# Reduction of QED to QM

1. Dec 31, 2008

### tim_lou

I've been spending some time to think about a really conceptually and mathematically clean way of showing how QED reduces to ordinary QM in low energy limit.

More specifically, I would like to consider a restatement of multi-particle QM (of electrons) based on creation and annihilation operators:

$$H_{\textrm{QM}}=\int d^3p \, E_p \,a^\dagger(p)a(p) + \iint d^3p' d^3p \, a^\dagger(p')\~ V(p'-p) a(p)$$

(neglecting sum of spins and anti-electrons, a couple 2π's might be missing)

Where $E_p=p^2/2m + m$ and $\tilde{V}(p)$ is the fourier transform of the Coulomb potential (or Yukawa potential with m->0). the creation and annihilation operators are only for electrons.

I believe that one can make certain assumptions about low energy QED and simplify that Hamiltonian to the form given above (hence explicitly demonstrating reduction of QED to QM).

I am however unsure about how to neglect antielectrons and treat the photon field. I think using the interaction picture and considering a certain subset of all Feynman diagrams would do the trick. However, I would like to find some reference on this matter. Any idea or comment is welcome.

Last edited: Dec 31, 2008
2. Dec 31, 2008

### meopemuk

The Hamiltonian you wrote is for electrons in the external potential $$V$$, which is not an isolated system. For comparison with QED it would be more appropriate to write a Hamiltonian for electrons interacting with each other via potential $$V$$

$$H_{\textrm{QM}}=\int d^3p \, E_p \,a^\dagger(p)a(p) + \int d^3p' d^3p d^3k\, a^\dagger(p-k) a^\dagger(p'+k)\~ V(p,p',k) a(p)a(p')$$

I would suggest to look at "dressed particle" (or "clothed particle") approach

O. W. Greenberg, S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim., 8 (1958), 378.

3. Dec 31, 2008

### olgranpappy

I think that it is usually a pain to start from QED and take a limit to find a non-relativistic theory of electrons interacting via the instantaneous coulomb force. The pain is basically because you have to get rid of the photons. But, if you start off by writing the full Hamiltonian of QED in the Coulomb gauge as
$$H_{\rm QED}=\int d^3r \psi^\dagger(r)\left(-i\bold{\nabla}+e\bold{A}/c\right)^2\psi(r)/2m +\frac{1}{2}\int\int \psi^\dagger(r)\psi^\dagger(r')\frac{e^2}{|r-r'|}\psi(r')\psi(r) +H_{0,\gamma}\;,$$
where $H_{o,\gamma}$ is the free hamiltonian of the electromagnetic fields and
$$\bold{A}=\sum_{k,\alpha}\sqrt{\frac{f_k}{V}}\left(a_{k\alpha}\epsilon_{k,\alpha}+a_{-k,\alpha}^\dagger\epsilon_{-k,\alpha}\right)e^{i\bold{k}\cdot\bold{x}}$$
where, the a's and a-daggers create photons and the epsilons are polaization vectors, and the V is a normalization volume and the f_k are some other numerical factors that I don't remember off the top of my head.

Then, just take $c\to \infty$ to obtain the non-rel expression; the eA/c term goes away and then we can ignore the H_{0,gamma} term since it is now just a "constant".

For more details you should check out the excellent old book by Heitler called "Quantum Theory of Radiation" (or something like that...it's from the 1950's). Good luck.

4. Dec 31, 2008

Staff Emeritus
But does it? How do you get the Lamb shift this way? There's a low-energy phenomenon that comes out of QED. g-2 is another example.

5. Dec 31, 2008

### meopemuk

In the "dressed particle" reformulation of QED the Hamiltonian is written as a perturbation expansion. In the second perturbation order the electron-electron interaction potential includes the usual Coulomb potential plus relativistic corrections (e.g, spin-orbit, spin-spin, etc). In the fourth perturbation order there are smaller terms, which are responsible for g-2 and for the Lamb shift among other things. All these terms can be obtained from the original QED Hamiltonian by the rigorous Greenberg-Schweber "dressing transformation".

This theory is just as simple as ordinary QM. The only significant difference is that there are also interaction terms that create/annihilate particles, e.g., describe the emission of photons.

6. Dec 31, 2008

### atyy

7. Jan 1, 2009

Staff Emeritus
I'm still confused. I have managed to convince myself that the low energy limit of QED is not NRQM + Maxwell. Like I mentioned before, that doesn't give you the Lamb shift - which is why it was so important and why Willis Lamb got the Nobel prize in physics.

I'm sure that one could cobble together an effective theory that looks like NRQM + Maxwell + some ad hoc pieces that represent one-loop QED corrections. But I am not so sure that such a theory is particularly valuable now that we have QED. It's certainly less predictive.

8. Jan 1, 2009

### atyy

NRQM+Maxwell+ad hoc bits works well sometimes. So QED should explain why our wrong theory works well, and when it should fail. I guess like the relationship between GR and Newtonian gravity.

9. Jan 1, 2009

### tim_lou

Thanks for the replies. olgranpappy's formula seem to be the in the right direction and I think a trip to the library is in order. I'll have to delve into this subject for a couple days and work out all the details.

10. Jan 1, 2009

### meopemuk

QED can calculate only S-matrix elements, which are static (time-independent) scattering properties. In order to get the time evolution of states and/or observables, you'll need a QM-like theory. It is true that so far nobody could observe time dependence in high-energy collisions. However, eventually, this experimental information will be available, and the traditional QFT approach will be found inadequate.

11. Jan 1, 2009

Staff Emeritus
Interesting point, one that I haven't considerd.

I'm not so sure I agree with that. As energies go up, the interacting particles become more and more time dilated, which means there is less and less time to observe any time dependence.

12. Jan 1, 2009

### tim_lou

Speaking of applications. I've seen it somewhere that says summing certain Feynman diagrams (in some low energy bound states resonances) is equivalent to solving the corresponding Schrodinger equations.

As for my motivation for considering this problem, I just feel that working through (the beginning portion of) Peskin's text merely enables me to do calculation without actually understanding why QFT works. I feel that explicitly showing the correspondence to ordinary QM will allow me to truely appreciate what QFT brings to the table and why certain formulas make sense.

13. Jan 1, 2009

### olgranpappy

For one thing, we should remember that QFT is not really anything "beyond" quantum mechanics; it is a convenient repackaging of "first-quantized" quantum mechanics (which involves fixed particle number) in terms of field operators which lets us more easily discuss particle creation and annihilation. But, all of quantum mechanics still applies.

For example, I wrote down a QED Hamiltonian (H) in terms of field operators. But, this is to be used in the same way as any other Hamiltonian. The time dependence of states is given by
$$\frac{\partial}{\partial t}|{\Psi}\rangle = -i\hat H|\Psi\rangle\;,$$
where H is the QED Hamiltonian and the state Psi characterizes all the electron and photons in the system.

14. Jan 1, 2009

### olgranpappy

Oh... and, to recover ordinary first-quantized expressions one simply checks that when the operators written in terms of field operators act on state with fixed numbers of particles, the same results are given as the usual first-quantized expressions.

E.g., when acting on a N-particle state the operator
$$\int d^3x \hat \psi^\dagger(\bold{x}) V(\bold{x})\hat \psi(\bold{x})\;,$$
has the same result as
$$\sum_{i=1}^N V(\hat r_i})\;.$$
I think a good reference for this stuff is Lowell Brown's QFT book.

15. Jan 1, 2009

### meopemuk

Suppose that you want to find the time evolution of a state that initially has just one electron. In QM you would use the Hamiltonian $$H = P^2/(2m)$$, or its relativistic generalization. Then the solution of the time-dependent Schroedinger equation is simple and makes perfect physical sense.

However, if you use the Hamiltonian of QED instead, you'll find that the state $$|\Psi(t) \rangle$$ at $$t > 0$$ is a terrible mess containing virtual photons, pairs, etc. It doesn't look like a single electron moving through space at all.

The dressed particle approach provides a rigorous way to rewrite the QED Hamiltonian in such a form that the time evolutions of the vacuum and one-particle states are physically acceptable, and, at the same time, the scattering amplitudes in states with 2 or more particles remain the same as in the standard QED.