# What is the quantum / Berry-Pancharatnam phase for a spin-j state with z-component m?

Quantum phase arises when a spin-j state is sent through a sequence of transitions that return it to its original position. For example with spin-1/2, a state picks up a complex phase of $\pi/4$ when it goes through three perpendicular (not orthogonal!) transitions:

$\frac{<+z|+y><+y|+x><+x|+z>}{|<+z|+y><+y|+x><+x|+z>|} = e^{2i\pi/8}$

where $|+y>$ means the state with spin-1/2 in the +y direction. Since the above path covers an octant, or one eighth of the surface of a sphere, the total quantum phase for the whole sphere in the spin-1/2 case is $2i\pi$.

More generally, for a spin state $|j,m,+z>$, we will have

$\frac{<j,m,+z|j,m,+y><j,m,+y|j,m,+x><j,m,+x|j,m,+z>}{|<j,m,+z|j,m,+y><j,m,+y|j,m,+x><j,m,+x|j,m,+z>|} = e^{2i n_{jm} \pi/8}$

where $n_{jm}$ is an integer. The same phase is picked up for a spin-1 state with m=1, but a spin-1 state with m=0 picks up no phase at all. (In fact the m=0 states do not depend on the orientation.) At the moment, I believe that the j=1,m=-1 state will pick up n=-1 (in analogy with the spin-1/2 case), but I haven't bothered to actually run the computation.

Yes, Dr. Motl found it. For a journal reference see "Geometric Phases" by Péter Lévay. The appropriate equation is (36) on page 19. Note that our $m$ is his $r-J$.

http://arxiv.org/abs/math-ph/0509064v1

Also, the method of looking at an infinitesimal rotation works because Berry-Pancharatnam phase doesn't depend on stuff like the arbitrary complex phases of spinors and of course there's no preferred region on the sphere of possible orientations for the spin axis.

As an aside, I was convinced that the j=1, m=+-1 case had the same coefficient as the j=1/2,m=+-1/2 case. This was because the geometric phase for light is the same as that for spin-1/2. But now it's clear that this is mixing apples and oranges.

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Hi @Carl, could you change the wording to include a question for folks to answer? – user346 Jan 20 '11 at 5:51
What is $n_{jm}$. – Carl Brannen Jan 20 '11 at 6:00

I tend to think that you're right that the phase - but not the absolute value - may be decomposed into infinitesimal pieces of solid angle. And for each tiny solid angle, the phase you get is the same.

It's not hard to see what happens with the $|jm\rangle$ eigenstate under a parallel transport around an infinitesimal solid angle $\Omega$. It simply gets transformed by the phase $$\exp(i m \Omega).$$ In your notation, $n_{jm}=m$.

It's because it reduces to exponentials of small rotations around $x,y,-x,-y$ axes, and the commutator boils down to $J_z$ for the usual reasons. For your example, $\Omega$ was $4\pi/8=\pi/2$ and $m$ was $1/2$. That's why you got $\exp(\pi i /4)$.

Of course, I realize that this answer seems to contradict the previous problem you posed because $\exp(4\pi i m)$ equals one for any conceivable value of $m$, so there should be no obstruction ever. So at this moment, I am not sure which answer is the right one. I would guess that my answer to your previous problem is completely wrong - there may exist bundles that can't be obtained simply as products of the non-existing low-spin bundles.

And there could also be an issue with the overall sign. Note that if the inner products between the two states remain real for any angle, you eventually hit the point where the inner product goes negative, but at this point, the absolute value changes the sign relatively to the inner product.

Cheers LM

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Oh. A silly error. There is no contradiction. The argument must vanish exactly, not just modulo two pi. It is because 2m vertices near the South pole cannot be unwound. So the global function only exists for vanishing m which only exists for integer j. – Luboš Motl Jan 20 '11 at 12:08

Just as a practice calculation, we can check the calculation for $j=1$.
Let $L_x = \sqrt{1/2}\left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right)$, $L_y = \sqrt{1/2}\left(\begin{array}{ccc}0&-i&0\\+i&0&-i\\0&+i&0\end{array}\right)$, $L_z = \left(\begin{array}{ccc}1&0&0\\0&0&0\\0&0&-1\end{array}\right)$.
Then the eigenvectors for spin $m=1$ in the +x, +y, and +z directions are:
$<+x| = (1\;\;\;\;\;\;\sqrt{2}\;\;\;\;\;\;+1)/2$,
$<+y| = (1\;\;\;\;\;\;i\sqrt{2}\;\;\;\;\;-1)/2$,
$<+z| = (1\;\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\;\;0)$.

Computing we find: $\frac{<+x|+y><+y|+z><+z|+x>}{|<+x|+y><+y|+z><+z|+x>|} = \frac{(i/2)(1/2)(1/2)}{(1/2)^3} = e^{i\pi/2}$ as expected, just twice the angle of the spin-1/2 case (which gives $\pi/4$).

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– Carl Brannen Jan 22 '11 at 23:45