I have recently been introduced to the idea of spectral decomposition of spin angular momentum operators in Quantum-Mechanics.

Out of curiosity I was wondering if the the spin angular momentum operator $\hat S_x$ could be written (in the standard basis given) in matrix form, given by $|↑\rangle = $$\begin{pmatrix}1\\ 0\end{pmatrix}$ and $|↓\rangle = $$\begin{pmatrix}0\\ 1\end{pmatrix}$

I have the spectral decomposition of the $\hat S_x$ operator:

\begin{align} \hat{S}_x=\frac{\hbar}{2}(|\downarrow\rangle\langle\uparrow|+|\uparrow\rangle\langle\downarrow|). \end{align}

And the matrix in the given standard basis would be as follows:

$$\begin{pmatrix} \langle\uparrow|\hat S_x|\uparrow\rangle & \langle\uparrow|\hat S_x|\downarrow\rangle\\ \langle\downarrow|\hat S_x|\uparrow\rangle\ & \langle\downarrow|\hat S_x|\downarrow\rangle\end{pmatrix}$$

Is this possible and if so if my representation of the matrix correct?


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

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


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

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

  • $\begingroup$ Thanks for your help and explanation Davide! $\endgroup$
    – GavinK14
    Feb 24 '20 at 19:42
  • 1
    $\begingroup$ Almost correct. But the pre-factor of $S_x$ should be $\frac{\hbar}{2}$ instead of $\frac{1}{\sqrt{2}}$. $\endgroup$ Feb 24 '20 at 19:50
  • $\begingroup$ This operator, given that it is a Pauli matrix is both unitary and Hermitian. Is it a projective operator? $\endgroup$
    – GavinK14
    Feb 24 '20 at 20:02
  • $\begingroup$ @ThomasFritsch Yes, of course, my bad! Thanks $\endgroup$
    – Quiver
    Feb 24 '20 at 20:26
  • $\begingroup$ @GavinK14 You should use the definition of a projective operator, if this is indeed a projection operator it'll follow all the conditions required $\endgroup$
    – Quiver
    Feb 24 '20 at 20:26

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

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

  • $\begingroup$ Thank you Nikita $\endgroup$
    – GavinK14
    Feb 24 '20 at 19:45

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