# Non-integral particle number

1. Jul 1, 2006

### pellman

What is the mathematical basis for integral particle number in quantum field theory?

In QFT we get integral particles in the following way. There exists some operator a dependent on a parameter which we associate with, say, momentum. This operator has the property that (for bosons)

$$\left[a(p),a^{\dag} (p')\right]=\delta(p-p')$$

The $$a^{\dag}(p)$$ acting on the vacuum state produces a state with one particle with momentum p. $$a^{\dag}(p)$$ and $$a^{\dag}(p')$$ both acting on the vacuum state produces a state of two particles, one with momentum p and one with p'. Of course, if p = p' then we have $$a^{\dag}(p)^2$$ on the vacuum state resulting in two particles with momentum p.

So what mathematically prevents us from considering the results of $$a^{\dag}(p)^k$$ where k is not an integer? Would that not result in a state of k particles? That is, if k = 0.3 would we not then have a state of 0.3 particles?

The creation and annihilation operators are usually presented as explaining how integral particle numbers "fall out" of the theory, but I don't see that.

Is restricting ourselves to integral particle number just one more empirically-driven assumption?

Last edited: Jul 1, 2006
2. Jul 1, 2006

### CarlB

Great question. I would think that the interpretation of such a fractional operator would be that it would have to correspond to a subparticle, a preon.

If one presumes to require that the raising and lowering operators be defined on an algebra, then by making an assumption that the states are "primitive", one can show that the raising and lowering operators cannot be made fractional.

Hmmm.

Maybe you could figure this out by looking at the simplest algebra of all, the Pauli algebra. The raising operator is
$$\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$$
Does the above matrix have a square root? Nope.

So I guess the answer to your question is that if you assume a particular algebra you can eliminate the possibility.

Carl

3. Jul 2, 2006

### vanesch

Staff Emeritus
No, the point in (free field) QFT is not that "creation and annihilation operators exist", the point is that the 4 operators H, Px, Py and Pz, which are supposed to form (apart from some independent spin degrees of freedom) a complete set of commuting observables, can be WRITTEN as a function of these creation and annihilation operators. Even better, the 4 operators H, Px, Py and Pz can be written as a function of operators a and a-dagger in such a way that when we work out the commutation relations of the field operators, they correspond to the commutation relations of the creation and annihilation operators of a harmonic oscillator as a function of the eigenvalues of E, Px, Py and Pz.
So you should only see the a(p) and a-dagger(p) as an intermediate mathematical trick in the construction of the set of common eigenstates of H, Px, Py, Pz, not as something "fundamental" or "postulated" in itself. They follow from the postulated commutation relations of the field operators and the way H, Px, Py and Pz are expressed.

So, the reasoning goes as follows:
we have a field operator (and its conjugate momentum operator) phi(x) and pi(x), which can always be written as a Fourier expansion in other operators a(p) and a-dagger(p). This is always possible. But the properties of a and a-dagger are still open.
From the commutation relations of phi(x) and pi(x) follows that a and a-dagger satisfy the same commutation relations as the creation and annihilation operators of a harmonic oscillator (hence they have the same spectrum).

From the expression of H, Px, Py, Pz as a function of the field operators and hence as a function of the a and a-dagger, follows, that the created states from the vacuum state through some number of actions of the creation operator a-dagger(p), are common eigenstates of H, Px, Py, and Pz, in such a way, that they have the same energy and momentum as a set of particles with mass m and the appropriate momenta given by the different a(p).
So it is the fact that we can build the entire set of common eigenstates of H, Px, Py, Pz using these created states from the vacuum through a-dagger, and the fact that these states have then eigenvalues corresponding to the energy and momentum of a set of N particles with mass m, that we can conclude that these creation operators "create a particle of mass m".
It is not put in by hand upfront.

4. Jul 3, 2006

### pellman

CarlB, interesting observation about the spin operator. But it may not apply to my example of momentum, the key difference being that momentum is a continuous variable and spin discrete and limited to a few values. But I don't have the final answer on it myself yet.

vanesch,

is a summary, I am afraid I don't see how that limits us to only integral powers of $$a^{\dag}(p)$$?

Before answering, consider that my question is not restricted to the momentum representation. It applies equally well to the position representation.

Here is another way to look at it.

The number density operator $$N(p)=a(p)a^{\dag}(p)$$ commutes with itself for different values of p. On the one hand that means, of course, that we can have states |p1,p2,p3,..,pN> which we refer to as a state of N particles. The eigenvalues of N(p) are then delta functions.

But the commutation of N(p) and N(p') for all p,p' also means that we could have non-zero eigenvalues for ALL values of p. This would be a state denoted by, say, |n> where $$n:R^3\rightarrow R$$ and $$N(p)|n>=n(p)|n>$$ for all values of p. That is, a finite density value associated with each value of the continuous momentum variable instead of delta functions. Obviously(?) this is not the case in nature. But where does the restriction to delta functions enter into the math?

Last edited: Jul 3, 2006
5. Jul 4, 2006

### samalkhaiat

Last edited: Jul 4, 2006
6. Jul 5, 2006

### pellman

Of course. Even in QM, a pure momentum (or position) eigenstate is not a realizable ideal. They are not normalizable.

I recognize this line of thought from the ladder operator method for the harmonic oscillator. But why do we only permit states for which applying $a$ results in a physical state? That is, why does "$a|\psi\rangle$ is not a physical state" imply that "$|\psi\rangle$ is not a physical state"?

btw, thanks for all the mathematical detail. It really helps.

Last edited: Jul 5, 2006
7. Jul 6, 2006

### vanesch

Staff Emeritus
Well, the point is that all states in Hilbert space are physical states (that's the superposition principle), so you cannot have a linear operator acting upon any state and result in a "non-physical state".
There are no "non-physical" states in Hilbert space, a priori.
So, or you have to declare the original state as being not in the domain (or not even in the closure of the domain) of the operator a, or to accept that the result is an element of hilbert space (and hence a physical state).

8. Jul 6, 2006

### pellman

I get it. Thanks, vanesch!

9. Jul 6, 2006

### vanesch

Staff Emeritus
I see your point and I have to say I don't know the answer. Most QFT calculations are always at one point or another taken to be some continuity limit of a discrete case (by imposing box boundary conditions or the like). Probably there is some way in which you could give a meaning to a continuous function n(p) starting from a discrete grid. Remember that, for a discrete grid, if you have a discrete set of (integer) values n(p_i), and you calculate the energy, you have now an energy density which is \sum_i n(p_i) hbar omega_i over a box with volume L^3.
If you take L to infinity and you want to keep the same energy density, you have to add more and more n(p_i), but this is ok, because you have also more p_i values in the same p -> p + dp interval
So you could interpret the continuous n(p) value as some averaged value of the different n(p_i) with p_i within p -> p + dp. It doesn't have to be an integer value because it is the average, over different p_i, of integer values.

my 2 cents.
Patrick.

EDIT: this is a bit as a continuous charge distribution rho(x), which is the smoothed version of delta functions which are all multiples of an elementary charge. Nevertheless, continuous charge distributions do not have to be, at any point, discrete multiples of electron charge (in classical EM).

Last edited: Jul 6, 2006
10. Jul 6, 2006

### pellman

Patrick, I'm not sure we're talking about the same thing here. In my post N is the number density operator whereas in your post you made it the number operator by integrating over momentum space.

As I defined it, $$n(p)$$ for a state that can be written $$a^{\dag}(p_{1})...a^{\dag}(p_{M})|0\rangle$$ has the form $$n(p)=\delta(p-p_{1})+...+\delta(p-p_{M})$$. In your notation this would correspond to $$n=M$$.

Just wanted to be clear about that.

But why can't n(p) be a smooth, finite-valued function which when integrated over all momentum space gives a finite integer value?

What this would amount to is a state which is a continuum limit of the form $$\prod{a^{\dag}(p_{j})^{\omega_{j}}}|0\rangle$$ where $$\omega_{j}$$ is such that in the continuum limit $$\omega_{j}\delta(p-p_{j})\rightarrow n(p_{j})$$ (or something like that). Sort of the product analog to a Riemann sum.

And, yes, I see now that the question of a smooth n(p) is independent of the question of non-integral particle number. For we could also have a $$n(p)=(0.3)\delta(p-p_{j})$$ which would correspond to a state with 0.3 particles with momentum $$p_{j}$$. I follow your argument though and recognize that such states are "not in our Hilbert space".

I suspect though that the state with smooth n(p) can be represented as a superposition of states of definite momenta. I'll play with it some more (maybe).

Last edited: Jul 6, 2006
11. Jul 7, 2006

### samalkhaiat

Last edited: Jul 7, 2006
12. Jul 10, 2006

### vanesch

Staff Emeritus
In fact, I don't know if it is related, but I have a similar conceptual problem with QFT. The usual way of quantizing a classical system, is by taking the classical configuration space, and assign an independent dimension in Hilbert space to each point of configuration space (basis of the Hilbert space is parametrized by the configuration space coordinates).
For instance, for a single particle classical system, where the configuration space is parametrized with x,y,z for all values of x,y,z in R, it is then assumed that to each point of the configuration space, with coordinates (x0,y0,z0), there is exactly one single independent basis element in Hilbert space, corresponding to |x0,y0,z0>.
(and then there are a few mathematical problems, which can be solved by introducing "smoothing functions" and so on).

But this is not explicitly what is done in QFT. In QFT, the classical system is a field over spacetime, and its configuration space is made up of all nice-enough field configurations over space. So one should in fact assign an independent basis vector in hilbert space to each configuration of the field in space (not of each space-TIME solution of course!). Starting from that, one should then construct the H, Px, Py and Pz operators in this basis (using some suggestion from their classical counterparts). But this is not exactly what is done, although it ressembles a bit.
In other words, one should start, a priori, with a basis in hilbert space, which has the form:
|phi0(x,y,z)>, the basis vector in Hilbert space corresponding to the classical configuration point given by the field configuration phi0(x,y,z), in the same way as the basis vector |x0,y0,z0> is corresponding to the configuration point (x0,y0,z0).

The field *operators* on a state looks a bit like the set of X, Y and Z operators for the point particle: it is supposed to reproduce the field function phi0(x,y,z) for the basis states |phi0(x,y,z)>, right ? Only with X,Y and Z there are 3 operators, and for the field, there are an infinity of operators (parametrized by the numbers x,y,z here).

And then of course one can combine them in a Fourier transform, to obtain the "field operators" parametrized in px, py and pz (instead of in x,y,z), which are then re-written as functions of a(px,py,pz) and a-dagger(px,py,pz). In the game, the original "founding basis" of the Hilbert space of states |phi0(x,y,z)> has been a bit lost, however.
Now, it is probably meaningful to say that the set of eigenstates of these combined field operators is also a basis, and from that moment on, we are in the "usual story". But the direct link to the original "hilbert space basis parametrized by the configuration space" is a bit obscure to me.
Of course, you can turn things around, and say that you parametrize directly the configuration space of fields with functions of px, py, pz, and then the field operators are directly written in their "Fourier" version, and hence one can introduce directly the a(px,py,pz) and a-dagger(px,py,pz) as simple linear combinations of the field operators, which then turn out to obey the SHO commutation relations.
But it is never explicitly done this way, and I don't know if there's a hick somewhere.

13. Jul 10, 2006

### pellman

Sam,

First, I mistakenly wrote $$N(p)=a(p)a^{\dag}(p)$$. Should have been $$N(p)=a^{\dag}(p)a(p)$$. Looks like you caught on to that though.

Really, we can just write $$\left[N(p),N(p')\right]=0$$. For $$p\neq p'$$ the deltas vanish. And for $$p=p'$$ we have $$N(p)^2 - N(p)^2 = 0$$.

Since $$\delta(x-a)f(x) = \delta(x-a)f(a)$$ let us be careless and write

$$N(p)|p_{1},p_{2}\rangle = \left(\delta(p-p_{1}) + \delta(p-p_{2})|p_{1},p_{2}\rangle$$

which is what I mean when I say this is an eigenstate.

Further, we have $$N(p) \prod_{k}a^{\dag}(p_{k})^{\lambda_k}|0\rangle = \sum_{j}\lambda_{j}\delta(p-p_{j}) \prod_{k}a^{\dag}(p_{k})^{\lambda_k}|0\rangle$$.

Now here is how I would proceed to construct a state such that $$N(p)|n\rangle = n(p)|n\rangle$$

First consider this relation as a limiting form using a finite nascent delta function and then allow the $$\lambda_k$$ to go to zero in such a way that the limit of $$\lambda_k$$ times the nascent delta function remains finite as the nascent delta function approaches the delta function ( and the parameter k becomes continuous).

The trouble here is that while using the nascent delta functions we no longer have eignestates. The relation $$N(p)|p_{1},p_{2}\rangle = \left(\delta(p-p_{1}) + \delta(p-p_{2})|p_{1},p_{2}\rangle$$ only holds because of the property of the delta function quoted above. That is, if we have limiting operators such that $$\right[a_{\epsilon}(p),a^{\dag}_{\epsilon}(p')] = \delta_{\epsilon}(p-p')$$ where $$\delta_{\epsilon}(p-p')$$ is the finite-valued nascent delta function, then our eigen-relation becomes

$$N_{\epsilon}(p) \prod_{k}a^{\dag}_{\epsilon}(p_{k})^{\lambda_k}|0\rangle = \sum_{j}\lambda_{j}\delta_{\epsilon}(p-p_{j}) a^{\dag}_{\epsilon}(p) \prod_{k\neq j}a^{\dag}_{\epsilon}(p_{k})^{\lambda_k}|0\rangle$$.

That is, not an eignstate at all. This pretty much screws up any limiting scheme, I believe.

So it seems that, basically, you are correct when you say that this really only makes sense after the integrations have been performed. I am at least convinced that pursuing a finite $$n(p)$$ is not worth the trouble. :-)

Last edited: Jul 10, 2006
14. Jul 10, 2006

### pellman

.

vanesch, I think the hick somewhere is me.

15. Jul 12, 2006

### samalkhaiat

16. Jul 13, 2006

### pellman

Sam, are you saying that "N(p), like any operator, commutes with itself" is nonsense without first performing the integration over p? Just because there are some states to which we may apply an operator that result in infinities does not mean that it no longer makes sense for an operator to commute with itself.

Keep in mind that we only get delta functions in the application of $$N(p)^2$$ when applying it to momentum eigenstates. On the other hand, for any normalizable function $$\phi(p_1,..,p_N)$$

$$N(p)\int\phi(p_1,...,p_N)|p_1,...,p_N\rangle dp_{1}...dp_{N}$$
$$= \sum_{j}\int\phi(p_1,...,p_N)\delta(p-p_j)|p_1,...,p_N\rangle dp_{1}...dp_{N} \newline$$
if we are dealing with bosons, then
$$= N\int\phi(p_1,...,p_N)\delta(p-p_1)|p_1,...,p_N\rangle dp_{1}...dp_{N}$$
$$=N\int\phi(p,p_2...,p_N)|p,p_2...,p_N\rangle dp_{2}...dp_{N}$$

No integration over p is necessary for this to make sense. And obviously $$N(p)^2-N(p)^2=0$$, as must be true.

Last edited: Jul 13, 2006
17. Jul 13, 2006

### samalkhaiat

18. Jul 13, 2006

### pellman

I agree that you have to be careful. But sometimes infinity minus infinity DOES equal zero. Your hesitancy could be likened to saying that differentiation doesn't make sense because it is zero divided by zero.

Your statement amounts to saying that $$N(p)^2|p'\rangle = N(p)^2|p'\rangle$$ doesn't make sense. Whatever else, shouldn't we at least hold onto the Reflexive Property that A = A?

But this difference aside (infinities are always troublesome) thanks for your help in sorting this out.

Last edited: Jul 14, 2006