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In a previous question I asked about the cartesian components of isospin, i.e. $\pi_1$, $\pi_2$, and $\pi_3$ and why only the $\pi^\pm$ and $\pi^0$ fields are physical.

The answer states that

The "cartesian" version isn't used, because the basis states aren't explicitly eigenstates of rotations about I3--the equivalent z-axis in isospin space.

I don't see why they need to be eigenstates of rotations. Spin is mathematically identical to isospin. Light for example can be linearly polarised, which would be cartesian eigenstates, or circularly polarised, which would be rotation eigenstates.

Why can we have both for spin, but only the latter for isospin?

infinitezero
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1 Answers1

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In principle, there is no reason why we cannot have any linear combination of the three pion states $\pi^+$, $\pi^-$, $\pi^0$. However, the instruments that we use to detect pions are sensitive to their electric charge. So the detection of a pion is a "measurement", in the quantum mechanical sense, that breaks a coherent linear combination into incoherent pieces.

Typically, electric charge conservation breaks the coherence of the $|\pi^1\rangle$ and $|\pi^2\rangle$ components. For example, the dominant weak interaction decays of the $\Lambda^0$ baryon are $\Lambda^0 \to \pi^- p$ and $\Lambda^0 \to \pi^0 n$. Here the pions must be in electric charge eigenstates.

Here is a more illustrative example, which I will work out very explicitly. The heavy meson $\psi'$ decays to the slightly less heavy $J/\psi$ meson plus a pair of pions. Isospin conservation tells us that the 2-pion state has isospin zero. Let's say that we have detectors to the left and to the right. The state that is created is $$ {1\over \sqrt{3}} \biggl[ | \pi^+\pi^- \rangle + |\pi^0\pi^0\rangle + | \pi^- \pi^+\rangle \biggr] $$ where the first pion is the one that goes to the left and the second pion is the one that goes to the right. This is the same as $$ {1\over \sqrt{3}} \biggl[ | \pi^1 \pi^1 \rangle + |\pi^2\pi^2\rangle + | \pi^3 \pi^3\rangle \biggr] $$ We could write the linear combination of states in either of these ways.

However, if the detector is a typical particle physics detector, it is built as a set of tracking chambers with a magnetic field and then an electromagnetic calorimeter that measures the energies of gamma rays.

Now let's think about the time evolution of this state. The first thing that happens is that the $\pi^0$'s each decay to two gamma rays. Now the signals from any detectors will be macroscopically different for $\pi^0$ vs, charged pions, so those states will be separated off. Next the positive and negatively charged pions will have slightly but macroscopically different trajectories in the magnetic field. So the pieces of the original coherent linear combination will be separated in the (+,-,0) basis. We will observe three distinct outcomes, all with probability 1/3: a $\pi^+$ to the left and a $\pi^-$ to the right, a $\pi^0$ to the left and a $\pi^0$ to the right, and a $\pi^-$ to the left and a $\pi^+$ to the right.

If there were no magnetic field (not even the earth's magnetic field, and the detectors were so thin that the charged pions did not scatter from the material in the detector, it might be possible to preserve the coherence of the state $$ |\pi^1> = {1\over \sqrt{2}}\biggl[ |\pi^+\rangle + |\pi^-\rangle \biggr] $$ I do not know any experiment in which this was possible in practice.

So, finally, it is the fact that our measurements are not ideal that picks out the electric charge basis.

Michael Peskin
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  • Thank you for this elaborate answer. My question now became: How does this basis come about? If I use $Q \propto I_z$ then the $\pi_i$ are eigenstates of the charge operator. So at some point, we need to choose a basis and stick to it. When does this happen? – infinitezero Aug 23 '21 at 12:33
  • As I said in the note, the $\pi^+$, $\pi^-$ and $\pi^0$ all have different behavior when they interact with a particle detector. So, when the pions interact with the detector, this is a "measurement" -- in the usual sense of quantum mechanics -- and that measurement chooses the basis. After the measurement, we no longer have a coherent linear combination. Instead, we have the separate charge eigenstates $\pi^+$, $\pi^-$ and $\pi^0$, with appropriate probabilities. – Michael Peskin Aug 24 '21 at 15:16