rangle \langle (+1/2) = \hbar/2 \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}$$ where they use the matrix representation of the operator. Then they apply the operator to the state vector $\mid (+1/2) \rangle$. $\endgroup$
    – Kyle Arean-Raines
    Commented Jun 16, 2015 at 19:32
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    $\begingroup$ Note that this equivalent to $$\hat S_{x}\mid (+1/2) \rangle = \hbar/2 \mid (+1/2) \rangle \langle (-1/2) \mid (+1/2) \rangle + \hbar/2 \mid (-1/2) \rangle \langle (+1/2) \mid (+1/2) \rangle $$ Since the $\mid (\pm1/2) \rangle$ vectors are orthonormal, $\langle (-1/2) \mid (+1/2) \rangle = 0$ and $\langle (+1/2) \mid (+1/2) \rangle = 1$, giving $$\hat S_{x}\mid (+1/2) \rangle = \hbar/2 \mid (-1/2) \rangle$$ I still don't know what the problem statement is, but hopefully this provides some clarity. You might find it useful to read about the Pauli matrices. $\endgroup$
    – Kyle Arean-Raines
    Commented Jun 16, 2015 at 19:33
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    $\begingroup$ Again, the solution your book provides is not the answer to the question you're asking. The book solution corresponds to "the outcome of a measurement along the x-axis of a particle that is spin up in the z basis." This is not the same as what you're asking, which is to write the expansion of spin up along the x-axis in the z basis, which is the result I've given. Your approach is entirely correct, you just made a sign error in the second coefficient. $\endgroup$
    – Kyle Arean-Raines
    Commented Jun 17, 2015 at 14:09
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    $\begingroup$ Okay, the problem statement seems pretty clear. No need to expand the kets into different bases. You can just multiply the ket in the $z$ basis by each of the Pauli matrices, which are the spin operator matrices in the $z$ basis. Hope that clears things up :) $\endgroup$
    – Kyle Arean-Raines
    Commented Jun 17, 2015 at 14:13
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I think you could work it like this:

$X_+ ={1 \over \sqrt{2}} (\begin{matrix} 1 \\ 1 \end{matrix}) =a (\begin{matrix} 1 \\ 0 \end{matrix} ) +b(\begin{matrix} 0 \\ 1 \end{matrix} ) $.where $X_+$ is the eigenvector on the positive axon of $S_x$

Solve and find a,b and there you are.

Note also that you can write a general spinor as $(\begin{matrix} cos\theta /2 \\ sin\theta /2 \cdot e^{ι \phi} \end{matrix} ) $ where $\theta$ is the angle in zy plane starting from z and $\phi$ is the angle at xy plane starting from x.

Hope this helps.

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  • $\begingroup$ Where does $X_+ = 1/\sqrt 2 \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ come from, please? $\endgroup$
    – a00
    Commented Jun 16, 2015 at 18:39
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    $\begingroup$ Hi. If I'm not mistaken that's the eigenvector of the operator $S_x $ on the positive x axon. On the the other side there is the vector at z axon. Here, have a look: quantummechanics.ucsd.edu/ph130a/130_notes/node267.html If I understood your question well, you want to express the spinor on x on the z basis. That's what my answer is showing, but you have to calculate the coefficients a and b. $\endgroup$
    – Constantine Black
    Commented Jun 17, 2015 at 7:21
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    $\begingroup$ Another approach: In general, if you you know the form of an operator A in a basis b, where the b's are eigenvectors some operator B and A and B don't commute, you can solve the eigenvalue problem $A a = \lambda a$ to find the a's in terms of the b's. In this problem you know the representation of $S_{x}$ in the $z$ basis - it's just the Pauli matrix $\sigma_{x}$. Solving the eigenvalue problem will give the eigenvectors of $S_{x}$ in the $z$ basis, and then it's a matter of finding the coefficients a and b as in Constantine's answer. This is done by requiring that $|a|^{2} + |b|^{2} = 1$ $\endgroup$
    – Kyle Arean-Raines
    Commented Jun 17, 2015 at 12:11
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    $\begingroup$ @user4127427 If you have time, you could have a look at the site I commented above. There you will see exactly how you find the eigenvectors of a spin operator-as you said. From there you do what my answer says and you have your solution. Note that there other exercises were you will have to find the the x spinor as a superposition of the y spinor and so on. Happy I helped. $\endgroup$
    – Constantine Black
    Commented Jun 18, 2015 at 8:41
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    $\begingroup$ Not quite. When you apply an operator to a ket, you can either express the operator in the ket basis or express the ket in the operator basis (expand the ket in terms of the eigenkets of the operator). Keep at it, these concepts can take a while to sink in :) $\endgroup$
    – Kyle Arean-Raines
    Commented Jun 18, 2015 at 17:21

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