# Second Quantization and Field Operators

1. Mar 10, 2008

### mkrems

When defining a field operator, textbooks usually say that one can define an operator which destroys (or creates) a particle at position r. What does this really mean? Are they actually referring to destroying (or creating) a state who has specific quantum numbers associated with the geometry that they call r. Any insight would be appreciated. Thanks.

Last edited: Mar 10, 2008
2. Mar 12, 2008

### Fredrik

Staff Emeritus
A creation operator is just a linear function that takes an n-particle state to an n+1-particle state, and an annihilation operator takes an n-particle state to an n-1-particle state (and the vacuum state to the 0 vector), so the really short answer to your question is "yes". A longer answer would explain stuff like what one-particle states are, and how the Hilbert space of n-particle states can be constructed from the Hilbert space of one-particle states.

Also, the term "second quantization" is kind of old-fashioned and useless.

Last edited: Mar 12, 2008
3. Mar 12, 2008

### Demystifier

Maybe in particle physics.
But not in solid state physics or even string theory.

4. Mar 12, 2008

### pellman

mkrems,

A theory which incorporates the fact that particles are "created" and "destroyed" necessarily has to include an operator which, when applied to a state of N particles, gives a state of N+1 particles.

At the theoretical level, what we really mean when we say particles can be "created" or "destroyed" is that states of different particle number are no longer orthogonal. If particle number is constant, then the inner product between a state of 2 particles and a state of 3 particles would necessarily have to be zero. That is, given a state of 2 particles, the probability of observing 3 particles would be zero. Letting particle number vary means we now have non-zero probability of observing a different particle number than what we started with.

Basically, the only way to do this is for the Hamiltonian to be built from an operator which changes particle as described above. If the Hamiltonian did not have such a term, then the time-evolution of a pure state of N particles will only, can only, evolve into states of N particles, forever and ever, amen.

Last edited: Mar 12, 2008
5. Mar 12, 2008

### vanesch

Staff Emeritus

Uh, states with different particle number ARE orthogonal. What you are referring to is the time evolution operator ; in other words, you are confusing final states and initial states. So it is correct that an n-particle FINAL state is not necessarily orthogonal to the TIME EVOLUTION of an m-particle INITIAL state.

However, an m-particle initial state is still orthogonal to an n-particle initial state, and similar for final ones.

In the end, we're talking about zero or non-zero elements of the S-matrix.

6. Mar 12, 2008

### pellman

You give me pause, vanesch. I have been corrected here before. Here is how I understand it.

If states A and B are orthogonal, then given state A, the probability of observing state B is zero. Period. The only way you could ever observe state B is if it is not orthogonal to A.

Consider a Hydrogen atom with the electron in an excited state. In the absence of the EM field, the excited state is orthogonal to the ground state and no transition can theoretically ever occur. When you perturb the system with the EM field, then you find that the state |electron excited + zero photons> is NOT orthogonal to the state |ground state electron + photon(s)>, i.e., a transition can occur.

What else do non-zero off-diagonal elements of the S-matrix represent? If the probability of putting m particles in and getting n particles out is non-zero, that is the same thing as saying the states are not orthogonal.

If phi is a state of m particles and psi is a state of n particles:
$$P=|\langle \phi(t_1)|\psi(t_2)\rangle|^2$$

7. Mar 12, 2008

### vanesch

Staff Emeritus
Indeed: given an initial state of 5 electrons, the probability to have an initial state of 6 electrons is zero!

A *time-dependent* transition! At t = 0, your "probability for transition" is 0. After a time t, you have a finite probability of transition, which, in the approximation of Fermi's Golden Rule, is linear with t, so the coefficient will give you the "probability per unit of time" to decay.

The only thing that it means is that |electron excited + 0 photon> is not an eigenstate of the full hamiltonian.

Do you notice that you use "in" and "out" ?

Do you notice that you have a t1 and a t2 ?

Orthogonal states means: "have distinguishable properties".

Consider (in NR QM), the "position states" |x1> and |x2> . I hope you agree with me that for x1 not equal to x2, these are orthogonal states, right ?

Now, consider that we start with an electron in position state |x1>. A bit later, we find a non-zero probability to find it in state |x2>. Does that now mean that they are, after all, not orthogonal ? Of course not. It means that the TIME EVOLUTION operator U(t1,t2) has mapped |x1> onto a state that is not orthogonal to |x2>.

< x2 | U(t1,t2) | x1 > is non-zero. But that doesn't mean that < x2 | x1 > is non-zero, it only means that U(t1,t2) has a component | x2 > < x1 |.

8. Mar 12, 2008

### pellman

Hey, I didn't say anything about time-independence.

9. Mar 12, 2008

### reilly

There's a lot of history here, started by Heisenberg's initial solution of the harmonic oscillator by matrix mechanics. What he found was that linear combinations -- sum and difference of position and momentum operators made the algebra easier. They transformed one state into another: going from x to x' in configuration space is a translation; a, the destruction operator, for example, moves a state by -1, from a state with energy wN to one with E =W(n-1). These operators were once referred to as ladder operators and step operators. Step operators are used extensively in angular momentum theory, and, more generally, in the study of Lie groups.

Prof. Fock developed so called Fock space, in which the step operators create a representation of a space containing from 0 an infinite number of oscillators. So quantum fields form the basis of a very efficient formalism for systems and interactions that do not conserve particle number. You could, if you wanted, do all of quantum field theory with configuration wave functions -- the two approaches are connected bya unitary transformation -- and you would quickly find out why the usual formulation of QFT is so much in vogue.

By the way, 2nd quantization is a mis-characterization of QFT, which is nothing more than ordinary quantum theory in a Fock basis. A quantum field is simply a highly useful mathematical construct, just like say, a Bessel function, a vector, resolvents, complex numbers, in fact.

Regards,
Reilly Atkinson

Last edited: Mar 12, 2008
10. Mar 12, 2008

### kdv

But you said that "states of different number of particles are not orthogonal". Usually one woul dinterpret that statement as saying you meant "states of different number of particles at a given time are not orthogonal" which is what Patrick is saying is wrong.
I guess you meant states at different times which was not clear from your initial statement ( and if we start discussing states at different times, the time evolution of the system most be specified). I think that's the whole point of Patrick's objection.

11. Mar 12, 2008

### reilly

1. As vanesch pointed out, states with different numbers of particles are always, repeat always orthogonal, quite independently of any time behavior. This is basic to the notion of Fock space;
<N| N+M> = 0 unless M=0 -- just put in your step operators, unless the number of a's on the left equals the number of a*'s on the right you get zero.

You might argue that as time goes on an initial state with fixed numbers of particles will generally evolve into a state with most any number of particles. This evolved state clearly then is no longer a state with a fixed N. Then you typically expand the state on the basis of the orthogonal |N> Fock states with N=0 to infinity.

The standard QFT interactions are built primarily on the 3-point interaction, which allows one particle to transform one to two, or two to one, or nothing to three, and three to nothing -- the two later types represent the vacuum creating, say, a electron positron pair along with a photon, or the other way around. Your statment about the Hamiltonian is correct
Regards,
Reilly Atkinson

12. Mar 13, 2008

### Count Iblis

It is an artifact of overcounting. The step where you consider the Schrödinger equation to be a classical field equation should be considered as quantizing minus one times, so overall it is 2 - 1 = 1.

13. Mar 13, 2008

### Count Iblis

Also, I've read that some people proposed calling quantizing gravity as "Third Quantization"

14. Mar 13, 2008

### Fra

Philosophical reflection

See this amusing reflection of John Baez about the n'th quantization.

http://math.ucr.edu/home/baez/nth_quantization.html

I always made loose philosophical connections between string theory as a constrained case of higher order quantization.

If we are talking about indistinguishable particles, it seems clear that we can not distinguish between a multiparticle system or wether the SAME particle seems to be all over the place, or beeing "smeared out" like an extended object?

IE. is a state smeared out as an extended object, or do we have several indistinguishable objects? And what's the difference?

The multiparticle interpretation vs the "second quantization". IMO it seems to be different "interpretation" of the same thing. I always actually liked the name "second quantization".

It even suggest here an inductive scheme. Which is what Baez reflects on. And others have done so as well.

/Fredrik

15. Mar 14, 2008

### reilly

Can you tell me where I might find the spectral resolution of Baez's K operator -- as in, for example does K^^N, as N goes to infinity, converge to a finite result? From Baez's description it seems to me that K is a unitary operator; but I'm far from certain about ascribing that characteristic to K.

Thanks,
Reilly Atkinson

16. Mar 15, 2008

### Fra

K is the map from the category of hilbert state vectors and linear operators on this space, to the category of "fock" state vectors and the linear operators on fock space.

K^inf would results in some infinite dimensional mess, increasing the degrees of freedom - this is why I don't think receipe along makes sense. It would not converge to anything useful IMO. But I still find the reflection interesting.

I make the following parallell association here...

Consider a distinguishable event x.

1) Either the event happens, or it doesnt {0,1}

2) Consider that we inflate our information capacity in on dimension, we can now consider that continous probability that this event occures {p(x)} ~ [0,1] ~ R

3) Consider again that we inflate our information capacity in another dimension, we can now consider a continous probability for a certain probability {p(p(x))} ~[0,1]x[0,1] ~ R^2

We go from point, to string, to plane; 0-brane, 1-brane, 2-brane.

I know this is fuzzy, but to try to formalize this isn't the interesting part IMO. It's the conceptual thing behind it.

This is connected to how I consider dimensionality to be dynamic, but I do not do it like above. Instead of considering a continous string, one can cnosider "string bits", and that way it's easier to understand how objects of different dimensionality can morph into each other, as part of what I consider to be an optimation problem.

The limiting physical information capacity is my main guide here. A continous string, may be recovered as an approximation but I've got a feeling that it contains far too much ghost degrees of freedom. I don't think the continuum is physically observable, and therefor I would prefer not to see it in the models either.

/Fredrik

17. Mar 15, 2008

### Fra

My original association is that if one like me, consider that the information has subjective reality in the observers microstructure then the state of knowledge of a point, can in fact look like a string. IE. the IMAGE of a point, can look like a string due to uncertainty. This is the coupling I make. But there seems to be more than one way to interpret this.

/Fredrik

18. Mar 15, 2008

### Fra

It seems most stringers consider the string real in another way, and rather thinkgs that the string is compactified and looks like a point. But that way of thinking is similar to the bird -> frog projection I don't think makes sense.

I prefer to say that the frogs uncertainty inflates the point into a string. And I think this can be understood without actually postulating the existends of strings.

That said I don't like the string theory foundations as it stands but it's still interesting reflections to compare views. So even though I disagree, I can see where the strings come from. It's just that I would choose to see if quite differently.

/Fredrik

19. Mar 15, 2008

### Fra

I think the preferred dimensionality might be understood as a generalization of the principle of maximum entropy. A too high dimensionality will decrease the certainty of information for obvious reasons, since the degrees of freedom we have no control of are inflated. A too low dimensionality will not be stable, since it keeps changing - ie the degrees of freedom is too low to describe the situation.

I'm sure there has been alot of work on this. But I see this as part of the problems that aren't yet solved to satisfaction. It's in this larger task I find the reflection on the quantization procedure as an induction step food for though, but not alone the solution.

Reillty, this is what I mean before when I don't think unitariy in the most general case can be maintained. To maintain it, I think we are forced to increas our degrees of freedom beyond what we can relate to. And I think the results is that we get lost in a landscape too large to relate to.

I think there may be another way, that instead of applying the standard QM procedure over and over again - tweak the procedure. I think there will still be a procedure, but then we can find a physical meaning of the procedural progress - time!

Thats my vision, but don't ask me to prove it, I can't. But it's the trac I'm tuned in on.

/Fredrik

Last edited: Mar 15, 2008
20. Mar 15, 2008

### strangerep

K is a functor, -- neither an operator, nor unitary. It maps from one Hilbert space to a
quite different one. E.g., from a 1-particle Hilbert space to a multiparticle Fock space.

That means it's not an operator, because operators act from one space to the same space.
Also, it can't be bijective, since (eg) the Fock space is larger than the 1-particle space.
Therefore, K can't have a well-defined inverse in the usual sense. Preserving an
Hermitian inner product also doesn't make sense here, since the spaces are different.