# Calculating $S_z$ operator in $S_y$ basis. What's wrong with my approach?

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.

## 2 Answers

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 $$-i$$.

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.