In quantum mechanics, amplitudes are represented by complex numbers $e^{i\phi}$, which have phase angles $\phi$. These phase angles are clearly not observable in absolute terms.

If I have two electrons, then it seems that their relative phase angle $\phi_1-\phi_2$ is observable, since the electrons could interfere, and this interference results in a probability that is observable, e.g., as a count rate in a detector.

Is the relative phase between non-identical particles not observable? Does it matter whether they interact?

For a particle at rest, I would think we'd have $\phi=\omega t=Et=mt$ (in units with $c=1$ and $\hbar=1$), so that particles at rest with different masses would oscillate at different rates. It seems unlikely that we could observe them coming in and out of phase with one another.

How would this play out for neutrino oscillations?

  • $\begingroup$ I suppose it ultimately boils down to superselection? $\endgroup$ – Jia Yiyang Nov 13 '14 at 5:56

Great question, which underlines a common misconception of phase in QM (in my opinion).

My short answer, before going into the detail, I would say: no, the phase is not observable and the relative phase between two particles is somewhat ill-defined.

I think the origin of this misconception about the individual phases of various particles from the analogy with classical waves, such as the ones on a water surface. Clearly, as long as we are in the linear regime, we could add water waves. Let's say we have some configuration of the water surface $\phi_1(t=0)$ and some other configuration $\phi_2(t=0)$. Both allow us to calculate $\phi_{1/ 2}(t)$ at a later time. Now our concept of interference is that if we have some superposition of the initial patterns, say $\phi_3(t=0)=\phi_1(t=0)+\phi_2(t=0)$, then the time-evolved configuration is simply the linear superposition of the individual time-evolved configurations $\phi_3(t)=\phi_1(t)+\phi_2(t)$. So far so good.

Now the way we commonly think about interference of two particles is to intuitively associate each particle with one such configuration like $\phi_{1,2}$. Although this analogy holds from a field theory perspective to describe the interference of two condensates in terms of bosonic coherent states (both initial condensates are required to be in an initial superposition of different number states), this analogy is wrong if we look at two particles.

Why is it wrong? Let us look at two particles, each can be individually described by a state $|\psi_1 \rangle$, $|\psi_2 \rangle$ living in the single-particle space $\mathcal H_{sp}$. If we now want a description of the two particles, we cannot (!) just look at what happens in $\mathcal H_{sp}$ and add the two states of the individual particles $|\psi_1 \rangle +|\psi_2 \rangle$ in any way. This contradicts the fundamental postulates of dealing with more than one particle in QM. Much rather, we have to introduce the two-body space $\mathcal H^{(2)} := \mathcal H_1 \otimes \mathcal H_2$, where $\mathcal H_1$ and $\mathcal H_2$ are the Hilbert spaces of the first and second particle respectively.

Now we know that any state for these two particles can be expressed as $$ |\psi \rangle = \sum_{j_1,j_2} c_{j_1,j_2} |j_1 \rangle_1 \otimes |j_2 \rangle_2, $$ where $j_1$ and $j_2$ both run over a complete basis of single-particle states. For bosons (fermions) the resulting state still has to be (anti-)symmetrized.

So now, let's finally address your question of the simplest possible state: particle 1 is in state $e^{i \phi_1}|\psi_1 \rangle_1$ and particle 2 in $e^{i \phi_2}|\psi_2 \rangle_2$. For electrons, the 2-particle state would be $$ \mathcal S_- e^{i \phi_1}|\psi_1 \rangle_1 \otimes e^{i \phi_2}|\psi_2 \rangle_2 = e^{i (\phi_1+\phi_2)}\frac{|\psi_1, \psi_2 \rangle- |\psi_2, \psi_1 \rangle }{\sqrt 2} $$ and the two phase only contribute to the overall phase of the many-body state, which is not measurable by itself. I hope I could present it in a way that makes sense.


If you assume free particles, then you can draw an analogy to wave interference (since the free-particle wave function is a plane wave). In that case, the parameters that matter are the frequency (i.e. energy) and the propagation direction. Different energies act like different frequency waves: you wouldn't see a stable interference pattern but there would be beats. The beats would be periodic rise/fall of probability at a particular location.

You mentioned particles at rest, in that case I think it would get less interesting as the particles wouldn't really interact if they stay put.

Neutrino oscillations are in fact a manifestation of such beats. The mass eigenstates are not the flavor eigenstates (yet the flavor eigenstates are what we measure). The flavors are coherent superpositions of mass eigenstates so as the phases of each mass eigenstate evolves, at its own rate, the flavor will oscillate.

  • $\begingroup$ You have to be careful not to confuse the correlation functions (such as $g^{(2)}$ for the beats or fermionic anti-bunching) with the effect of the 'choice of phases' for the single particle states. The latter have no effect on the former. $\endgroup$ – ulf Nov 13 '14 at 5:23

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