Is intregral particle number a theoretical assumption?

[pellman]

Clearly integral particle number is an experimental fact. But within quantum theory, does integral particle number follow as a mathematical necessity? or is it itself an additional assumption?

In QM it is built into the mathematical infrastructure: the basic entity is the wave function and the number of position variables must be an integer. If $$\psi(x_1)$$ is a wave-function for one particle and $$\psi(x_1,x_2)$$ is a wave function for two particles, there simply is no way to write down a wave-function for 3/2 particles.

But in QFT we apply a raising (creation) operator $$a^\dag (p)$$ to the vacuum state $$|0\rangle$$ to get multi-particle states. So for example $$a^\dag (p_1)^4 a^\dag (p_2)^3|0\rangle$$ gives a state of 4 particles with momentum p_1 and 3 with momentum p_2.

Is there any mathematical or theoretical reason why we cannot consider non-integer powers of the creation operator?

 

[alxm]

In QM it is built into the mathematical infrastructure: the basic entity is the wave function and the number of position variables must be an integer. If $$\psi(x_1)$$ is a wave-function for one particle and $$\psi(x_1,x_2)$$ is a wave function for two particles, there simply is no way to write down a wave-function for 3/2 particles.

Well, the configuration space isn't necessarily immediately related to the particles; but I don't see any way around this, no. But if I may drag in my quantum-chemical perspective, there's still the Hohenberg-Kohn theorem, i.e. you can still (in principle) determine the ground-state energy of a system (of electrons, say) solely from its particle density distribution.

There's nothing in this formalism which stops you from having a non-integer number of particles. But http://prl.aps.org/pdf/PRL/v49/i23/p1691_1" , is that you then end up with a non-differentiable energy with respect to particle number - the derivative has discontinuities at integer values.

Is there any mathematical or theoretical reason why we cannot consider non-integer powers of the creation operator?

Well, how would you do it?

 

[pellman]

Well, how would you do it?

I don't know. Could be the answer is "non-integer powers of the creation operator cannot be meaningfully defined." Could just be that simple.

 

[Parlyne]

The answer is that there's actually nothing wrong with the idea of states with non-integer numbers of particles. Consider the coherent states of light. These are eigenstates of the photon annihilation operator, rather than the photon number operator. The coherent state $|\alpha \rangle$ can be represented as
$$|\alpha \rangle = e^{-\frac{|\alpha |^2}{2}}e^{\alpha \hat{a}^\dagger}|0\rangle$$.
This is clearly a state of indefinite photon number.

This sort of state will generally only be seen for particles which carry no conserved charges, as the creation processes for such particles will need to involved definite numbers of such charges.

 

[pellman]

Parlyne, coherent states have indefinite particle number in the sense that they are not eigenstates of the particle number operator, i.e. the probability of observing the system to have any specific number of photons is less than 1. But (as far as I know) it is still a superposition of only integral photon number states. The probability of observing, say, 1.3 photons is zero.

I am assuming that a state of 1.3 photons is orthogonal to all the integral number states the way that the integral number states are orthogonal to each other.

 

[haael]

I am assuming that a state of 1.3 photons is orthogonal to all the integral number states the way that the integral number states are orthogonal to each other.

What you write here is pretty cool, I mean it touches the very roots of QM.

I have 2 answers:
1. The particle count is integer by definition. I mean, if you ever registered 1.3 of a photon, then you could change your definition of particle to 0.1 of a photon ("subphoton"), and say that you got 13 subphotons.
The spectrum of a particle count operator of bosons looks like N, while for fermions looks like Z_1. Kinda deep for me.
2. When constructing QM, at some point you normalize the states to 1:
$$<a|a> = 1$$
I think that's the source of integer particle count. If you chose some other number, it would become minimal nonzero particle count.

 

[haael]

The spectrum of a particle count operator of bosons looks like N, while for fermions looks like Z_1.

I remember now: it comes from a harmonic oscillator, boson or fermion respectively. A harmonic oscillator has a discrete set of states.

 

[Parlyne]

I remember now: it comes from a harmonic oscillator, boson or fermion respectively. A harmonic oscillator has a discrete set of states.

Right. The point here is that the Fock (number) states are a complete set of states for the free field. So, any state that does not have an integer number of particles still must be a superposition of Fock states. That, in turn, means that it shouldn't be possible to have a definite non-integer number of particles.

 

[pellman]

Right. The point here is that the Fock (number) states are a complete set of states for the free field. So, any state that does not have an integer number of particles still must be a superposition of Fock states. That, in turn, means that it shouldn't be possible to have a definite non-integer number of particles.

This makes sense. But it is another way of saying that integral particle number is an assumption, because by saying that Fock number states are a complete of set of states is the same as saying we are only considering states which are superpositions of integral particle number states.

This is fine since particles are an experimental fact. But let us be clear that within the theory integral particle number is a starting assumption and not a consequence of more fundamental principles.

 

[sheaf]

This is fine since particles are an experimental fact. But let us be clear that within the theory integral particle number is a starting assumption and not a consequence of more fundamental principles.

It's a starting assumption , but it seems like a reasonable assumption, since we observe in nature countable quanta satisfying $E=h\nu$. I wouldn't know what to do with the square root of a creation operator.