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Why integral particle number?

  1. Nov 14, 2007 #1
    In a free field theory, at least, we have mathematically a state of N particles with momenta [tex]k_j,j=1,..,N[/tex] being constructed from the vacuum state according to

    [tex]|k_1,...,k_N\rangle\propto a^\dag(k_1) \cdot\cdot\cdot a^\dag(k_N)|0\rangle[/tex]

    where [tex]a^\dag(k)[/tex] is the suitable creation operator. (I'm not sure yet about what interactions do to this. I'm not that far along yet.)

    So from a theoretical perspective, what prevents us from applying the creation operator a non-integral number of times?

    To put it another way, the particle density operator (in the momentum representation) acting on a state of N particles yields something like

    [tex]n(k)|k_1,...,k_N\rangle\propto \Sigma_j\delta(k-k_j)|k_1,...,k_N\rangle[/tex]

    Why are the eigenvalues here restricted to delta functions? Why not something like

    [tex]n(k)|\rho\rangle\propto \rho(k)|\rho\rangle[/tex]

    where [tex]\rho(k)[/tex] is a continuous function? I realize a continuous number of particles is an even bigger jump than just a fractional number. But still why not? Why are "partices"--excitations of the field--discrete?

    Constructing such a state out of the creation operators is a little tricky. It would involve something in the way of an infinitely continuous product of the creation operators, the product analog to the passing from Reimann sums to integration. But just starting from the vacuum state and the creation & annihilation operators, I don't see why it would not be allowed.

    Of course, observation requires integral particles. But where does it come into the theory? As an analogy, there is integral spin, which is demanded by the algebra of the generators of rotation. Is there something like that for particle number? Or is it one more postulate tacked on?
    Last edited: Nov 14, 2007
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  3. Nov 14, 2007 #2


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    The answer is kinematical and mathematical rather than dynamical and physical.
    The point is that the basis consisting of all states with an integer "particle" number is COMPLETE. Any QFT state at given time can be written as a linear combination of such states.

    It is illuminating to compare it with the Hermite polinomials. They represent a discrete basis for any function of a real variable x. Besides, they are energy eigenstates for the quantum harmonic oscillator. But they are a good basis for any other quantum interaction as well.
  4. Nov 14, 2007 #3


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    What type of Lie algebra do the creation and annihilation operators form ? Could you compare it to some similar algebra met at a simpler level ?
  5. Nov 14, 2007 #4


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    I'm not sure what you mean with this question. The relevant fact about the particle number operator (for simplicity consider the 1D harmonic oscillator) is that it is proportional (up to the vacuum contribution) to the energy of the corresponding eigenstate of the Hamiltonian. Subtracting the vacuum energy and dividing by twice the vacuum energy to obtain the "particle number" is just a matter of convenience. You could divide N by 3 to get a fractional particle number. Does this change anything for you ? I hope not. The energy levels remain equidistant.

    Maybe one could actually define what it means to apply the creation operator an arbitrary number of times, probably by something like

    [tex](a^\dagger)^\lambda = e^{\lambda \ln a^\dagger}[/tex]

    but this doesn't give you an eigenstate of the harmonic oscillator (which is what you actually want).

    On the other hand, if you have a many photon state, nobody can keep you from defining a state density [itex]\rho(k)[/itex] as a continuous form of the occupation number representation. Of course you can represent any continuous function as a "sum" (i.e. integral) of delta-functions:

    [tex]\rho(k) = \int d k^\prime \rho(k^\prime) \delta(k-k^\prime)[/tex]

    Yes, I guess this is what you really wanted to know. But this has nothing to do with fractional or even continuous particle number but rather with discretely vs. continuously distributed wave vectors k.
  6. Nov 14, 2007 #5
    Demystifier, I understand that the space that we consider is precisely that spanned by integral-particle states, and thus any allowed state is going to be expressible as a sum of integral particle states. But why? What is the theoretical justification?
  7. Nov 14, 2007 #6
    I guess what you are asking for is [tex][a(k),a^\dag(k')]\propto \sqrt{k^2+m^2}\delta^3(k-k')[/tex]
  8. Nov 14, 2007 #7
    Hi pellman,

    The question you are asking indeed looks non-trivial if you treat quantum fields as fundamental objects and derive particles as "field excitations".

    However, this question doesn't make sense if QFT is formulated as a theory of particles (the Fock space is built as a direct sum of tensor products of 1-particle Hilbert spaces, creation and annihilation operators are defined in this Fock space, particle number operators explicitly have integer eigenvalues,...) and quantum fields are defined simply as certain linear combinations of particle creation and annihilation operators. In this case, integral particles are built into the theory from the beginning. This is how QFT is approached in Weinberg's "The quantum theory of fields" vol. 1., and this approach looks more physical to me.

    Last edited: Nov 15, 2007
  9. Nov 14, 2007 #8
    OOO, what I am refering to here is this.

    If we define the particle density (or number) operator as [tex]\rho(k)=a^\dag (k)a(k)[/tex], then we find that [tex]\rho(k)[/tex] and [tex]\rho(k')[/tex] commute for all values of k,k'. Therefore we can in principle find states that are simulaneous eigenstates for [tex]\rho(k)[/tex] all values of k. That is, there is some function n(k) and state [tex]|n\rangle[/tex] such that

    [tex]\rho(k)|n\rangle=n(k)|n\rangle[/tex] for all values of k.

    But in standard QFT, n(k) is a sum of delta functions if we are dealing with a state of definite momentum. If we have N particles, n(k) is a sum of N delta functions. If we do not have definite momentum, then n(k) takes some other form but presumably its integral over all k-space must still be N, where N is an integer.

    why is N--the integral of n(k) over all k-space--restricted to be an integer?

    I probably should not have volunteered all the math in my opening post like I know what I'm talking about. All I am really asking is where does discrete particles enter into the theory? Is it a fundamental assumption to make the theory fit with observation? Or is it a consequence of prior mathematical assumptions?

    In non-relativistic quantum mechanics it is clearly an assumption, though it is usually not explicitly stated.

    The way the QFT is presented, it looks like integral particles go hand-in-hand with the existence of the creation/annihilation operators. But I don't see that a good QFT type theory could not be formulated which allows for non-integral number of particles (which I guess would be a continuum theory then and no longer quantum theory, of course).
  10. Nov 14, 2007 #9


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    But you can certainly have situation where you do NOT have a well defined number of excitations in a system. AFAIK Fock states are quite rare in the "real world" since they are quite difficult to create.
    A coherent state (e.g. light from a laser) can be created by operating with the unitary displacement operator [itex] D(\alpha)=\exp(\alpha a^\dag -\alpha^* a) [/itex] - alpha being some complex number- on the vacuum.

    The coherent states contain an indefinite number of excitations (in the case of light photons) and form a two-dimensional continuum of state. They are also complete.

    Hence, while you can expand the coherent states in terms of number states this does not neccesarily imply integral particle numbers.
  11. Nov 15, 2007 #10


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    It comes into the theory this way: you start with a non-interacting theory, because these are the easiest to understand. A theory without interaction corresponds to a classical field equation which is linear in the fields. A linear classical equation of motion corresponds to a Lagrangian which is square in the fields and its derivatives. But such a Lagrangian is nothing but a large set of coupled harmonic oscillators. You decouple them by making a spatial Fourier transform, and after that, the amplitudes for the respective k are not coupled to one another and behave like 1D harmonic oscillators. Now you quantize them which means you're dealing with the eigenstates of the 1D quantum harmonic oscillator. You notice that, mathematically, you can "generate" the eigenstates by hand with creation operators (but nothing is ever created or destroyed during the actual time evolution of the linear theory !!!).

    After that you try to understand more complicated theories, the ones with interaction. They are nonlinear (the most simple example in the textbooks being the phi4 theory). Here the eigenstates are not harmonic oscillator states any more. This is good because if they were, no particles would be created or destroyed (since eigenstates of the Hamiltonian are stationary). At first sight it seems your hard earned knowledge about the harmonic oscillator is useless here. But remember: often the interaction/nonlinearity is comparatively weak, so you can do perturbation theory. Thus you start again with the well-known noninteracting theory and treat the nonlinearities as small perturbations. In this approach you can expand the time evolution of the interacting theory in terms of the (complete set of) eigenstates of the noninteracting theory (the harmonic oscillator), which you can construct from creation and annihilation operators. Here you are !
    Last edited: Nov 15, 2007
  12. Nov 15, 2007 #11


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    I don't think this is really the case. The integral over a density could be anything. See post #4. The density is written as an integral (the continuous representation of a sum) of delta functions. Nevertheless it is an arbitrary function.

    You would get an integer particle number if you had restricted the problem to a spacetime cube instead of all spacetime. Then the wave number k would become discrete, the integral would become a sum and the delta function would become a kronecker delta.

    PS: As far as I can see the whole confusion comes from calling N(k) the particle number operator. To clear this up, take a look at what Ryder says after equation 4.20:

    As an illustration let's suppose you really had a one particle state of wave vector k. This means the system is in the first level of a quantized, infinitely extended plane wave. This is utterly unrealistic. In the real world, waves and detectors aren't infinite. So what you actually have is a continuous density of field quanta with say wave vectors [itex]k\in \left[k_1,k_2\right] [/itex]. In my opinion this is exactly what you were asking for. The question is only, how does one create such a continuous distribution by multiplication of the discrete creation operators ? I don't know, but f95toli's post #9 is part of the answer.
    Last edited: Nov 15, 2007
  13. Nov 15, 2007 #12
    i think this is a little confusing
    I guess pell mann is asking why the eigenvalues of the number operator acting on Fock space is restricted to integral values.
    And why not a real quantity say like 12.5637
    Well now does it it make any sense to say that so and so fock space is an eigenket of N(k) having eigenvalue of 12.5637.
    In other words the field is associated with 12.5637 quanta of momentum k.
    In that case QFT would be a CFT
  14. Nov 15, 2007 #13


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    Hm, actually its quite more easy. Just consider the state proportional to [tex] |\psi\rangle = 3 a^\dag(k_1) |0\rangle + 7 a^\dag(k_1)a^\dag(k_1) |0\rangle[/tex]. How many particles are in this state ? You can answer this question by calculating the mean number of particles:

    [tex]\frac{\langle\psi| N |\psi\rangle}{\langle\psi|\psi\rangle} = \frac{\langle\psi| \int dk a^\dag(k) a(k) |\psi\rangle}{\langle\psi|\psi\rangle}[/tex]

    which is obviously non-integer in this state. But when you actually do any measurement that involves detection the number of particles (i.e. almost any measurement made using particle detectors) the state should collapse to one of the eigenstates of N which are the states with integer number of particles (1 and 2 with propabilities ~0.15 and ~0.84 for this state). Similarly any physical state with non-integer number of particles can be written as a sum of the states with integer number of particles and after measurement of N will collapse to one of them.

    Why all eigenstates of N has integer number of particles ? The reason is just the same as for why oscillator has discreet energy level and can be obtained algebraically from commutation rules [tex][a^\dag(k) a(k')]=\delta(k-k')[/tex]. And here it doesn't matter how you formulate you QFT: starting from fields and then doing canonical quantization or starting from particles and building Fock space.
  15. Nov 15, 2007 #14


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    See the proof that the hermite polinomials represent a complete basis for the space of functions of x. The integral-particle states are nothing but a generalization corresponding to an infinite-dimensional harmonic oscillator. To see this explicitly, you need to work in the functional Schrodinger picture of QFT. See e.g. the book
    B. Hatfield, QFT of point particles and strings
    If you do not have this book, I can also recommend you some papers.
  16. Nov 15, 2007 #15

    Not exactly. The example you give here is a superposition of two states of integral particle number. What I am talking about is a pure state of non-integral particle number. If you make an observation of your example state you would either observe one particle or two particles. what I mean is a state which represents an observable value of particle number which is non-integer. You don't get that by summing over states of differing integral particle number. The only way to get it would be by applying the creation operator to the vacuum state a non-integral number of times. Something like

  17. Nov 15, 2007 #16
    No. This is just the difference between a free particle and a particle in a box. Or ten free particles vs ten particles in a box. It doesn't have anything to do with the particle number.

    There is a problem inherent in the notation [tex]|k_1,...,k_N\rangle[/tex]. It already assumes a integral number of particles. If particle number was continuous, you would have to have a different notation.

    Just to clarify, in my post I was using [tex]\rho(k)[/tex] for the number density operator and [tex]N = \int\rho(k)d^3k[/tex] is the number operator. Correct me if I am wrong but the eigenvalues of N are necessarily integers. Superpositions of states of different numbers of particles might be allowed but then these states are no longer eigenstates of N.

    Let's take another look at what I am asking about. This time in a finite box. The number density operator and the number operator are essentially the same entity in this case and now takes on values only at the discrete values of k. Acting on a usual momentum eigenstate looks like


    If then we sum over the allowed values of k, we get N (the number, not the operator, I mean.)

    So the question is, why are the eigenvalues of the number density operator restricted to Kronecker deltas in the finite box case and Dirac deltas in the infinite case?

    To avoid the notation problem, denote a generic state by [tex]|g\rangle[/tex] where g(k) is some function defined for the allowed values of k appropriate to the box. g gives the eigenvalues of the number density operator according to

    [tex]\rho(k)|g\rangle\propto g(k)|g\rangle[/tex].

    Why are the allowed values of g(k) only 0,1,2,3,...? And in the infinite-box case, why is g(k) restricted to a sum of delta functions?
  18. Nov 15, 2007 #17
    Yep. That's it.

    One possibility might be that if we allow non-integral application of the annihilation operators in our theory, then we end up with states representing a negative number of particles (apply the annihilation operator to a state of 0.5 particles and you get -.5 particles). Since this is meaningless, we only allow integral particle number states.

  19. Nov 15, 2007 #18


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    I think the operator [tex](a^\dag)^\frac{4}{5} [/tex] does not exist. As a check, try to find its matrix representation, given the matrix representation of [tex] (a^\dag)[/tex].
  20. Nov 15, 2007 #19


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    Using the analogy with the hermite polinomials, a task analogous to yours is to find a way to apply a derivative operator d/dx a non-integer number of times. It should be clear to you that you cannot do that.

    In fact, there is a formal way to do that by using the Cauchy formula of complex analysis. But in fact this formula leads to troubles when generalized to non-integer parameters.

    Yet another analogy, closely related to the observations above is the fact that a function analytic around x=0 can be represented by a series
    [tex] f(x)=a+bx+cx^2+ ... [/tex]
    Note that only non-negative integer powers of x appear.
  21. Nov 15, 2007 #20
    Well, of course it is not going to have a matrix representation if we are only allowing states that represent integral particle number. ;-)

    It appears to me--though I am not 100% sure--that integral particle is built into the theory from the start by our choice of what mathematical entities we build the theory from, which are, namely, integral particle states. It is not that the theory forbids non-integral particle number; it is that non-integral particle states do not even make sense from within the theory. Like the square-root of -1 makes no sense if we are only working on the real line.

    The question is then, could we, by broadening the sort of mathematical entities we start with, produce a theory which otherwise is identical to the existing theory but allows non-integral particle number states? The answer might be "no". It might be like asking to describe the geometry of a space with 3.14 dimensions. It might not make sense no matter how you look at it.
    Last edited: Nov 15, 2007
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