I want to find the matrix representation of $S_z$ operator in terms of the eigenkets of $S_y$ operator.

( It's Problem 1.24 (b) of Sakurai Second Edition )

One simple way to solve this problem is to map (relabel) axes [x,y,z] to [z,x,y] ( note that the order of the axes is still a right-handed system ). With these new axes, $|$1 0 $\rangle$ and |0 1 $\rangle$ are eigenkets are $S_y$ operator. And so ( by comparison ), the matrix representation of operator $S_z$ is $\frac{\hbar}{2} \sigma_x$

Please correct me if the above solution is wrong. I verified this solution in an online solution manual of the textbook. ( This is not my main question though.)

However, I am trying to get the same solution by another approach, but I am getting a different ( wrong ) answer. The other approach is as follows:

Let $S_z$ be the abstract operator for measuring the z-component of spin ( whose matrix representation we want to find ). I can insert the identity operator to the right and the left side of $S_z$ to get:

$ S_z = \mathbb{1} S_z \mathbb{1} $

Now we can expand $\mathbb{1}$ in terms of the eigenkets of $S_y$ operator:

$ S_z = | S_y;+ \rangle \langle S_y;+ | S_z | S_y;+ \rangle \langle S_y;+ | \quad + $

$ \qquad | S_y;+ \rangle \langle S_y;+ | S_z | S_y;- \rangle \langle S_y;- | \quad + $

$ \qquad | S_y;- \rangle \langle S_y;- | S_z | S_y;+ \rangle \langle S_y;+ | \quad + $

$ \qquad | S_y;- \rangle \langle S_y;- | S_z | S_y;- \rangle \langle S_y;- | $

So, using $S_y$ basis, the $S_z$ operator can be written in the following matrix representation:

$$S_z^{(rep. \thinspace in \thinspace y \thinspace basis)} = \begin{bmatrix} \langle S_y;+ | S_z | S_y;+ \rangle & \langle S_y;+ | S_z | S_y;- \rangle \\ \langle S_y;- | S_z | S_y;+ \rangle & \langle S_y;- | S_z | S_y;- \rangle \end{bmatrix}$$

I just need to compute these elements of the matrix to get the required matrix representation.

Now I will use a concept here about which I am quite skeptical ( my main question is regarding the validity of this concept). I think that we can calculate these matrix elements in ANY BASIS ( not necessarily in $S_y$ basis ) because $\langle \alpha| O | \beta \rangle$ will give me the same value irrespective of which basis I choose to calculate it. After all, $ \alpha, \beta $ are abstract states and O is an abstract operator. So $\langle \alpha| O | \beta \rangle$ should give the same value is all basis/representations.

So, I proceed ahead and try to calculate these matrix elements in the standard $S_z$ basis ( ideally I should be calculating it in $S_y$ basis).

In standard z basis, we have :

$$S_z = \begin{bmatrix} \hbar/2 & 0 \\ 0 & -\hbar/2 \end{bmatrix}$$ $$| S_y;+ \rangle = \begin{bmatrix} 1/\sqrt{2} \\ i/\sqrt{2} \end{bmatrix}$$ $$| S_y;- \rangle = \begin{bmatrix} i/\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix}$$

Using the above operators and states, I get this value for the representation of $S_z$ operator in $ | S_y ; \pm \rangle basis$ :

$$S_z^{(rep. \thinspace in \thinspace y \thinspace basis)} = -\frac{\hbar}{2} \sigma_y $$ which is $incorrect^*$.

I have spent good amount of time on this problem. Any help will be appreciated. To make my question concrete, here are the actual questions whose answers I want:

  1. Is my assumption ( given in boldface above ) correct?
  2. If it's incorrect, can you please tell me what's wrong with my understanding? Shouldn't the overlap of two states ( with a sandwiched operator ) be independent of basis?
  3. If you can find any other flaw in my understanding, that would also be very helpful.

[*]: As has been clarified in the answers below, this matrix representation is also correct. So the statement of my Stackexchange question is ( slightly ) wrong.


Your assumptions are not wrong, they are just incomplete. Let's start with the standard picture which is representing everything in the eigenbasis of $S_z$, meaning that $S_z = \hbar/2 \sigma^3$ using the Pauli matrices. The point here is that this is not enough to uniquely define $S_x$ and $S_y$. The reason is that any rotation about $S_z$ is also a valid choice for $S_x$ and $S_y$, while still respecting the right-handedness of the axes. That is, I can always define $S'_j = e^{i\theta S_z}S_j e^{-i\theta S_z}$ which will leave $S_z$ unchanged, but will change how the $x$ and $y$ axes are oriented. More explicitly, I can always do $$S_x = \frac{\hbar}{2} [\cos(\theta) \sigma^1 + \sin(\theta) \sigma^2]$$

$$S_y = \frac{\hbar}{2} [\cos(\theta) \sigma^2 - \sin(\theta) \sigma^1]$$ and everything remains ok.

This freedom of choice is encoded into the eigenstates by the arbitrary phase we can assign them. Indeed, a rotation about the $z$-axis will assign a phase to the eigenstates, which is physically insignificant.

If you want the "correct" result (the result you got is also correct, of course) you just have to choose $|S_y ; -\rangle = (|\!\uparrow\rangle - i |\!\downarrow\rangle)/\sqrt{2}$, which is multiplying your choice by $-i$.

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Take, to avoid needless complication, $\hbar=2$, and observe the isomorphism of the algebra of the Pauli matrices, $$ \sigma_x \equiv \sigma_x ', \qquad \sigma_y \equiv -\sigma_z ', \qquad \sigma_z \equiv \sigma_y ', $$ That is, the σ' matrices obey the very same Lie algebra (commutators) as the σ ones, and we could use the standard representation of one set to (unconventionally) represent the other set.

So $\sigma_z$ in the basis of eigenvectors of $\sigma_y$ will be the same expression as $-\sigma_y$ in the basis of eigenvectors of $\sigma_z$.

Note that $\sigma_y \sigma_z=i \sigma_x=i \sigma_x'= \sigma_y' \sigma_z' $ . How did you conclude your fine answer is "incorrect"? You just rotated by -π/2 around the x-axis.

As @yu-v 's fine answer illustrates, any other algebra isomorphism will also do, like the cyclic permutation one you started with.

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