# Berry phase and the U(1) gauge symmetry

I discussed Berry phase back in February, (Berry or Pancharatnam-Berry or Quantum phase) from a vector / density matrix point of view, but I thought it would be nice to describe Berry phase from the point of view of the U(1) gauge symmetry of quantum mechanics. From a density matrix point of view, the U(1) gauge symmetry is what arises in the state vector formalism, from the requirement that all physical observables be capable of being written in the density matrix formalism.

Let’s consider functions of three normalized spin-1/2 SU(2) spinors: |a), |b), and |c) in the bra-ket notation. (And I’m using round brackets to avoid a WordPress screwup.) Since the state vectors of quantum mechanics are unchanged by phase, we can also choose three arbitrary fixed real numbers, $\alpha, \beta, \gamma$ to respectively multiply these spinors. The bras and kets take the complex conjugate phases so the arbitrary phase transformation on these objects is as follows: Quantum mechanic’s physical predictions will be unchanged by the substitutions. This symmetry is the U(1) symmetry of the SU(3)xSU(2)xU(1) symmetry of elementary particles and it was the first gauge symmetry. In short, letting the phase depend on position, and requiring that the theory of electrons be unchanged, defines a new field which turns out to be the photon.

This makes a limitation on how the spinors can be combined to produce physical observables. Since our spinors are normalized,(a|a) = 1. This fact is preserved in the transformed spinors because the bra and ket take opposite signs, which cancel. The alternative product of a spinor with itself, |a)(a|, is the density matrix that I am so fond of. If we were talking about a spinor valued wave function, rather than the simple qubit, would be a function of position: the probability density.

For two different mixed spinors, there are only two possible products ( a | b ) , and ( b | a ). On transforming the phases, these get multiplied by $\exp(-i(\alpha-\beta))$ and $\exp(+i(\alpha-\beta))$, respectively. Since these depend on the phases, neither of the products nor can be a physical observable. However, their phases cancel each other, and therefore the product of these two unphysical products: (a|b)(b|a), does not depend on the arbitrary phases and can be a physical observable. This is the transition probability between the two states.

As it turns out, (a|b) and (b|a) are complex conjugates of each other and so their product is a non negative real number. Sometimes students are told that we must convert the complex numbers into squared magnitude because physical observables must be real numbers but I think it is better to explain it as necessary to preserve the ungauged U(1) symmetry. The reason for this slant is because of what happens when you consider physical observables that are functions of three spinors which we discuss next.

Functions of three Spinors

The six bras and kets have phases that sum to zero so any function that uses each of them once will be unchanged by a phase transformation and so can be a physical observable. The available players are (a|b), (a|c), (b|a), (b|c), (c|a), and (c|b) and we need to include one each of the six bras and kets. There are two possibilities:
BP = (a|b)(b|c)(c|a),
BP* = (a|c)(c|b)(b|a).
I’ve arranged these so that the first is the reverse of the second. This shows that they are complex conjugates and so their product is real. But each of these products has one ket and one bra for each of a, b, and c, and therefore the various phases will cancel. Since these products are unchanged by the U(1) symmetry transformation they are possible physical observables. Can they be observed? Yes! These are Berry phases.

Note that I also arranged the bras and kets so that the b’s and c’s are adjacent to each other: . This makes the conversion into density matrix formalism easier. Instead of three bras and three kets, in density matrix formalism, there are only three objects are |a)(a|, |b)(b|, and |c)(c|. The Berry phase is the trace of their product:

BP = tr ( |a)(a| |b)(b| |c)(c| ),
= tr ( (a| |b)(b| |c)(c||a) ), [property of traces]
= (a|b)(b|c)(c|a) .

Another way of defining the U(1) symmetry of QM is that it is the symmetry that results when the observables of the theory are restricted to those available in the density matrix formalism. This is another indication that the density matrix formalism is more fundamental than the state vector formalism.

Calculating Berry Phase

The three spinors |a), |b), and |c) define three points on the unit sphere. In the density matrix formalism, this is sometimes called the Bloch sphere. Three points on a sphere define an oriented spherical triangle. The orientation depends on the order that the three points are given and changing the orientation negates the Berry phase. The three orientations (a,b,c), (b,c,a), and (c,a,b) are all the same, the other three are opposite.

It turns out that the Berry phase is proportional to the area of the spherical triangle. Since density matrices do not carry arbitrary phase it is fairly easy to prove this and I did so in the previous post on the subject, (Berry or Pancharatnam-Berry or Quantum phase). The constant of proportionality is 1/2, so the Berry phase is half the area of the spherical triangle.

The reader is invited to verify this. Let a, b, and c be the states for spin in the x, y, and z directions. Their spinors are: One finds = (x|y) = 0.5 + 0.5i, and (y|z) = (z|x) = sqrt{0.5}. Remember that each of these three products carries an arbitrary phase, it is only their product that does not depend on phase. And that product is $(1 + i)/4 = \sqrt{1/8} \exp(i\pi/4)$ so the Berry phase is $\pi/4$ and the area of the spherical triangle bounded by the x, y, and z axes is twice this: $\pi/2$, which is an octant, or 1/8th of the full spherical area of $4\pi$.

Theories: How to make a Bilinear into a Linear

One multiplies a ket by a bra to make a density matrix out of a spinor so density matrices are quadratic objects rather than linear. From the point of view of density matrices, it is perfectly natural that spinors end up with arbitrary phases. Such things always arise when a quadratic theory is rewritten as a linear theory. For example, the quadratic polynomial $x^2 + 9$ can be split into two linear polynomials, $x^2 + 9 = (x+3i)(x-3i)$. If one had a theory where a physical observable could be naturally written as $x^2 + 9$, one could instead obtain the same physical observable by computing the squared magnitude of x+3i. But one could also use x-3i; the sign of the imaginary part is arbitrary. Analogously, with spinors, the phase of the spinors is arbitrary. In either case, leaving a theory which is inherently quadratic (or more accurately, bilinear) that way makes it a lot easier to do certain things.

Now the above gives a good explanation of the U(1) gauge symmetry. The gentle reader might wish to know where the SU(2) and SU(3) symmetries come from. For that, they will have to go on a fishing expedition to find my other papers on the subject.

Density matrix theory is quite powerful on defining quantum bound states. In this context, this month I plan on discussing the energy levels of the hydrogen atom as an example of the method I use for applying Feynman diagrams on bound states one might more usually model using quantum mechanical potentials.

The three states chosen for the Berry phase calculation given above, +x, +y, and +z, are special in that these three axes form a MUB (mutuallly unbiased bases) set for the Pauli algebra. The resulting Berry phase, $\sqrt{1/8}\;\exp(i\pi/4)$, after a bit of algebra, ends up in the Koide mass formulas for the leptons. In converting 3×3 matrices of complex numbers into 3×3 matrices of Feynman diagrams describing the bound states that make up the leptons, one must take the cubed root of this product. This introduces a factor of $\sqrt{1/2}$ in the off diagonal terms. The Berry phase is cancelled in the charged leptons but shows up in the neutrinos as a factor of $\exp(i\pi/12)$ in the off diagonal elements.