Time evolution in quantum field theories

[meopemuk]

Arnold,

before doing any specific exercises I would like to understand the logic of what's going on.

It seems that we both agree that we are doing quantum mechanics in a certain Hilbert space (Fock space or non-Fock space, whatever). One of quantum postulates is that the time evolution is described by the Hamiltonian H as

|\psi(t) \rangle = \exp(iHt) |\psi(0) \rangle......(1)

If I understand correctly, the axiomatic Wightman approach does not use this formula and even refuses to provide an explicit form of the Hamiltonian H. Instead, it is suggested to use a set of Wightman functions in order to do the time evolution. I can believe that such Wightman functions can be calculated (perturbatively or non-perturbatively, exactly or approximately, it doesn't matter now). I can even believe that using these Wightman functions one can reproduce the same time evolution as given by equation (1).

If this is so, then given the full set of Wightman functions one should be able to recreate the Hamiltonian H. Presumably, this Hamiltonian should be formulated in terms of observables of physical (rather than bare) particles. I would like to know whether such a Hamiltonian has been constructed in simple QFT models? I haven't seen explicit formulas in the references that you've provided earlier. Is it because

(a) the Hamiltonian formulation is theoretically unacceptable for some reason?

(b) this is difficult to do, but people are working on it?

(c) I've just missed this piece of information?

(d) people working on these models dislike Hamiltonians for some reason, or just don't care to derive them?

(e) other?

Thanks.
Eugene.

 

[A. Neumaier]

It seems that we both agree that we are doing quantum mechanics in a certain Hilbert space (Fock space or non-Fock space, whatever). One of quantum postulates is that the time evolution is described by the Hamiltonian H as

|\psi(t) \rangle = \exp(iHt) |\psi(0) \rangle......(1)

Yes. This is the meaning of the statement that H=P_0 is the generator of time translations. It is as true in QFT as in QM, and therefore also holds in the Wightman representation.

If I understand correctly, the axiomatic Wightman approach does not use this formula and even refuses to provide an explicit form of the Hamiltonian H.

This is incorrect. I gave you the explicit formula

H \psi(x_1,t_1,...x_N,t_N) = -i\sum_k d/dt_k \psi(x_1,t_1,...x_N,t_N)

If this is so, then given the full set of Wightman functions one should be able to recreate the Hamiltonian H. [...] I would like to know whether such a Hamiltonian has been constructed in simple QFT models? I haven't seen explicit formulas in the references that you've provided earlier.

(c) I've just missed this piece of information?

(c) is correct. I just repeated the formula for your convenience.
To understand its meaning, I invite you to do the exercises I recommended.

 

[meopemuk]

This is incorrect. I gave you the explicit formula

(c) is correct. I just repeated the formula for your convenience.
To understand its meaning, I invite you to do the exercises I recommended.

It looks like we are going in circles, aren't we? Let me make another attempt. Suppose we are doing a low-dimensional theory (phi^4 or something else, doesn't matter). Suppose also that I've prepared a well-separated couple of particles at time 0. I have a wave function for this state. I would like to know how this wave function evolves in time. I know how to do that with the Hamiltonian expressed in terms of a/c operators of particles.

However, your Hamiltonian does not have this form. So, which steps should I take in order to obtain the wave function at time t by using the Hamiltonian you've wrote? I am not asking for exact and complete algorithm. I just want to understand the idea.

Eugene.

 

[A. Neumaier]

It looks like we are going in circles, aren't we?

We are going in circles because you misunderstand something that I can clarify only through worked examples.

However, your Hamiltonian does not have this form. So, which steps should I take in order to obtain the wave function at time t by using the Hamiltonian you've wrote?

You should do the exercises so that I can explain. I can't explain the later steps without the former steps done, because I can see already from the way you respond that you wouldn't understand.

 

[meopemuk]

You should do the exercises so that I can explain. I can't explain the later steps without the former steps done, because I can see already from the way you respond that you wouldn't understand.

In a different thread https://www.physicsforums.com/showthread.php?t=474666&page=8 we've agreed that our views on quantum mechanics are completely different. We even disagreed on whether the square of a wave function can be interpreted as a probability density. So, if you want to teach me something you'll need to do it starting from the kindergarten level, indeed. I don't think this is worth your time and effort.

Thank you.
Eugene.

 

[A. Neumaier]

we've agreed that our views on quantum mechanics are completely different. We even disagreed on whether the square of a wave function can be interpreted as a probability density. So, if you want to teach me something you'll need to do it starting from the kindergarten level, indeed.

That's why I start with the anharmonic oscillator. We don't need to agree on the interpretation of the wave function, since everything will happen on the formal mathematical level. Thus the ''shut up and calculate'' attitude is good enough.
You can interpret it your way and I interpret it my way.

I don't think this is worth your time and effort.

I offered to do it, if you do the exercises.

 

[meopemuk]

We don't need to agree on the interpretation of the wave function, since everything will happen on the formal mathematical level. Thus the ''shut up and calculate'' attitude is good enough.
You can interpret it your way and I interpret it my way.

I don't think so. I don't know how to do quantum mechanics without interpreting |\psi(x)|^2 as the probability density for finding the electron at point x. This would be too formal for me. I think that the root of our disagreements about QFT goes back to much deeper disagreements about basic quantum mechanics. They should be resolved first, if we want to have a meaningful discussion.

Eugene.

 

[A. Neumaier]

I don't think so. I don't know how to do quantum mechanics without interpreting |\psi(x)|^2 as the probability density for finding the electron at point x. This would be too formal for me. I think that the root of our disagreements about QFT goes back to much deeper disagreements about basic quantum mechanics. They should be resolved first, if we want to have a meaningful discussion.

The power of QM stems from the fact that people with very different interpretations of the formalism can still agree on the formal part. Usually, interpretation issues cannot be resolved by agreement since they are rooted in mutually incompatible but deeply seated philosophies. At best, one can learn to think in multiple interpretations, and switch from one to the other if there is need for it. (We are already working on that in the photon thread.)

In any case, the anharmonic oscillator has a well-defined position representation; so you can think all the time of that while you do the exercise. Think of the Fock space as just being the position representation expressed in a basis of Hermite polynomials, and everything will have meaning for you.

The exercise will give you a representation of the standard anharmonic oscillator in the position representation in terms of Wightman states - at present very unfamiliar objects to you, but they will become familiar through the exercises. At the present stage, you don't need to think of what is done as a field theory - why it is one will emerge at the end of the whole sequence of exercises.

This done, we have the basis for understanding how the Wightman representation works in a very special case - one never considered in the literature but specially developed by me for you to pave a road to the real thing.

After all misunderstandings are cleared up on this level, we shall proceed to the standard charged multiparticle systems in R^3 with Coulomb interaction, and repeat the same exercise - but only in outline, since now actual computations would become somewhat messy. One of the reasons to do the anharmonic oscillator first is to keep the work down to a reasonable amount. We'll end up with a nonrelativistic version of the stated used by Weinberg in Chapter 10. Then we'll connect it to field equations and renormalization issues.

So please work out to least nontrivial order the following:
1. The ground state Omega;
2. The Heisenberg operators q(t);
3. The Wightman functions W(t_1,...,t_N) for N<=4 (at first, N<=2 is already sufficient), doing the free case for arbitrary N would also be useful;
4. The Wightman states |t_1,...,t_N> for N<=2 (at first N<=1 is sufficient);
Then we have material to play with, and to see how it relates to the Wightman Hamiltonian.

 

[meopemuk]

The power of QM stems from the fact that people with very different interpretations of the formalism can still agree on the formal part. Usually, interpretation issues cannot be resolved by agreement since they are rooted in mutually incompatible but deeply seated philosophies. At best, one can learn to think in multiple interpretations, and switch from one to the other if there is need for it. (We are already working on that in the photon thread.)

I disagree that probabilistic interpretation of the wave function has anything to do with philosophy. This interpretation is actually the essence of quantum mechanics. I can accept that people may have different philosophical views on the wave function collapse: Copenhagen, multi-world, pilot wave etc. Yes, these are philosophical issues, which can be debated. But the idea that |\psi(x)|^2 is the probability density of finding the particle at point x is not debatable. If you don't accept this idea, you are not doing quantum mechanics. Period. At least, you are not doing quantum mechanics that I am familiar with.

I don't think there is any point in discussing QFT before we resolved this foundational issue. Otherwise we will not be able to understand each other in anything related to state vectors, Hamiltonians, Fock space, etc.

We cannot even agree on the description of the simplest QM experiment - the double slit. I insist that when the electron is passing through the slits its state is described by the wave function \psi(x), and the probability (density) of this electron to land at the point x on the photographic plate is given be the square of this function.

You are saying that the electron can be described by a kind of classical field or by a quantum field, whish does not have the probability interpretation. Then the image on the photographic plate is the result of interaction of this field with the fields of atoms in the plate.

So, our understandings deviate starting from page 1 of quantum mechanics textbooks.

Eugene.

 

[A. Neumaier]

I disagree that probabilistic interpretation of the wave function has anything to do with philosophy. This interpretation is actually the essence of quantum mechanics.

The probability interpretation says that _if_ you can set up an experiment that measures a self-adjoint operator for a system in state psi then the probability of observing the k-th eigenvalue is psi^* P_k psi, where P_k is the projector to the k-th eigenspace. It says _nothing_ at all about which particular operators are observable in this sense.

Everything beyond that is interpretation, and hence (at the current state of affairs) a matter of philosophy. In particular, which operators can be measured is not part of the probability interpretation but a matter of theoretical and experimental developments.

Regarding what is arbitrarily _precisely_ measurable, there is a no go theorem by Wigner (I can give references if you want to check that) that states that _only_ quantities commuting with all additive conserved quantities are precisely measurable. The position operator is not among these.

But the idea that |\psi(x)|^2 is the probability density of finding the particle at point x is not debatable. If you don't accept this idea, you are not doing quantum mechanics. Period. At least, you are not doing quantum mechanics that I am familiar with.

Nobody comparing QM with experiments is making use of this particular assumption, and as the comparison of your formula with that of Mandel and Wolf show, it is false for a position measurement via detector clicks.

I don't think there is any point in discussing QFT before we resolved this foundational issue. Otherwise we will not be able to understand each other in anything related to state vectors, Hamiltonians, Fock space, etc.

Since what we discussed in this thread only concerns the time development of pure state, and no measurements at all, it can be resolved on the basis of mathematics alone. No interpretation in terms of real life is needed - hence it doesn't matter whether you apply QM with your interpretation or whether I apply QM with my interpretation. The mathematics is exactly the same.

We cannot even agree on the description of the simplest QM experiment.

In this thread, we are not discussing descriptions of experiments but the time evolution in quantum field theories. The latter is defined in a purely algebraic way, independent of any reference to experiment or double slits.

If you want to understand the meaning of time in QFT, you'd better listen to those who understand it, rather than claiming in your ignorance that there is none (as you do in your book). My exercises are designed to lead to such an understanding, and I'll take care that for the anharmonic oscillator, everything done is valid with the interpretation of |psi(x)|^2 as the probability density of detecting the oscillator at x. (But it is completely irrelevant for understanding the math and the meaning of the Wightman representation.)

 

[Avodyne]

So please work out to least nontrivial order the following:
1. The ground state Omega;
2. The Heisenberg operators q(t);
3. The Wightman functions W(t_1,...,t_N) for N<=4 (at first, N<=2 is already sufficient), doing the free case for arbitrary N would also be useful;
4. The Wightman states |t_1,...,t_N> for N<=2 (at first N<=1 is sufficient);
Then we have material to play with, and to see how it relates to the Wightman Hamiltonian.

What is your point? These are all standard problems, though (as I'm sure you know) #3 can be done with Feynman diagrams without needing to do #1, #2, or #4 explicitly. And, in 0+1 dimensions, there is no issue of which Hilbert space the operators live in, no Haag's theorem, etc.

 

[Avodyne]

Nobody comparing QM with experiments is making use of this particular assumption, and as the comparison of your formula with that of Mandel and Wolf show, it is false for a position measurement via detector clicks.

Mandel and Wolf are talking about photons; the wave-mechanics formalism only applies to nonrelativistic particles. For nonrelativistic particles, absolutely everyone comparing QM with experiments does make use of the wave-mechanics interpretation of |\psi(x)|^2 as a probability density.

 

[meopemuk]

OK, let us take an anharmonic oscillator - this is simpler than the hydrogen atom, which has an electron in a Coulomb potential: Scattering states are absent, and neither spin nor angular momentum need to be accounted for.

I'll guide you into doing the computations yourself.

Consider the anharmonic oscillator with Hamiltonian H=p^2/2m+V(q), where
V(q)=kq^2/2+aq^3/3+bq^4/4
and b>0, acting in the standard Fock representation with a single mode. The anharmonic oscillator is a 1+0 dimensional QFT in which space consists of a single point only (namely the oscillating mode), and everything happens in the simplest Fock space F, that you understand very well.

Let q(t) be the time-dependent position operator in the Heisenberg picture, and define the Wightman functions
W(t_1,...,t_N):=Omega^*q(t_1)...q(t_N)Omega,
where Omega is the ground state of the Hamiltonian, N=0,1,2,..., and the t_k are arbitrary times.

Please work out q(t), Omega, and W(t_1,...,t_N) in perturbation theory to the least nontrivial order, assuming that a, b are tiny. If you like, you may put m=1, k=1, and either a=0 or b=0 to simplify the formulas. (b=0 is simpler than a=0 but is a nonphysical limiting case.)

If this is sucessfully completed, I'll tell you how to use the Wightman functions.

OK, let's do that.

However, I am not going to do these calculations formally. At each step I would like to understand the physics of it, the nature of approximations etc. My first problem is with your choice of the model. In fact you are considering a multiparticle system confined at one space point. This can be imagined as a tiny box with impenetrable walls placed at x=0. One can add/remove particles to/from the box, thus changing the total energy of the system.

If the Hamiltonian was non-interacting H=p^2/2m+kq^2/2 then we would get the equidistant spectrum of the harmonic oscillator. In particle interpretation this means that particles, which are placed in the box, are non-interacting. By adding a new particle (this can be done by using the creation operator a*, which is a linear combination of your p and q) the energy of the system is raised by a fixed amount Mc^2 (note that the "particle mass" M here is different from the parameter m in the Hamiltonian), which is independent on how many particles already present in the box. Zero particles in the box corresponds to the ground state Omega of your Hamiltonian. The energy of this state is non-zero, but this can be fixed by shifting the energy scale. Operators p and q have no relationship to position and momentum, because nothing can move in the 0-dimensional box. So far so good.

Now you suggest to change the Hamiltonian by adding non-harmonic terms

H=p^2/2m+kq^2/2+aq^3/3+bq^4/4

This means that particles in the box interact with each other. The energy spectrum is not equidistant anymore. So, adding a new particle disturbs all other particles already present in the box. This interaction is very peculiar as it changes the energy of the zero-particle and one-particle states. The energy of a single particle in the box is not the same as the 1-particle state before. This means that single particle interacts with itself. This is kind of strange. It is even stranger that the energy of no-particle vacuum state has changed as well. So, there is some non-trivial interaction even if there are no particles at all and it is difficult to imagine what can interact with what in order to cause this energy shift?

In summary, I find your model very artificial and unrealistic. I don't understand what kind of physics is described there. Please clarify.

Eugene.

 

[meopemuk]

The probability interpretation says that _if_ you can set up an experiment that measures a self-adjoint operator for a system in state psi then the probability of observing the k-th eigenvalue is psi^* P_k psi, where P_k is the projector to the k-th eigenspace. It says _nothing_ at all about which particular operators are observable in this sense.

Yes, this is true. QM does not talk about the specifics of observations and measuring devices. For example, P_k can be a projection on the k-th eigenvalue of the position operator. Then psi^* P_k psi is the probability (density) for measuring position k in the state described by psi. QM tacitly assumes that some ideal precise measuring device can be constructed, which does exactly that - determinining whether the particle is located at point k or not. In QM we don't care how exactly this device is made. This can be a simple ruler or a photographic plate or a wire detector - doesn't matter. The only requirement is that this is a "device measuring particle position". All the worries about how to make such devices efficient, precise, etc are delegated to experimentalists.

In QM formalism this ideal position-measuring device is represented by a Hermitian operator of position.

Of course, you can take a further step and decide that in order to be more accurate you will make a quantum mechanical model of the measuring device. You may build an extended Hilbert space, which includes states of both the particle and the original device. You may describe states of this huge combined system as vectors in this Hilbert space or as wave functions with multiple arguments. But this is not enough, because the wave function alone does not give you probabilities of measurements. You still need some other projection operator Q_k, which is a part of some other observable Q, which is a quantum-mechanical model of some ideal measuring device appropriate to your newly specified experimental setup. For example, if the extended quantum system was built as "particle+photographic plate", then the new measuring device Q may be the experimenter's eye, which determines which grain of the photoemulsion was blackened. So, you'll need to represent the experimenter's eye by a Hermitian operator in this case.

But this complicated treatment will not give you any more insight than the simple QM model with 1-particle wave function and the ideal position-measuring device represented by the 1-particle position operator. The simple 1-particle model describes the distribution of blackened spots on the photographic plate just as well as the complicated "particle+device" model.

Eugene.

 

[A. Neumaier]

However, I am not going to do these calculations formally. At each step I would like to understand the physics of it, the nature of approximations etc. My first problem is with your choice of the model. In fact you are considering a multiparticle system confined at one space point. This can be imagined as a tiny box with impenetrable walls placed at x=0. One can add/remove particles to/from the box, thus changing the total energy of the system.

Such a system is approximately realized by a so-called quantum dot. http://en.wikipedia.org/wiki/Quantum_dot

On the other hand, you can as well consider it as a standard anharmonic oscillator in one space dimension, since the mathematics is exactly the same.

Thus you get two interpretations for the same mathematics - a common textbook example and a modern technology application.

If the Hamiltonian was non-interacting H=p^2/2m+kq^2/2 then we would get the equidistant spectrum of the harmonic oscillator. In particle interpretation this means that particles, which are placed in the box, are non-interacting. By adding a new particle (this can be done by using the creation operator a*, which is a linear combination of your p and q) the energy of the system is raised by a fixed amount Mc^2 (note that the "particle mass" M here is different from the parameter m in the Hamiltonian), which is independent on how many particles already present in the box. Zero particles in the box corresponds to the ground state Omega of your Hamiltonian. The energy of this state is non-zero, but this can be fixed by shifting the energy scale. Operators p and q have no relationship to position and momentum, because nothing can move in the 0-dimensional box. So far so good.

Yes.

Now you suggest to change the Hamiltonian by adding non-harmonic terms

H=p^2/2m+kq^2/2+aq^3/3+bq^4/4

This means that particles in the box interact with each other. The energy spectrum is not equidistant anymore. So, adding a new particle disturbs all other particles already present in the box. This interaction is very peculiar as it changes the energy of the zero-particle and one-particle states. The energy of a single particle in the box is not the same as the 1-particle state before. This means that single particle interacts with itself. This is kind of strange.

This is called mass renormalization - replacing the particle by an effective quasi-particle. This is very common in condensed matter physics. For example, electrons in a conduction band are quasiparticles, not free electrons. http://en.wikipedia.org/wiki/Effective_mass_(solid-state_physics ) . Since a quantum dot is a kind of miniature solid state, one has it there as well.

On the other hand, interpreted as particle in a quartic potential, this is nothing strange at all. Keep thinking about both interpretations!

It is even stranger that the energy of no-particle vacuum state has changed as well.

This is called vacuum renormalization (done by normal ordering). You also have it everywhere in condensed matter physics. This is harmless since only energy differences are physically interesting, and corresponds to measuring energies from the ground state.

 

[A. Neumaier]

What is your point? These are all standard problems, though (as I'm sure you know) #3 can be done with Feynman diagrams without needing to do #1, #2, or #4 explicitly. And, in 0+1 dimensions, there is no issue of which Hilbert space the operators live in, no Haag's theorem, etc.

Of course, this is standard. I give a series of exercises to build up intuition about Wightman functions. The absence of Haag's theorem makes the 1+0D case an ideal toy object, in which all difficult things are still absent. Once someone understands that, it is much easier to discuss the advanced stuff, such as inequivalent representations.

 

[A. Neumaier]

Mandel and Wolf are talking about photons; the wave-mechanics formalism only applies to nonrelativistic particles. For nonrelativistic particles, absolutely everyone comparing QM with experiments does make use of the wave-mechanics interpretation of |\psi(x)|^2 as a probability density.

Please show me a comparison with experiment that does this.

Nonrelativistic particles have no different interpretation than relativistic ones.

Particle detectors respond to the momentum of a particle, not to its position.
Scattering experiments are interpreted in the momentum picture. Nobody is interested in the position of particle tracks, only in their momentum (which tells about masses).

 

[Avodyne]

Particle detectors respond to the momentum of a particle, not to its position.

This is completely wrong. Experimenters have been recording particle tracks in position space with cloud chambers, bubble chambers, spark chambers, and drift chambers for many decades. The experiments are typically done in a strong magnetic field, which allows for measuring the momentum of charged particles by measuring the curvature of a track in position space. Modern experiments also have calorimeters at the boundaries of detectors that measure energy deposited; this does give a direct measurement of a particle's energy, but not its momentum.

For some recent pictures of particle tracks in position space from the LHC see

http://public.web.cern.ch/press/pressreleases/Releases2010/PR15.10E.html

 

[meopemuk]

Such a system is approximately realized by a so-called quantum dot. http://en.wikipedia.org/wiki/Quantum_dot

On the other hand, you can as well consider it as a standard anharmonic oscillator in one space dimension, since the mathematics is exactly the same.

OK, let us stick to the "quantum dot" model. We have a box where we can place our non-interacting particles. The energy of the system is trivially E=nmc^2, where n is the number of particles in the box and m is the particle's mass. The Hamiltonian can be written through particle a/c operators as

H = mc^2 a^{\dag}a.....(1)

Rather formally one can introduce "position" q and "momentum" p operators by

q = \frac{1}{\sqrt{2}} (a^{\dag} + a)
p = \frac{i}{\sqrt{2}} (a^{\dag} - a)

Then, again formally, the Hamiltonian (1) can be re-written in the "harmonic oscillator" form H=p^2/2m+kq^2/2 (I am not paying attention to correct coefficients here, just trying to formulate the ideas). The next logical step would be to allow the particles in the box to interact. Interaction takes at least two particles to participate. So, addition of interaction cannot have any effect on zero-particle and one-particle states and energies. Only 2-particle, 3-particle, etc. states and energies can be affected by interaction. From this condition it is clear that any reasonable interaction that can be added to the free Hamiltonian (1) must have the form (I drop numerical coefficients in front of operator symbols as I am interested only in the general operator structure of the terms)

V = a^{\dag} a^{\dag}aa + a^{\dag}a^{\dag}aaa + a^{\dag}a^{\dag}a^{\dag}aa + \ldots...(2)

The characteristic feature of this operator is that in the normally-ordered form it has at least two annihilation operators and at least two creation operators in each term. The ellipsis at the end indicates that more complex terms with these features can be added there.

Now, you are suggesting something completely different. Your interaction V= aq^3/3+bq^4/4 being expanded in a^{\dag}, a, does not have the form (2). This means that 0-particle and 1-particle states and energies are affected by your interaction. This means that interaction has changed the definition of particles. Your new 1-particle state (which can be defined as the state with the 2nd lowest total energy value) is a linear combination of eigenstates of the old H. So, by introducing interaction V= aq^3/3+bq^4/4 you have changed the physics of your quantum dot in a very dramatic way. Your new physical vacuum is different from the old (bare) vacuum. Your new physical particles are different from the old (bare) particles. All your theory is formulated in terms of bare (=meaningless) operators a^{\dag}, a which do not correspond to any physical thing anymore. You have created a lot of problems by introducing a completely unphysical interaction operator. These problems will lead you to the need of doing renormalizations and other headaches down the road.

I don't think that quantum dots with interacting particles behave in this way. So, all these complications introduced by you are completely artificial. You are thinking that interaction V= aq^3/3+bq^4/4 is simple and innocent. In fact, this interaction is very complex, unphysical and dangerous.

I would be more interested to consider your example with the simple and physical interaction (2). Then there will be no unrealistic vacuum polarization and particle self-energies.

Eugene.

 

[Avodyne]

In d+1 dimensions with d>0, your form (2) is ruled out by Lorentz invariance; to maintain Lorentz invariance, interactions must be written in terms of q. See Weinberg vol.I or chapter 4 of Srednicki.

 

[meopemuk]

In d+1 dimensions with d>0, your form (2) is ruled out by Lorentz invariance; to maintain Lorentz invariance, interactions must be written in terms of q. See Weinberg vol.I or chapter 4 of Srednicki.

No. There is no proof that with the Hamiltonian (2) you cannot build a relativistically invariant theory. It is true that existing relativistic quantum field theories do not use form (2), but neither Weinberg nor Srednicki nor somebody else has proven that there is no other way. There are actually counterexamples in which the Hamiltonian of the form (2) satisfies approximately the necessary commutation relations, and there are good reasons to believe that it is possible to satisfy these relations exactly (i.e., in all orders of perturbation theory). For mode discussions of this issue you can visit an Independent Research thread https://www.physicsforums.com/showthread.php?t=474666 [/URL].

Eugene.

 

[A. Neumaier]

In d+1 dimensions with d>0, your form (2) is ruled out by Lorentz invariance; to maintain Lorentz invariance, interactions must be written in terms of q. See Weinberg vol.I or chapter 4 of Srednicki.

We are discussing here a quantum dot, which is 1+0D QFT. The 1-dimensional Lorentz group is trivial, hence Lorentz invariance poses no restriction.

 

[A. Neumaier]

The next logical step would be to allow the particles in the box to interact. Interaction takes at least two particles to participate. So, addition of interaction cannot have any effect on zero-particle and one-particle states and energies. Only 2-particle, 3-particle, etc. states and energies can be affected by interaction. From this condition it is clear that any reasonable interaction that can be added to the free Hamiltonian (1) must have the form (I drop numerical coefficients in front of operator symbols as I am interested only in the general operator structure of the terms)

V = a^{\dag} a^{\dag}aa + a^{\dag}a^{\dag}aaa + a^{\dag}a^{\dag}a^{\dag}aa + \ldots...(2)

The characteristic feature of this operator is that in the normally-ordered form it has at least two annihilation operators and at least two creation operators in each term. The ellipsis at the end indicates that more complex terms with these features can be added there.

Now, you are suggesting something completely different. Your interaction V= aq^3/3+bq^4/4 being expanded in a^{\dag}, a, does not have the form (2). This means that 0-particle and 1-particle states and energies are affected by your interaction. This means that interaction has changed the definition of particles. Your new 1-particle state (which can be defined as the state with the 2nd lowest total energy value) is a linear combination of eigenstates of the old H. So, by introducing interaction V= aq^3/3+bq^4/4 you have changed the physics of your quantum dot in a very dramatic way. Your new physical vacuum is different from the old (bare) vacuum. Your new physical particles are different from the old (bare) particles. All your theory is formulated in terms of bare (=meaningless) operators a^{\dag}, a which do not correspond to any physical thing anymore. You have created a lot of problems by introducing a completely unphysical interaction operator. These problems will lead you to the need of doing renormalizations and other headaches down the road.

That's precisely one of the points of the exercises - you'll learn to understand the meaning of renormalization in a simple case where there are no divergences. Everything is harmless. The point is that in solid state physics (and hence in a quantum dot), interactions turn free particles into effective particles that are _different_ from the original ones. (For example, effective photons in glass are slower than free photons in vacuum.)

But in the present context you'd use the term ''free'' in place of bare, since the interaction can be switched on and offf (by changing the confining magnetic fields); so they have a physical interpretation (quite unlike the case in relativistic QFT, where switching off the interactions is impossible).

I am discussing the anharmonic oscillator in the Fock representation, or, equivalently, the quantum dot, a 1+0-dimensional QFT defined by a quartic Lagrangian. Both are well-defined quantum systems, and the fact that one has two different interpretations of the same abstract model is a big advantage, which I intend to exploit didactically.

There the interaction is given by a quartic polynomial, expanded into c/a operators. This produces interaction terms that are not only of the form you want, but are at most quartic. These interaction terms are physically important, as they generate
(i) in the anharmonic oscillators the physically measurable line shifts if an interaction is turned on;
(ii) in the quantum dot the renormalization of the vacuum and the particle mass. In 1+0D, these effects are finite, hence mathematically and physically respectable. And by doing the exercises you'll learn to interpret these things correctly.

 

[A. Neumaier]

This is completely wrong.

I replied in the thread https://www.physicsforums.com/showthread.php?p=3185296

 

[meopemuk]

There the interaction is given by a quartic polynomial, expanded into c/a operators. This produces interaction terms that are not only of the form you want, but are at most quartic. These interaction terms are physically important, as they generate
[...]
(ii) in the quantum dot the renormalization of the vacuum and the particle mass. In 1+0D, these effects are finite, hence mathematically and physically respectable.

I don't see anything "physically respectable" in the renormalization of the vacuum and the particle mass. No matter how strong is interaction between two or more particles, this cannot have any effect on a single particle (which has nothing to interact with). There can be no effect on the vacuum at all, because there are no particles and no interactions in the vacuum, by definition.

In your example you've introduced a Hamiltonian, which is mathematically simple (quartic), but physically unacceptable. This Hamiltonian violates the important rule that interaction can affect only states with two or more particles. So, you've created a huge problem for yourself, which you are then going to solve with "heroic" renormalization efforts.

Why do you find it necessary to define interactions so that the vacuum and 1-particle states are affected? Why don't you like my proposal with interaction a*a*aa, which would make everything much simpler?

Eugene.

 

[meopemuk]

The point is that in solid state physics (and hence in a quantum dot), interactions turn free particles into effective particles that are _different_ from the original ones. (For example, effective photons in glass are slower than free photons in vacuum.)

I don't find this reference to solid state physics convincing. In solid state physics you have a "medium", e.g., a crystal. In the quasiparticle picture you don't treat this medium explicitly, but model its presence implicitly by renormalizations. If I am interested in an electron moving through empty space, there is no "medium" of any kind. Likewise, there is no "medium", when I am studying particles placed in a small box. These two cases should not involve renormalizations.

Eugene.

 

[A. Neumaier]

I don't find this reference to solid state physics convincing. In solid state physics you have a "medium", e.g., a crystal. In the quasiparticle picture you don't treat this medium explicitly, but model its presence implicitly by renormalizations.

A quantum dot cannot exist except in a medium that constrains the system to a dot. Thus the medium is essential. In the approximation considered here, the medium is represented by the nonquadratic terms in the potential. This causes the renormalizations. The renormalizations are needed - otherwise we are no longer solving the original Lagrangian system.

If I am interested in an electron moving through empty space, there is no "medium" of any kind. Likewise, there is no "medium", when I am studying particles placed in a small box. These two cases should not involve renormalizations.

First things first. One cannot understand the quantum field theory of an electron on a deeper level without first having understood the quantum dot.

We can do the electron later, after having progressed from the quantum dot to nonrelativistic solid state physics and then to relativistic field theory.

 

[A. Neumaier]

I don't see anything "physically respectable" in the renormalization of the vacuum and the particle mass. No matter how strong is interaction between two or more particles, this cannot have any effect on a single particle (which has nothing to interact with). There can be no effect on the vacuum at all, because there are no particles and no interactions in the vacuum, by definition.

Our quantum dot is a solid state device with a confining potential, defined by a quartic Lagrangian (and hence the Hamiltonian I gave), which has an interaction that can be tuned by changing the factor g in front of a and b. The ''vacuum'' is the ground state of this interacting system. Thus , in a quantum dot, the vacuum is simply the unoccupied dot - far from the vacuum you imagine. The free vacuum is the empty dot in which the confining potential is harmonic (g=0). The dressed vacuum is the emptied dot in which a perturbation is switched on; of course this changes ground state and hence renormalizes the vacuum.

In your example you've introduced a Hamiltonian, which is mathematically simple (quartic), but physically unacceptable.

A real quantum dot can probably have a fairly arbitrary function of a and a^* as the Hamiltonian; certainly there is nothing that would forbid the terms that you don't want to have. They are essential for modeling the change in the ground state energy of the combined system (quantum dot + confining matrix) when the interaction is switched on.
If the quantum dot is formed by arbitrarily many bosons with only two energetically accessible states psi_0 (unexcited) and psi_1 (excited) - both functions of 3-position or 3-momentum -, the state space is the space spanned by the states |k> consisting of the symmetrized tensor product of k bosons in state psi_1 and all other bosons in state psi_0. Thus the N-particle state is the state of N excited bosons, and a ''particle'' is simply an excitation. The effective quantum dot Hamiltonianl is
H = sum H_jk (a^*)^ja^k,
where H_jk=<j|H_full|k> is computed by taking matrix elements of the full many-particle Hamiltonian in the Fock space over L^2(R^3). There is no reason why some of the H_jk should be zero. Thus a general quantum dot has no restrictions on the form of the potential.

Why do you find it necessary to define interactions so that the vacuum and 1-particle states are affected? Why don't you like my proposal with interaction a*a*aa, which would make everything much simpler?

The main purpose of the exercise is not to do computations for a realistic quantum dot, but to have illustration material for the Wightman representation. The latter is a representation for field theories defined by Lagrangians that are at most quadratic in the derivative of the fields.

In order to serve our purpose, our quantum dot must therefore correspond to a Lagrangian L(q)= m qdot^2/2 -V(q). (One could also add first order terms in qdot; but this makes things only more complicated.) Doing the usual Legendre transform results in H=p^2/2m+V(q).

If you want to have things as simple as possible, take V(q) = k q^2/2+g q^3/3. While this is unphysical, it is the limiting case c->0 of the quartic potential V(q) = k q^2/2+g (q^3/3+c q^4/4) (which has unbroken symmetry for 0<g<4kc), and has therefore a fully adequate perturbation theory (where g is infinitesimally small). Note that some wellknown textbooks on renormalization start for the same reason with the unphysical phi^3 theory!

 

[meopemuk]

A quantum dot cannot exist except in a medium that constrains the system to a dot. Thus the medium is essential. In the approximation considered here, the medium is represented by the nonquadratic terms in the potential. This causes the renormalizations. The renormalizations are needed - otherwise we are no longer solving the original Lagrangian system.

Please resolve my confusion. I am still not sure about the formulation of this problem. Assume that the "quantum dot" is prepared as a vacancy in semiconductor or something of that sort. We allowed to put one or more bosons into this vacancy. The vacuum in this case corresponds to the empty vacancy. So, the energy of this "vacuum" is the total energy of the crystal with empty vacancy. Now we start to put bosons into the vacancy one-by-one. Of course, the bosons interact with the crystal (through the confining potential of the vacancy), but they don't interact between themselves. So, with each new boson the total energy of the system increases by fixed energy E. This is expressed by writing the model Hamiltonian as

H = Ea*a = p^2/2m + kq^2/2...(1)

My question is: where the non-quadratic terms in the Hamiltonian come from? I thought that these terms were supposed to express the (originally neglected) mutual interaction between bosons. Then these additional terms cannot have any effect on the 0-boson (=empty vacancy) and 1-boson (vacancy+one boson) states. So, these terms *cannot* have the structure ~q^3 + q^4. I propose interaction ~a*a*aa as a reasonable alternative. If the non-quadratic terms are some corrections to the boson-crystal interactions, then why didn't we took them into account in our original formulation of the Hamiltonian (1)?

Eugene.

 

[A. Neumaier]

Assume that the "quantum dot" is prepared as a vacancy in semiconductor or something of that sort. We allowed to put one or more bosons into this vacancy. The vacuum in this case corresponds to the empty vacancy. So, the energy of this "vacuum" is the total energy of the crystal with empty vacancy. Now we start to put bosons into the vacancy one-by-one. Of course, the bosons interact with the crystal (through the confining potential of the vacancy), but they don't interact between themselves. So, with each new boson the total energy of the system increases by fixed energy E. This is expressed by writing the model Hamiltonian as

H = Ea*a = p^2/2m + kq^2/2...(1)

My question is: where the non-quadratic terms in the Hamiltonian come from? I thought that these terms were supposed to express the (originally neglected) mutual interaction between bosons.

No. As always in perturbation theory in QM, the Hamiltonian is assumed to be parameterized by a parameter g, denoting for example the strength of an external electric or magnetic field. Thus
<br /> H(g)= \sum_{j,k} H_{jk}(g) (a^*)^ja^k,~~~ H_{jk}(g)=&lt;j|H_{full}(g)|k&gt;.<br />
Since the e/m field interacts linearly, H_jk(g) is linear in g, hence H=H(0)+gV where V has (the in general arbitrary) matrix elements V_jk=dH_jk(g)/dg. On the other hand, H(0) - the Hamiltonian when the e/m field is switched of f - was used to construct the basis for the reduced description, hence has the form of a harmonic oscillator counting the number of bosons.

If we'd want to consider a real quantum dot, we'd have to remain general and treat all low order V_jk as parameters - we cannot simply put the V_jk with j<2 or k<2 to zero as you propose, since these terms describe how the empty dot and the dot filled with one particle respond to the e/m field.

But the purpose of the exercise is to learn about standard Lagrangian field theory in a very simplified setting. Therefore we keep things as simple as possible subject to the constraint that we have an associated Lagrangian. Thus, in order to be able to use the quantum dot as an example for illustrating the dynamics of Wightman fields, we choose the interaction artificially so that it corresponds to a system derived from a standard quartic (or if this seems too complicated for you, cubic) Lagrangian.

Then these additional terms cannot have any effect on the 0-boson (=empty vacancy) and 1-boson (vacancy+one boson) states. So, these terms *cannot* have the structure ~q^3 + q^4. I propose interaction ~a*a*aa as a reasonable alternative. If the non-quadratic terms are some corrections to the boson-crystal interactions, then why didn't we took them into account in our original formulation of the Hamiltonian (1)?

Instead of simply solving the exercise and learning from it, you turn it into a long discussion about the modeling of a real quantum dot. This is like refusing to solve an exercise in classical mechanics on the anharmonic oscillator because a real oscillator has an extension, friction, and all that stuff, while the purpose of the exercise is simply to get practice in a certain way of thinking.

I want to teach you intuition about Wightman fields, not about quantum dots!