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Let $b_k^\dagger ,b_k$ represent the creation and annihilation operators for an electron in state $k$. Let $d_j^\dagger ,d_j$ represent the same for a positron in state $j$. And let $|0\rangle$ represent the vacuum.

Is it possible to have a state described by $ \left( b_k^\dagger + re^{i\theta} d_k^\dagger \right)|0\rangle $? I include the $re^{i\theta}$ for generality.

How do I interpret such a state? If I make measurements of the number of particles, the energy, the momentum, the charge, etc... what would I observe?

The question of how many particles is easy. The answer is 1. (On that note, can we have superpositions of states with differing number of particles?)

What the energy and momentum are depends on what the labels $k$ and $j$ mean.

But what about charge? What would I measure for charge? If we enclosed the system in a box that measured the electric field, what would we get for $\oint{\vec{E}}\cdot d\vec{A}$?


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I can't formulate a concise answer on this, but I am sure someone will. In the meantime you can read about Super-selection-rules: – DJBunk Jul 11 '12 at 12:20

2 Answers 2

The Vacuum sector Hilbert space is generated by the action of $U(1)$ invariant operators $\overline{\hat\psi(x)}\Gamma\hat\psi(x)$ (and their Fourier transforms, where $\Gamma$ is an arbitrary Dirac matrix) on the Lorentz and translation invariant vacuum vector, which does not allow the state you ask about to be constructed because $U(1)$ invariant operators do not change the charge. One can also discuss this in terms of superselection rules, which is, however, an alternative presentation of $U(1)$ (or other) invariance of observables. [Strictly speaking, note that $\hat\psi(x)$ and its Fourier transform are operator-valued distributions, not operators, which introduces issues that anyone who might be concerned about such details can fill in.]

The question of how many particles there are in a given state is best answered in aggregate: what is the aggregate number of electrons minus the number of positrons, which for the vacuum sector is zero. One can create other Hilbert spaces, with different aggregate numbers of electrons/positrons, which can then be used to construct mixed states, but not superpositions.

There is, however, a lot else that could be said.

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We can have mixed states of different numbers of electrons and positrons, but not superpositions of such states? Please explain. I don't see why we cannot have an arbitrary superposition of many body states. – ChickenGod Jul 18 '12 at 8:09

Yes, you can make a physical system for which the Hilbert space contains states which are superpositions of electrons and positrons.

Mathematically, this is because the Hilbert space depends on the boundary conditions at spatial infinity, and these boundary conditions may be themselves be quantum superpositions. (Note, this generalizes rather than contradicts Peter Morgan's answer. He is considering the vacuum sector, which means trivial boundary conditions at infinity.)

Gedankenexperimentally, you trap electron-positron pairs in separate jars and then use a binary observable derived from an atomic system to decide which to mail off to Alpha Centauri. The Hilbert space which describes your remaining jar will be in a superposition of states.

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Since your example includes the act of observation (that is, the amplification of a binary system sufficient to separate a particle out to infinity and beyond), I believe that you will have a classical (statistical) mixture rather than a quantum mixed state. My understanding is that superselection rules imply that these sorts of electron positron states can always be thought of as merely statistical mixtures. – Carl Brannen Jul 12 '12 at 1:44
@Carl Brannen: Perhaps the example is broken, but are you so sure that separation to infinity necessarily causes decoherence? Suppose we catch one particle in a magnetic trap and let the antiparticle sail off to infinity. – user1504 Jul 12 '12 at 2:17
I'm not an expert in this; it just seems intuitive that sending a particle to infinity is the type of amplification that accompanies a measurement process. I'd like to see an example which doesn't dispose of the body that way. – Carl Brannen Jul 12 '12 at 17:11

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