# Matrix representation of the operator $\hat{S}_x$ in the standard basis

From the form you've given of the operator $$S_x$$ and the basis vector you've given, you can easily calculate the matrix representation of it. Since

$$S_x = \frac{\hbar}{2}(|\uparrow\rangle\langle\downarrow|+|\downarrow\rangle\langle\uparrow|)$$

and

$$|\uparrow\rangle = \left(\begin{matrix}1\\0\end{matrix}\right)\implies\langle\uparrow| = \left(1\;\;\;\;0\right)\\ |\downarrow\rangle = \left(\begin{matrix}0\\1\end{matrix}\right)\implies\langle\downarrow| = \left(0\;\;\;\;1\right)$$

using simple matrix multiplication you get

$$|\uparrow\rangle\langle\downarrow| = \left(\begin{matrix}1\\0\end{matrix}\right) \left(0\;\;\;\;1\right) = \left(\begin{matrix} 0&1\\ 0&0 \end{matrix}\right)\\ |\downarrow\rangle\langle\uparrow| = \left(\begin{matrix}0\\1\end{matrix}\right) \left(1\;\;\;\;0\right) = \left(\begin{matrix} 0&0\\ 1&0 \end{matrix}\right)$$

and so that the matrix representation of the operator is just

$$S_x = \frac{\hbar}{2}\left(\begin{matrix} 0&1\\ 1&0 \end{matrix}\right)$$

which is just one of the Pauli matrices. This also gives the result of elements of the matrix you gave which are indeed correct.

## Bonus matrix multiplication

I find that many people don't get how to do matrix multiplication with simple vectors so i wanted to give an explanation to everybody that found this answer in a colorful way. I'll evaluate only one of the two matrices in the answer

$$|\uparrow\rangle\langle\downarrow| = \left(\begin{matrix}1\\0\end{matrix}\right) \left(0\;\;\;\;1\right) = \left(\begin{matrix} 0&1\\ 0&0 \end{matrix}\right)$$

The multiplication is done using row by column multiplication. First step we take the first row element of the first vector and multiply it by the first element of the first column of the second vector

$$|\uparrow\rangle\langle\downarrow| = \left(\begin{matrix}\color{red}{1}\\0\end{matrix}\right) \left(\color{red}{0}\;\;\;\;1\right) = \left(\begin{matrix} \color{red}{0}&1\\ 0&0 \end{matrix}\right) \qquad \text{First row - first column}$$

and so on for the remaning elements

$$|\uparrow\rangle\langle\downarrow| = \left(\begin{matrix}\color{green}{1}\\0\end{matrix}\right) \left(0\;\;\;\;\color{green}{1}\right) = \left(\begin{matrix} 0&\color{green}{1}\\ 0&0 \end{matrix}\right)\qquad\text{First row - second column}\\ |\uparrow\rangle\langle\downarrow| = \left(\begin{matrix}1\\\color{blue}{0}\end{matrix}\right) \left(\color{blue}{0}\;\;\;\;1\right) = \left(\begin{matrix} 0&1\\ \color{blue}{0}&0 \end{matrix}\right)\qquad\text{Second row - first column}\\ |\uparrow\rangle\langle\downarrow| = \left(\begin{matrix}1\\\color{orange}{0}\end{matrix}\right) \left(0\;\;\;\;\color{orange}{1}\right) = \left(\begin{matrix} 0&1\\ 0&\color{orange}{0} \end{matrix}\right)\qquad\text{Second row - second column}$$

Hope it'll be useful to somebody

Yes,it is possible. Your representation is correct. It is easily to check using explicit form of $$\hat{S}_x$$:

$$\begin{pmatrix} \langle\uparrow|\hat S_x|\uparrow\rangle & \langle\uparrow|\hat S_x|\downarrow\rangle\\ \langle\downarrow|\hat S_x|\uparrow\rangle\ & \langle\downarrow|\hat S_x|\downarrow\rangle\end{pmatrix} = \frac{\hbar}{2} \begin{pmatrix} 0 & 1\\ 1& 0 \end{pmatrix} = \frac{\hbar}{2} \sigma_x$$