Quantum Field Theory - variational principle

[JustinLevy]

Quantum Field Theory -- variational principle

In non-relativistic quantum mechanics, the ground state energy (and wavefunction) can be found via the variational principle, where you take a function of the n particle positions and try to minimize the expectation value of that function with the hamiltonian.

In relativistic quantum mechanics (but fixed particle number, like the dirac equation), the same can be said.

Is there something equivalent in quantum field theory?
What exactly would I be varying as there doesn't really seem to be a wavefunction anymore? Could one write the state as a sum of a wavefunction in front of each possible number of particles?

Something like:
$$\Psi = \left( \phi(\mathbf{r}_1)a^{\dagger}(\mathbf{r}_1) + \phi(\mathbf{r}_1,\mathbf{r}_2)a^{\dagger}(\mathbf{r}_1)a^{\dagger}(\mathbf{r}_2) + \phi(\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3)a^{\dagger}(\mathbf{r}_1)a^{\dagger}(\mathbf{r}_2)a^{\dagger}(\mathbf{r}_3) + ... \right) |0\rangle$$

Is there an exact solution to the hydrogen atom in quantum field theory? Or do they use the relativistic quantum mechanics solutions are a starting point and do perturbations around that?

 

[Fredrik]

I don't know the answer to your question, so I'll just mention that I don't consider relativistic wave equations like Klein-Gordon and Dirac to be "relativistic quantum mechanics". I think of "QM" as all that stuff about Hilbert spaces, not including the Schrödinger equation, and the way to define a theory of a single non-interacting particle in that framework, is to impose the requirement that the Hilbert space is the vector space associated with an irreducible representation of the symmetry group of spacetime. If you take that group to be the Galilei group, you have non-relativistic QM. If you use the Poincaré group, you have special relativistic QM. (Both include a Schrödinger equation, because the time translation subgroup is represented by operators that satisfy U(t)U(t')=U(t+t'), which implies U(t)=exp(-iHt) and therefore idU/dt=HU).

I think of the field equations and the procedure of "canonical quantization" as a way to explicitly construct those representations.

 

[Physics Monkey]

There is definitely something like it in quantum field theory. One of the most famous variational states of all time is the BCS state for superconductivity. It has the form (conventionally written in momentum space) $$| \text{BCS} \rangle \sim \prod_k (u_k + v_k \, c^+_{k \,\text{up}} c^+_{-k \,\text{down}} ) | \text{vac} \rangle$$ where c^+ is an electron creation operator. If you expand out the product you will see that this state is exactly a sum of terms of the form you wrote. This is a variational quantum state of the quantum field theory describing the non-relativistic electrons in the superconductor.

 

[meopemuk]

Could one write the state as a sum of a wavefunction in front of each possible number of particles?

Something like:
$$\Psi = \left( \phi(\mathbf{r}_1)a^{\dagger}(\mathbf{r}_1) + \phi(\mathbf{r}_1,\mathbf{r}_2)a^{\dagger}(\mathbf{r}_1)a^{\dagger}(\mathbf{r}_2) + \phi(\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3)a^{\dagger}(\mathbf{r}_1)a^{\dagger}(\mathbf{r}_2)a^{\dagger}(\mathbf{r}_3) + ... \right) |0\rangle$$

I would say that this is a pretty good representation of the most general state in QFT.

Is there an exact solution to the hydrogen atom in quantum field theory?

In a good approximation the state vector of the hydrogen atom belongs to the 2-particle sector "1 proton + 1 electron". The full Hamiltonian of QFT does not conserve the number of particles, so that a more exact state vector has admixtures from higher sectors "1 proton + 1 electron + N photons + M electron-positron-pairs".

However, in order to see hydrogen as an eigenstate of the QED Hamiltonian H you'll need to use the "dressed particle" version of H. The presence of the bound state electron+proton is not obvious when looking at the "traditional" QED Hamiltonian.

Eugene.

 

[TriTertButoxy]

Is there something equivalent in quantum field theory?
What exactly would I be varying as there doesn't really seem to be a wavefunction anymore? Could one write the state as a sum of a wavefunction in front of each possible number of particles?

There is most definitely a wavefunction in relativistic quantum field theory. It's just that everyone has been hiding it from you. See Samalkhaiat's lengthy https://www.physicsforums.com/showthread.php?t=388556&page=2" for details.

There was a program in the '80s when people tried to apply variational principles to quantum field theory; in fact, if I remember correctly Richard Feynman was quite involved. One can write an ansatz for the wavefunction of the ground state (vacuum state) of an interacting quantum field theory. In fact, this is what the gaussian effective potential is based on.

 

[meopemuk]

One can write an ansatz for the wavefunction of the ground state (vacuum state) of an interacting quantum field theory.

What the xxxx is "the wavefunction of the vacuum state"? Vacuum, by definition, is a state with no particles. Any experiment performed on vacuum yields null result. So, vacuum wavefunction must be just plain zero.

Eugene.

 

[Haelfix]

People have in mind QCD there, where the vacuum is most assuredly not empty.

 

[meopemuk]

People have in mind QCD there, where the vacuum is most assuredly not empty.

What experiment can be performed in the "QCD vacuum" to demonstrate that it is "not empty"?

Eugene.

 

[Haelfix]

You know that what you ask is impossible of course.

However I can tell you what would instantly falsify the proposition-- The discovery of an isolated quark in nature.

Anyway without delving into semantics at this point, there is no such thing as a Fock vacuum for physical QCD (it is unstable), whatever 'it' is, is something else and essentially the definition of color confinement.

 

[meopemuk]

You know that what you ask is impossible of course.

I just wanted to draw your attention to this curious situation: In QCD we first assume existence of particles (quarks, gluons) which have not been observed. Then, in order to justify their non-observability, we introduce some strange vacuum, whose properties cannot be observed either. Then we postulate the existence of some "confinement mechanism", which, as far as I know, is still hypothetical.

This piling up of unprovable assumptions looks rather suspicious to me. I can believe that QCD makes accurate predictions for certain observations. But still ...

Eugene.

 

[TriTertButoxy]

I just wanted to draw your attention to this curious situation: In QCD we first assume existence of particles (quarks, gluons) which have not been observed. Then, in order to justify their non-observability, we introduce some strange vacuum, whose properties cannot be observed either. Then we postulate the existence of some "confinement mechanism", which, as far as I know, is still hypothetical.

This piling up of unprovable assumptions looks rather suspicious to me. I can believe that QCD makes accurate predictions for certain observations. But still ...

Eugene.

There is a widespread misconception that for every quantum field in the theory, there must correspond a particle. I believe this misconception has been the source of many conceptual problems for students of field theory. One should view the quarks and gluons not as particles, but only as fields. And the quantum dynamics are so severe that the states hardly resemble the elementary excitations of these fields. Because phi-4 theory and QED exhibits a milder form of this, quantum excitations seem to resemble the field content of the theory.

 

[Physics Monkey]

There is a widespread misconception that for every quantum field in the theory, there must correspond a particle. I believe this misconception has been the source of many conceptual problems for students of field theory. One should view the quarks and gluons not as particles, but only as fields. And the quantum dynamics are so severe that the states hardly resemble the elementary excitations of these fields. Because phi-4 theory and QED exhibits a milder form of this, quantum excitations seem to resemble the field content of the theory.

Indeed, I like this sentiment very much. In condensed matter physics we suffered from a similar kind of confusion for many years. The high energy (a few eV) description of basically any terrestrial condensed matter system is in terms of electrons and Coulomb interaction (plus nuclei). In my opinion, the field seemed to think for a long time that this somehow meant that the low energy physics always contained objects (quasiparticles) that basically resemble electrons. It was a great moment when we understood clearly that the low energy physics was potentially much richer including things like "deconfinement" where the low energy degrees of freedom resemble fractions or pieces of the electrons.

QCD is quite similar. The high energy description is in terms of nearly free quarks and gluons, but the low energy description of dilute qcd matter is in terms of totally different objects.

 

[samalkhaiat]

meopemuk;2649840]What the xxxx is "the wavefunction of the vacuum state"?

It is the lowest (ground state) energy solution of Schrodinger equation, often written as

$$\Psi_{0}(x) = \langle x | 0 \rangle \equiv \langle x |\Psi_{0} \rangle$$

For harmonic oscillator, it is given by the following everywhere-positive function;

$$\langle x|0 \rangle = ( \frac{m\omega}{\pi \hbar})^{1/4} \exp (-m\omega x^{2}/2 \hbar)$$

Vacuum, by definition, is a state with no particles.

Indeed, it is a state with no real quanta. However, this does not mean no energy. Again the vacuum oscillator energy is $1/2 \omega \hbar$.

Any experiment performed on vacuum yields null result.

This is plain wrong! Even in the vacuum, a cupfull of Helium (which is a quantum object) does not solidify (at sufficiently low temperatures at atmospheric pressure). Sir. it is already an industry.

So, vacuum wavefunction must be just plain zero.

I find your "logic" rather strange! I have just given you an example where it is NON-ZERO EVERYWHERE.


regards

sam

 

[meopemuk]

Indeed, it is a state with no real quanta. However, this does not mean no energy. Again the vacuum oscillator energy is $1/2 \omega \hbar$.

I don't think that there is a good analogy between vacuum and harmonic oscillator. (Yes, I've read QFT textbooks where such an analogy is proposed, but I am not convinced). So, if the ground state of oscillator has a non-zero energy, this does not mean that vacuum energy must be non-zero (actually, infinite!) too.

IMHO (measurable) energy can be associated only with particles. Vacuum is a state without particles. So, vacuum energy is exactly zero.

Even in the vacuum, a cupfull of Helium (which is a quantum object) does not solidify (at sufficiently low temperatures at atmospheric pressure). Sir. it is already an industry.

Sorry, I missed your point. By "vacuum" I mean empty space. If there is a "cupfull of Helium" then it is not vacuum.

There are no physical objects in the real vacuum. So, any measurement performed there should yield null result. That's my point.

Eugene.