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I understand the pauli matrix $\sigma_z = \bigl( \begin{smallmatrix}1 & 0\\ 0 & -1\end{smallmatrix}\bigr)$ rotates a state around $z$-axis by angle $\pi$ in $SO(3)$. We can see it works by thinking about rotating eigenstate of $\sigma_x$: $$|+\rangle_x = \frac{1}{\sqrt{2}} \bigl( \begin{smallmatrix}1 \\ 1\end{smallmatrix}\bigr)$$ $$|-\rangle_x = \frac{1}{\sqrt{2}} \bigl( \begin{smallmatrix}1 \\ -1\end{smallmatrix}\bigr)$$ and applying $\sigma_z$, we have $\sigma_z|+\rangle_x = |-\rangle_x$ and vice versa.

But consider rotation of eigenstate of $\sigma_z$ itself, conceptually it shouldn't change the state but obviously we have different phase for $|+\rangle_z$ and $|-\rangle_z$ after the rotation. This kind of makes sense because otherwise, this operator won't rotate $|+\rangle_x$ properly. But formally, I don't quite know how to think about this rotation. I think it due to the fact that $SU(2)$ is the double cover of $SO(3)$.

Can someone give some clarification on exactly why this has to be the case? Or give some intuition on how to think about this kind of rotation in $SU(2)$ rather than $SO(3)$?

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Actually there’s a clarification needed here. The transformation that takes $\vert+\rangle_x$ to $\vert -\rangle_x$ is a finite transformation, whereas $\sigma_z$ is a generator of infinitesimal transformations. It just so happens that the finite unitary transformation $$ e^{i\pi \sigma_z/2}=\left(\begin{array}{cc} i&0\\ 0&-i\end{array}\right) $$ will transform $\vert +\rangle_x$ to $\vert -\rangle_x$ up to an overall phase, like $\sigma_z$ does.

You can also transform $\vert +\rangle_x$ into $\vert -\rangle_x$ by a finite rotation of the type $e^{i\theta\sigma_y}$; this is obvious geometrically but not so clear if you try to repeat your trick of acting on your states with $\sigma_y$. This illustrates that your trick of acting with $\sigma_z$ is a coincidence.

It turns out you are using a property of the Pauli matrices: they satisfy $$ e^{ia\hat n\cdot\vec\sigma}= I\cos(a)+\hat n\cdot\vec\sigma\sin(a)\, , $$ Here, the left hand side is a finite rotation by an angle $a$ about an axis along the unit vector $\hat n$, and $\hat n\cdot \vec\sigma= n_x\sigma_x+n_y\sigma_y+n_z\sigma_z$, but the right hand side can still be written as a sum of Pauli matrices.

The Pauli matrices have the unusual property that $$ i\sigma_z=\left(\begin{array}{cc} i&0\\ 0&-i\end{array}\right)\, ,\quad i\sigma_x=\left(\begin{array}{cc} 0&i\\ i&0\end{array}\right)\, ,\quad \sigma_y= \left(\begin{array}{cc} 0&1\\ -1 &0\end{array}\right) $$ (note the extra “$i$” factor in front of $\sigma_z$ and $\sigma_x$) are not only generators of $\mathfrak{su}(2)$ but also elements of a finite subgroup of $SL(2,\mathbb{C})$ (which contains in fact $\pm I, \pm i\sigma_z,\pm i\sigma_x$ and $\pm \sigma_y$), so there’s a number of situations (particularly in quantum information science) where one uses de facto the Pauli matrices as equivalent (up to a phase) to finite transformations, as you have done.

If you are interested in more on generalized Pauli matrices (there are also $n$-dimensional ones with the same properties of being elements of a finite subgroup and spanning a Lie algebra) see

Patera, J., and H. Zassenhaus. "The Pauli matrices in n dimensions and finest gradings of simple Lie algebras of type A n− 1." Journal of Mathematical Physics 29.3 (1988): 665-673.

or

Patera, J. "The four sets of additive quantum numbers of SU (3)." Journal of mathematical physics 30.12 (1989): 2756-2762.

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  • $\begingroup$ Plugging into the general rotation formula by $\pi$ shows $\sigma_i$ rotates a state by axis-i in general. I'm not sure what's not clear about $\sigma_y$. Could you clarify? $\endgroup$ – Histoscienology Mar 9 '19 at 2:44
  • $\begingroup$ Not sure I understand. I suppose the issue is: $\sigma_z$ is NOT a rotation as these must have det=1. $\sigma_i$ are infinitesimal generators and it just so happens that multiples of them coincide with rotations. $\endgroup$ – ZeroTheHero Mar 9 '19 at 4:15
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The factor you're talking about is a global phase shift, which has no physical meaning.

$$ \sigma_z|-\rangle =\left( \begin{matrix}1 & 0\\ 0 & -1\end{matrix}\right)\left( \begin{matrix} 0 \\ 1 \end{matrix}\right)= \left(\begin{matrix} 0 \\ -1 \end{matrix}\right) = (-1)\cdot|-\rangle $$

Recall that $-1=e^{i\pi}$ is a $\pi$ phase shift, but has unit modulus, so it has no physical meaning.

When you measure the state after the rotation, you'll also get "minus", as it should be, and with the same 100% probability, as expected.


And, so, is this phase shift just a junk factor which appears unavoidably, or is it actually real? Well, it is real. It has been found in lab that it is real.

If you rotate it, you get the phase shift, but that is undistinguishable from the original particle. However, if you make it interfere with an identical oriiginal particle, the interference shows that the sign exists. IF it didn't exist, the interference would be different.

So, in conclusion, the sign exists, but it doesn't mean anything when dealing with it alone.

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  • $\begingroup$ Well, thanks for your answer. I agree with the probability statement. But my point is this global phase shift is different for spin up and down. Similar to the interference effect you mentioned, this difference effectively gives the rotation of x-eigenstate as I included in the question. So I don't think this doesn't mean anything. I believe there is a way to derive this shift using $SU(2)$ algebra. $\endgroup$ – Histoscienology Mar 4 '19 at 0:33
  • $\begingroup$ I don't understand your question then. What are you looking for? $\endgroup$ – FGSUZ Mar 4 '19 at 11:32

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