# The role of wave function in QED

1. Dec 26, 2006

### Quantum River

I am just learning QED and could not understand the role of wave function. Is the basic equation in QED the Schrodinger Equation? Is the difference between Quantum mechanics and QED just they have different Hamiltonians.

I have tried to read the original paper of Tomonaga in 1946 Progress of Theoretical Physics, but just not could get the original paper. I have no access to Progress of Theoretical Physics. And this is the only paper that concerns the foundations of QED in my knowledge. Dyson's review The Radiation of Tomonaga, Schwinger, and Feynman gives a rather short description of Outline of the Theoretical Foundations(Dyson's words). Could anyone recommend me some papers in this direction.

What I am confused about is the role of wave function in QED. In QM, the wave function means the distribution probability in the space (Born)? Then what concept or quantity in QFT means such things (distribution probability in the space) or corresponds to the wave function ? Is there some correspondence principle between QM and QED? Could the basic QED equation retreat to QM Schrodinger equation or Dirac equation under some conditions?

For example, in the calculation of Lamb thift, what is the use of wave function (1s state of hydrogen)? It seems the Lamb shift has some connections with the wave function value at the r=0 point. Look at Baranger, Bethe, and Feynman's calculation. (Phys. Rev., Vol. 92, NO. 2,482) They use the wave function phi(r=0) to calculate the Lamb shift. But there is no wave function (of course no 1s wave function) in Quantum field theory? When calculating scattering, it is easier to understand the role of QFT, but when considering the Bounded states, I just could not understand how to calculate the bounded state wave function from QFT.

Quantum River

Last edited: Dec 26, 2006
2. Dec 26, 2006

### Demystifier

To understand a difference between two different quantum theories one should first understand the difference between the corresponding classical theories. While QM is a theory of particles, QED is a theory of fields. If you understand the difference between particles and fields in classical mechanics, the difference in the quantum case is essentially the same.

However, in reality, things are not that simple. QED is not only a theory of fields, it is also a theory of particles (photons and electrons). It does not have an analog in the classical case. The correspondence between particles and fields is essentially the correspondence between wave functions and field operators. Wave functions are certain matrix elements of the field operators. For textbooks where this is explained to some extent see the textbooks of Ryder or Schweber. You may also see
http://xxx.lanl.gov/abs/quant-ph/0609163

3. Dec 26, 2006

### Quantum River

But in Quantum field theory, particle and field are one thing. Electron is a particle in classical physics and Quantum mechanics (Schrodinger Equation), but it is a field in Quantum field theory. On the first thought, QM corresponds to classical physics (such as Newtonian physics), while Quantum field theory corresponds to classical field (such as Maxwellian electromagnetic field). But in Quantum field theory we not only quantize electromagnetic field, but also electron field (Does it exist?). I think the first thought is somewhat naive.

Wave functions are certain matrix elements of the field operators.
Could you explain the sentence above more concretely?

4. Dec 27, 2006

### marlon

That's hardly a decent answer to the OP's question. There is far more going on and you are missing out on the most essential part : how particles are connected to quantum fields in QFT.

TO THE OP :

In QFT (eg QED), particles arise due to vibrations of the quantum fields.

Think of a mattress (ie the quantum field) on which "YOU" jump in one place. The surface of the mattress vibrates and due to this vibration, there is energy coming free from the mattress (ie the vibrational energy).
If you take into account that energy is the same as mass (E=mc^2) you can see how the vibration of a quantumfield can mimic a certain particle with mass m and momentum p.

All particles (photons, electrons, etc) are born this way. For example, suppose we have two electrons (which are vibrations of a quantumfield themselves) "sitting" on the mattress. The two electrons cause the mattress to vibrate and the resulting vibrations give us the energy for a particle (a photon for example), as explained before. This is how two electrons interact with each other via a photon. Keep in mind that i am giving a simplified picture here with some loss of accuracy (like the necessary conditions for the quantum fields and conservation laws).

But essentially, in QFT, we have vibrating quantum fields that give us the particles we need. Such quantum fields are written in terms of creation and annihilation operators that acto onto a quantum state.

Finally, let's compare to QM. The main difference between QFT and QM are the basic ingredients :

1) In QM the basic ingredient are wavefunctions.
2) In QFT the basic ingredient are quantum fields.

In QFT, the "old" QM wavefunction is now the quantum field written in terms of creation and annihilation operators. In other words, the QM wavefunction has become a field operator in QFT.

regards
marlon

Last edited: Dec 27, 2006
5. Dec 27, 2006

### hellfire

You can indeed define a "wavefunction" in QFT proceeding in a similar way than for (non relativistic) QM. In QM the variables that define the configuration space are the positions. The wavefunction is a function of the positions in configuration space. In QFT the variables that define the configuration space are the fields. You can define a functional of the fields in configuration space. This "wavefunctional" would obey a Schrödinger-type equation and would be analogous to the wavefunction in QM. Take a look to section 5.2 of this lecture notes.

What you cannot do in QFT, as far as I know, is to define a wavefunction for a particle as in QM. The expression of a localized single particle $\phi(x) \vert 0 \rangle$ "looks like" the expansion in terms of momenta of the eigenstate of postion, however, the product of two of them $\langle 0 \vert \phi(x) \phi(y) \vert 0 \rangle$ is not equal to zero (it is equal to the propagator), which would be required in order to define a position basis and a wavefuction as such.

Last edited: Dec 27, 2006
6. Dec 27, 2006

### vanesch

Staff Emeritus
This is entirely correct. Only, the Schroedinger equation in QFT is quite a complicated beast, as it is an equation for a functional (and not a partial differential equation of a wavefunction over a finite-dimensional configuration space, as in QM). So nobody actually knows how to deal with it directly. Hence, the Schroedinger equation doesn't get much attention in QFT, as we don't know how to use it. In fact, the only thing we can do with it, is to write down the Schroedinger equation in integrated form, and even in an approximated way, and then even only for the case of t= - infinity to t = + infinity. The result of this is the S-matrix, which is nothing else but the solution to the Schroedinger equation for initial condition t = - infinity taken at the point t = + infinity: its elements are the individual integrals of the Schroedinger equation for specific initial conditions ("incoming particles").

The only relationship we have is that of the Born rule for the final solution of the Schroedinger equation (t = + infinity) when we have initial conditions at t = - infinity. It are hence the squares of the elements of the S-matrix, and when properly put into context, it gives you the cross sections for certain reactions.

You can use the superposition principle as well in QFT as in QM. If you have the solutions (the matrix elements of the S-matrix) for plane waves, you know that the solution to an initial state which is a superposition of plane waves will be the superposition of the corresponding final states.

What is usually done (as far as I know, I'm no expert) is to start with a specific bound state (usually obtained in NR QM), then use the superposition principle to "transfer" this to a superposition of final states, and eventually "project" this on other bound states (using the Born rule again, in a basis of bound states). This gives you then the transition probability from the initial bound state to the final bound state.

7. Dec 27, 2006

### Demystifier

http://xxx.lanl.gov/abs/quant-ph/0609163
Also, read Secs. VIII and IX in it, as well as the post of vanesch above.

8. Dec 28, 2006

### reilly

I'ts important to recognize that any quantum field theory in use is equivalent to a standard quantum theory -- that is, anything that is characterized in terms of particle creation and destruction operators, can be reformulated in the ordinary QM use of Fock space. This is very apparent in non-relativistic many-body theory, and is discussed in any number of publications. Note also that particle creation and destruction can occur, via transitions between subspaces with different numbers of particles.

With the imposition of relativity, life gets more difficult. While we regularly use a momentum operator in QFT, we do not use a position operator. Rather, we use the space-time parametrization (x,y,z,t) as if the spatial coordinates are equivalent to position eigenstates. But we do use wave functions in momentum space, and consider their Fourier transforms as spatial wavefunctions. Yes, we use wavefunctions and state vectors in relativistic QFT -- that's how we compute scattering crossections for example.

Regards
Reilly Atkinson

9. Dec 28, 2006

### CarlB

I agree completely and think that this is very important. I have a couple QFT books that hint at this, but I have no textbooks or other references that spell it out. Do you know of any other sources?

10. Dec 29, 2006

### Truth Finder

11. Dec 30, 2006

### vanesch

Staff Emeritus
12. Jan 1, 2007

### Truth Finder

Wrong!!!!!!
These equations have been absolutely believed of since feynmann......... Thanks, to revise your knowledge about QFTs.

Thanks ........... and Brgrds.
Nuclear Scientist!!!!!!!

13. Jan 3, 2007

### dextercioby

The eqns need to be checked, as they are written in noncovariant language. But the assertion "Quantum electrodynamics is a generalization of quantum mechanics to include special relativity" is definitely wrong.

Daniel.

14. Jan 3, 2007

### Truth Finder

Dexter,
You are right. Dirac included SR in QM. The other 2 equations included Field Quantization (May be:{DUE TO QUNIZATION OF SOURCE CHARGES}).
You can check them as much you can. But a question arises: What is the original? Is this form or the other or another one "if there is no equivalence"?
By the way, I was just asking about the meaning of its implication for Dirac's Equation for a nonconservative field. But I am sure that they are right and already believed of. I already studied this long before and discussed with many professors. And I think that it is time now for me to understand its meaning and how QED founders made it like this.

Schwartz Vandslire
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Either to work correctly as required, or to leave it.

15. Jan 3, 2007

### Truth Finder

Of Course Dexter, Lagrangian Form and Hamiltonian Form are different faces to the same Theory; QED.

Schwartz Vandslire
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Either to work correctly as required, or to leave it.

16. Jan 3, 2007

### Amr Morsi

T. Finder,

That's right, these equations are absolutely right.
But the problem is that many some scietists now are talking about the origin of field quantization to be due to the probability of photon itself. What do you think?

No doubt, the wave function in QFTs are that of the charges (and particles in general). But, isn't strange to have Dirac's Equation put explicitly like this (in its general forum)? Got me!

I think, as you said below, this is a very good starting point to reach the whole picture, sir.

But, isn't it some what strange that QM and QFTs in general don't speak about individual events, but about a number of same-system events?
Got you! Yaaiih!!!!

Engineer\ Amr Morsi.

17. Jan 4, 2007

### Truth Finder

OOPS!!

The issue of quantum field theories is a puzzle for me till now. But, it is quite apparent that there is some differences in scientists' complete consideration for it. However many of them is talking about the theory as a total, but still away from the details. Why not and it was not till Born who explained the right meaning of the wave function...... I think you pointed to that in one of your previous answers.

The last fact is absolutely right. It cannot describe only one trial. But, this doesn't have to do with the uncertainty principle? Does it! Scientists are investigating this problem in the mean time. But, we cannot deny that Quantum Mechanics still get more exact solutions? But still can't be applied to one experiment!

Schwartz Vandslire

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Either to work correctly as required, or to leave it.

18. Jan 5, 2007

### Anonym

CarlB:” I agree completely and think that this is very important. I have a couple QFT books that hint at this, but I have no textbooks or other references that spell it out. Do you know of any other sources?”

I think that one may be helpful too:
Herman Hesse's book "The Glass Bead Game".

19. Jan 5, 2007

### reilly

Once again, I appeal to history and note that this issue, that QED or QFT can be written in terms of ordinary (relativistic) QM was pretty much recognized some 80 years ago. First, recall that Heisenberg's matrix solution to the harmonic oscillator problem, used so-called step operators, now generally called creation and destruction operators. The rest is history.

I presume that Dirac can be cited as a reasonable authority in this matter. His groundbreaking paper on QED (1927) basically uses wavefunctions. Many papers on QED and QFT proior to WWII used wavefunction, albeit in creative ways. See, for example, Wiesskopf's paper on the electron's self energy of 1939. (Many of the important papers in QED, from inception to renormalization, can be found in Schwinger's Dover Books collection, Quantum Electrodynamics. (I'm no expert, but I suspect wave functions may be used in Stat Mech problems in which the chemical potntial is required.)

When I taught this stuff, I used a class to discuss this very issue, just to show that there are many ways to frame a problem, but that some are more equal than others.

Regards.
Reilly Atkinson

20. Jan 5, 2007

### Haelfix

There is far more to field theory than wavefunctions, so im of the point of view that the Dirac equation and say the relativistic generalization of the Schrodinger equation are *not* correct. For instance they miss radiative corrections like the lamb shift, that only field theory will see.

Of course they are retrieved in suitable limits or regimes where such things are not important, and for instance in looking at leading order terms in the dynamics.

Indeed there are many examples where we have absolutely no idea what the wavefunction even *is*, yet we can still calculate most quantities of interest. Take QCD as an example.