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This is from "Quantum Mechanics in Simple Matrix Form" by Thomas F. Jordan:

$(-i\Sigma_1)(-i\Sigma_2)=-i\Sigma_3$. This corresponds to the fact that the product of rotations by 180 degrees around the 1 and 2 axes is a rotation by 180 degrees around the 3 axis.

Axes 1, 2, and 3 are, of course: x, y, and z, and $\Sigma$'s are the Pauli matrices. But I can't picture why this statement is true. Any help? Is it just true in matrix multiplication?

Michael T Chase
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    Pick up a rectangular parallelipiped or similar you have lying around, like a six-sided die or a book (or even a cup with handle, or yourself if you're feeling acrobatic). Then perform said rotations on that object and see what happens. – Arthur Jun 15 '21 at 06:42
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    @Arthur Thanks. I had to ask the question to actually play it out, and answered it while you commented. – Michael T Chase Jun 15 '21 at 06:43
  • Take (with your actual physical hands) an object where you can designate a top and a front (for example, a toy car or a stapler) and perform 180° rotations around two axes. – Eike Schulte Jun 15 '21 at 06:44

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Just put your arm out to the side and put the other arm straight up in the air. Now turn the side arm (by spinning the body) 180 degrees, then imagine turning the arm that is straight up to straight down with the whole body. You'll see that the third axes, the line of sight, is rotated 180 degrees.

Michael T Chase
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    Good (+1)! Often we represent 3D-rotations with $3\times3$ matrices. 180 degree rotations about one of the coordinate axes happens to be diagonal because all the vectors perpendicular to the axes of rotation are mapped to their negatives. In other words, the three rotations become $$R_1=\pmatrix{1&0&0\cr0&-1&0\cr 0&0&-1\cr},$$ $$R_2=\pmatrix{-1&0&0\cr0&1&0\cr 0&0&-1\cr},$$ and $$R_3=\pmatrix{-1&0&0\cr0&-1&0\cr 0&0&1\cr}$$ respectively. From this it is clear that the product of any two of them is the third! The Pauli matrices describe rotations in a bit more complicated way. – Jyrki Lahtonen Jun 15 '21 at 07:25
  • I was resting on my answer, but in reality the 3rd axis, the line of sight, is right back to where it was. So in reality the spin changes both the side arm and the line of sight, which are two rotations, leaving the up-stretched arm unchanged. So it might make sense in matrix-form, but not in reality. – Michael T Chase Jun 15 '21 at 09:23
  • Maybe the paradox has to do with adding 180 degree turns rather than multiplying. – Michael T Chase Jun 15 '21 at 09:34
  • $-i\Sigma_1=\begin{pmatrix}0&-i\-i&0\ \end{pmatrix}$ and $-i\Sigma_2=\begin{pmatrix}0&i\-i&0\ \end{pmatrix}$, so $(-i\Sigma_1)(-i\Sigma_2)=\Sigma_2$ for Pauli matrices, and $-i\Sigma_3=\begin{pmatrix}i&0\0&i\ \end{pmatrix}$. I still don't get then how $\Sigma_2=-i\Sigma_3$ ???? – Michael T Chase Jun 15 '21 at 10:21
  • I see I did those calculations of $i$ wrong. But it boils down to how $(-i\Sigma_1)(-i\Sigma_2)=\begin{pmatrix}0&1\-1&0\ \end{pmatrix}$ and $-i\Sigma_3=\begin{pmatrix}-1&0\0&1\ \end{pmatrix}$ – Michael T Chase Jun 15 '21 at 10:35
  • One of your $\Sigma_1$ and $\Sigma_2$ is wrong. IIRC $i\Sigma_j$ is Hermitian for all $j$. This would suggest that your $\Sigma_1$ is wrong. Please check. I cannot check for there are several equally useful variants of Pauli matrices. – Jyrki Lahtonen Jun 15 '21 at 10:38
  • Okay, you're so right. Shown work: $-i\Sigma_1=\begin{pmatrix}0&1\1&0\ \end{pmatrix}\cdot -i=\begin{pmatrix}0&-i\-i&0\ \end{pmatrix}$; $-i\Sigma_2=\begin{pmatrix}0&-i\i&0\ \end{pmatrix}\cdot -i=\begin{pmatrix}0&-1\1&0\ \end{pmatrix}$. – Michael T Chase Jun 15 '21 at 10:59
  • If I did that is right, $(-i\Sigma_1)(-i\Sigma_2)=\begin{pmatrix}0&i\-i&0\ \end{pmatrix}$. Could $-i\Sigma_3=\begin{pmatrix}1&0\0&-1\ \end{pmatrix}\cdot-i=\begin{pmatrix}-i&0\0&i\ \end{pmatrix}$? How do these two answers relate? – Michael T Chase Jun 15 '21 at 11:08
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    $$\pmatrix{0&a\cr b&0\cr}\pmatrix{0&c\cr d&0\cr}=\pmatrix{ad&0\cr 0&bc\cr}.$$ – Jyrki Lahtonen Jun 15 '21 at 11:55