Explaining Quantum States: Global Phase

[supernano]

TL;DR Summary

What's a good way to explain global phase of a quantum state?

I was teaching the basics of quantum states and was showing the students that an arbitrary state in a quantum two-level system could be written as $|\psi\rangle = C_1 |+\rangle + C_2 |-\rangle = R_1 |+\rangle + R_2 e^{i \alpha} |-\rangle$, with {$C_i$} complex and {$R_i$} real.
Then someone asked me why I only included a phase in one of the terms, and as I started to explain about global phase, I found it really hard to give them an intuitive explanation.
How would you explain it?

 

[Demystifier]

Summary:: What's a good way to explain global phase of a quantum state?

How would you explain it?

Show them that probabilities (given by the Born rule) do not depend on the global phase.

 

[vanhees71]

I'm also still struggling how to teach QT right (particularly as I have to challenge to teach it to high-school teacher students and I have only about 1/2 a semester time for it). It's difficult to teach both some intuition about quantum mechanics as well as enough math.

I think it is important to stress that pure states are not represented by normalized vectors but by rays, i.e., all physically observable content in pure states are the probabilities (or probability distributions) for the outcome of measurements, and these are given by Born's rule,
$$P(a)=|\langle a|\psi \rangle|^2,$$
where $|a \rangle$ are the eigenvectors of the self-adjoint operator $\hat{A}$, representing some observable $A$, and I assume that all eigenvalues are non-degenerate. Both the eigenvalues and the state vector should be normalized to 1, and it's clear that you can multiply both with arbitrary phases without changing the probabilities.

If you consider $|\psi \rangle$ written in terms of an arbitrary basis, of course you have the freedom to choose only one overall phase, and thus in your example you can arbitrarily choose one of the coefficients as real and the other with an arbitrary complex number to get all states in your two-level system.

Since (unit) rays are somewhat complicated, an equivalent alternative is to stress that states are in fact represented not by state vectors at all but by statistical operators, i.e., a self-adjoint positive semidefinite operator with trace 1. Then the pure state is just a special case, where the statistical operator is a projection operator an thus of the form $\hat{\rho}=|\psi \rangle \langle \psi|$ with a normalized vector $|\psi \rangle$. It's immediately clear that for a given pure state $\hat{\rho}$ the vector $|\psi \rangle$ is only defined up to an arbitrary phase factor.

 

[WernerQH]

Wave functions need not be normalized. (A normalization factor is a technical detail that you can add to your equations for observable quantities explicitly.) The crucial point is that you can multiply a wave function by any number different from zero and it still represents the same state. That's why we can call states that vary as $e^{i E t / \hbar}$ stationary.

 

[vanhees71]

Sure, you can also define a pure state simply as a ray in Hilbert space. Then you have to take care of the normalization explicitly:
$$P(a)=\frac{|\langle a|\psi \rangle|^2}{\langle a|a \rangle \langle \psi|\psi\rangle}.$$
That doesn't change the math of this important freedom to choose a phase factor.

That's more important than most treatments of QT suggest:

(a) it enables the well-known non-relativistic quantum mechanics to begin with, because for a physically meaningful dynamics you have to use a non-trivial central extension of the Galilei group with mass as the central charge, i.e., a true unitary ray representation rather than a proper unitary representation of the "classical" Galilei group.

(b) it enables half-integer angular-momentum representations (spin), i.e., the use of the covering group SU(2) instead of SO(3) as subgroup of the Galilei or Poincare group for non-relativistic and relativistic Q(F)T, respectively

 

[Demystifier]

$$P(a)=\frac{|\langle a|\psi \rangle|^2}{\langle a|a \rangle \langle \psi|\psi\rangle}.$$

I would like to add that it doesn't work when $a$ is continuous, because then you have something like $\langle a|a'\rangle=\delta(a-a')$

 

[vanhees71]

Sure, I referred to state vectors, and those are elements of the Hilbert space and not the dual of the nuclear space.