# Representation of phase in quantum mechanics

[Note: My discussion of the three answers can be found just after the question.]

Imagine three points in space that differ only by a phase angle of "something" (what doesn't really matter).

One way to represent the difference in phase angle between the three points is assigning each one a specific angle designation, e.g. $0^{\circ}$, $90^{\circ}$, and $180^{\circ}$, or in the equivalent complex number forms, $1$, $i$, and $-1$. [1]

However, assigning phase angles assumes that the starting angle of 0 degrees is a universal constant across the entire space in which such points could exist. For many phenomena, especially in quantum mechanics, it turns out that how you select this universal or global $0^{\circ}$ setting doesn't really matter. The ability to select the $0^{\circ}$ point arbitrarily -- or equivalently, the ability to add any arbitrary angle or phase to each point in the space -- is typically described as a symmetry of physics.

If you don't mind exploding the number of mathematical quantities involved, you can instead express the relationships without assuming any global constants to exist. More specifically, you can use a Mach-like approach in which phase angles only exist between pairs of locations in the space. With this approach and assuming transitive phase relationships, you can use $n$ phase vectors (that's a phrase I made up, so don't bother looking it up) to specify all phase relationships between all particles, where $n$ is the number of points (or particles) in space that have phase angles:

$p_1 \rightarrow p_2 = i = 90^{\circ}$

$p_1 \rightarrow p_3 = -1 = 180^{\circ}$

Notice that in the above example, setting all phase vectors to start at the same location provides a close equivalent to assuming a universal $0^{\circ}$. The only difference is that here you have explicitly stated which point will serve as the universal $0^{\circ}$ value by referencing all other particles to it.

The above two phase vectors fully define the phase relationships of the three points I described earlier, but there are many ways to capture the phase relationships. The full expansion of phase vectors implied by $\{(p_1=1), (p_2=i), (p_3=-1)\}$ is:

$\{(p_1 \rightarrow p_2 = i), (p_2 \rightarrow p_1 = -i),$

$(p_2 \rightarrow p_3 = i), (p_3 \rightarrow p_2 = -i),$

$(p_1 \rightarrow p_3 = -1), (p_3 \rightarrow p_1 = -1)\}$

Now, at last, my questions. Feel free to answer any or all, or create your own. Answers of the type "Terry you dunce this is just [...] phrased badly!" are not only welcome but hugely appreciated, if accurate.

1. Is phase symmetry real, or is it just an artifact of using mathematical methods that assume the existence of a universal $0^{\circ}$ value that has no real physical meaning, since phase can only be measured by using pairs of locations?

2. Phase arrows have their own symmetry, since any set of $n-1$ phase arrows [2] that fully connects $n$ particles can be used to represent their phase relationships. So, do phase arrows just obscure phase symmetry by representing it in an oddly particle-centric way?

3. Or is it the other way around: Is the equivalence of all fully connected sets of phase arrows the deeper symmetry, since it reflects an entanglement-like preservation of phase across measurements of many, many particles?

4. If phase symmetry is just an artifact of assuming a global $0^{\circ}$, what does that imply for expressing electromagnetism as a phase symmetry? What would change?

2012-09-02 - Assessment of the three answers

[1] This note is for any readers who may not familiar with complex-plane phase angles. You may be wondering why @Kostya's complex-plane angle notation looks so different from my assertion that $1$, $i$, and $-1$ represent "angles." We really are talking about the same thing, I assure you!

Angles are very often represented and calculated in physics disciplines using a compact exponential notation with remarkable computational properties. That notation works like this: If you raise the mathematical constant $e$ ($\approx 2.71828$) to an imaginary exponent -- and yes, once upon a time and many years ago, a lot of deep mathematical thinking went into figuring out what it even means to raise a number to an imaginary exponent -- then a most curious relationship emerges.

The first part of what happens is the magnitude $\phi$ of the imaginary exponent -- that is, the real number by which $i$ is multiplied in the exponent -- ends up specifying a complex value that is always one unit away from the origin ($0$) of the complex plane.

Next, if you interpret the origin $0$ of the complex plane as the vertex of an angle, and the line from $0$ to $1$ as one side of that angle, then the line from $0$ to the unit-length complex value generated by $e^{i\phi}$ will define the other side of that angle. To make that work correctly, the angle $\phi$ must be expressed in units of radians rather than degrees, where $180^\circ = \pi$ (radians). (Side note: The truly amazing relationship between $e$, $i$, $\pi$, and $-1$ that this relationship produces is known as Euler's Identity.)

So, if you start out with a real angle $\phi=\pi/2=90^\circ$, then what the function $e^{i\phi}$ does is "translate" that somewhat abstract real angle into a tangible and easily visualized angle formed by the complex-plane vertex $0$, the unit value $1$ (which defines the "always assumed" horizontal side of the angle), and the output of $e^{i\phi}$, which in this case would be $i$. Since $i$ is straight up from the vertex $0$, the resulting angle is $90^\circ$.

So, I sort of cut to the chase and spoke only of a few carefully selected examples of how $1=0^\circ$, $i=90^\circ$, $-1=180^\circ$, and $-i=270^\circ$ show the potential for complex numbers to represent angles. That potential is realized in the $e^{i\phi}$, which enables unit-circle complex angles to be represented simply and elegantly.

Still, using complex numbers alone for angles has advantages too. In particular, simply multiplying two unit complex numbers is the same as adding two angles -- a neat trick indeed, and a lot faster than doing a pile of trig functions! It also provides a different way of understanding why $i^2=-1$. If those numbers are instead interpreted as angles, you are simply adding two $90^\circ$ angles (represented by the two $i$ values) to produce a new angle of $180^\circ$ (the -1 result).

[2] Corrected from $n$; thanks @NickKidman!

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Representations:

Let $n$ be the number of phases. As $p_1\rightarrow p_1$ is trvial because there is no difference between a phase and itself, you need exactly $n-1$ relations

$$((p_1\rightarrow p_2),(p_1\rightarrow p_3),(p_1\rightarrow p_4),\dots,(p_1\rightarrow p_n)),$$

to specify all the phases. If you specify our representation this way, the entry $(p_1\rightarrow p_1)$ as first component

$$((p_1\rightarrow p_1),(p_1\rightarrow p_2),(p_1\rightarrow p_3),(p_1\rightarrow p_4),\dots,(p_1\rightarrow p_n)),$$

so that e.g. $$x=(\Delta\varphi_{11},\Delta\varphi_{12},\Delta\varphi_{13},\dots,\Delta\varphi_{1n}),$$ is a phase vector, then it can be viewed as a point of an n-dimensional $2\pi$-periodix box $\sim S^n$. Of course the phase arrow

$$((p_1\rightarrow p_2),(p_2\rightarrow p_3),(p_2\rightarrow p_4)\dots,(p_2\rightarrow p_n)),$$

is equally good at describing the system. Dropping the first component gives vectors $$x'=(\Delta\varphi_{12},\Delta\varphi_{13},\dots,\Delta\varphi_{1n})$$ in a space $\sim S^{n-1}$. The representation $$((p_n\rightarrow p_1),(p_1\rightarrow p_2),(p_2\rightarrow p_3),(p_3\rightarrow p_4),\dots,(p_{n-1}\rightarrow p_n)),$$

i.e.

$$x''=(\Delta\varphi_{n1},\Delta\varphi_{12},\Delta\varphi_{23},\dots,\Delta\varphi_{(n-1)n}),$$

also contains one entry too much, but uses relative phases only, in a way which avoids $(p_i\rightarrow p_i)$, i.e. an angle, which is necessarily $0°$.

The global symmetry is an 1-dimensional translation by

$$c\ (1,1,\dots,1),\ \ \ c\in \mathbb{R}\ \text{mod}\ 2\pi \sim S.$$

or the individual $\varphi$ angles, which you use to compute the differences.

Counting phase vectors:

In any case, you must be able to compute all the relative phase differences by adding or subtracting the pairwise phase difference, which means there must be a path between any two different vertices $n$. Adding additional phase information / edges is superfluous, and so there should only be one such path between two vertices. Obviously, we don't need any edges $(p_i\rightarrow p_i)$, which are self-loops. This means that, apart from the phase sign which indicates the direction of the edge, the phase vectors for the $n$-angle system are isomorphic to all the trees of $n$ vertices, and this further implies that there are $n^{n-2}$ at least equally valid phase vectors. More specifically, as said above, each angle corresponding to $(p_1\rightarrow p_2)$ is really a directed edge and, apart from a minus sign, equals the reversed one $(p_2\rightarrow p_1)$. With the exact number of $n-1$ edges trees always have, there are clearly $2^{n-1}$ configurations for one tree, so $2^{n-1}n^{n-2}$ phase vectors total.

The transformations between these representations of the relative angles will be inherited from graph theoretical morphisms as well, and these then also include the $(\mathbb{Z}_2)^{n-1}$ switching symmetries. The angle values are independent from the transformations. And so as the minimal amount of information (= tuple of numbers to specify) only depends on the number of particles. Naming the transformation only tells you the number of phases.

In computation:

Stating the fact that you have a global symmetry says that there is a phase parameter which can be varied (and that parameter gives the possibility for relative phases) and what the structure of the theory definitely does not look like. From a linguistic point of view, I expect it's not necessary to mention a fixed phase $0^\circ$ for an initial angle, if the particles you work with are labeled already. This gives the formulation:

"Is there a (maybe quite complicated and cumbersome) way of doing field theory computations, where for all particle pairs, need only use the variables $\Delta\varphi_{ij}?$ to compute the (eventually necessarily phase value independent) observables?"

Knowing one phase vector is knowing all information (and also all phase vectors) and so you could just label the particles and associate all the relative angles between particles. In a way the question above is trivial as $0°=\Delta\varphi_{ii}$ and we can just say the $\Delta\varphi_{ij}$'s are only defined up to a global shift (which will necessarily drop out at the end of the computation), but I guess we want to exclude that trick. We could think about if there is a representation of field theory with $\varphi_{ij}$'s and $i\ne j$ or, which would be even harder, if there is one that only uses $n-1$ fixed $\Delta\varphi_{ij}$'s.

As a side note, the operations like averaging of angles $\int_\text{circle} f(\phi)\text d\varphi$ don't even require you to specify a base point.

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@TerryBolinger: Here is the formalization of the phase computation: Choose one of the $N$ particles, indexed by $r$ say. Provide the $N-1$ initial data $i:=\Delta\phi_{ri}$. Compute the symbols $[i,j,k]:=i-j+k$. You'll now find $[i,j,r]=\Delta\phi_{ij}$, see this wikipedia article. – NikolajK Aug 7 '12 at 15:46

Let me refine the math a little bit. So you have a set $\Omega$ of "points". And you have a number of phase differences between those points: $$P(x,y) = (p_x\to p_y) = e^{i\phi_{xy}}\,,\quad \mbox{for}\quad x,y\in \Omega$$ These differences must satisfy a set of axioms, like, I guess: $$\forall x\in\Omega, P(x,x) = 1$$ $$\forall x,y\in\Omega, P(x,y) = P(y,x)^*$$ $$\forall x,y,z\in\Omega, P(x,y)P(y,z) = P(x,z)$$

That looks a lot like fundamental group stuff from homotopy theory. And I think for a reason...

1 . Is phase symmetry real, or is it just an artifact of using mathematical methods that assume the existence of a universal $0^\circ$ value that has no real physical meaning, since phase can only be measured by using pairs of locations?

Well, you can select any point $x_0\in\Omega$ and then every element in $\Omega$ will have an associated phase $\forall y\in\Omega, P(y)=e^{i\alpha}P(x_0,y)$ with arbitrary extra phase $\alpha$. I can't say that it is "not real", but I'd say that this construction is "redundant" (while it is usually more useful for concrete calculations).

2 . Phase arrows have their own symmetry, since any set of $n-1$ phase arrows that fully connects $n$ particles can be used to represent their phase relationships.

Right, but this construction doesn't represent all possible things that could happen. Remember Aharonov-Bohm effect? You can have different phase differences depending on your path.

So, you (we) have made a description of only a small connected patch of a whole set of "points with phases", which will have to be "glued" from such patches with possibly nontrivial topology.

4 . If phase symmetry is just an artifact of assuming a global $0^\circ$, what does that imply for expressing electromagnetism as a phase symmetry? What would change?

But you are talking about global phase symmetry, while electromagnetism requires arbitrary local phase changes. Also you need a continuous manifold of "points" to set up a connection.

Actually, it seems that this discussion inevitably goes to all this concepts: Manifolds, connections, homotopy and holonomy and stuff.

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I think the question doesn't really involve space. The group structure is only relevant as the thing which acts on the phases and you have to introduce the connection tools (photon field etc.) to change them for each point in spacetime. But what Terry probably wants to get rid of are the actual phases in the numeric value represenation, such that there is a base point. If you talk about the transformation group representation, then you're not there and if you represent the phases themselves like a group, then the phase which corresponds to the unit element is the singled out one, i.e. $0°$. – NikolajK Jul 26 '12 at 19:00

If you make the "phase relations" come from fixed phases at different points, and you say that each link has a phase $\theta_{ij}$ (the phase acquired when you go from point i to point j) then you have the property that the sum of all the phases around a closed loop is zero (modulo $2\pi$).

If you make a grid, give phases $\theta_{ij}$ between neighboring points, and drop the constraint that the product of all the complex numbers around a small square (a plaquette) is zero, you get the discrete lattice U(1) gauge theory. This is in some sense the most primitive formulation of electrodynamics.

The statement that electromagnetism is a gauging of a phase symmetry is saying that the phases are relative, and further, the phases are inconsistent, in that you can have a loop where you can't choose the phase definition to make the relative phase consistent.

This is the standard definition of lattice gauge theory, and I think it would be good for you to read about it, to avoid using nonstandard notation. The gauge invariance in the fields corresponds exactly to your freedom to redefine the zero phase value at every point arbitrarily, and the "relative phase values" are the gauge field values on neighboring lattice points. The continuum gauge theory just comes from making the lattice infiniteismal.

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