# Potential energy in hydrogen atom

1. ### qsa

362
Can somebody elabroate on how the P.E. term e^2/r is thrown in shrodinger's(also Dirac) eqution for the hydrogen atom by assuming such term does describe the effects of the negative and the positive charge. It enters into the equation like mass, yet quantum theory( a good one) should calculate such energies, wouldn't you think. The theory should at least give us a function with some constant that we can supply experimentally. That does not sound good for a "fundamental " theory which we have to be keep feeding it ansatz. I can understand external potentials applied but not potentials generated by the system itself.

2. ### arunma

907
If I'm reading you right, you're basically asking where the Coulomb potential comes from. Am I correct? There's actually a good discussion on this in Peskin and Schroedter's quantum field theory book (I don't have it handy at the moment, otherwise I'd supply the page number). I'll try to summarize it as best as I can.

In some sense, it's a bit weird to even have a potential in quantum theory. Potential energy is a classical idea, but it appears in non-relativistic quantum mechanics for some reason. In relativistic quantum mechanics we discard wavefunctions and instead talk about fields (which are defined at every point in space and time), and we actually get rid of the notion of potentials too. Instead we have Lagrangians which describe how various fields interact with each other. Surprisingly, you can actually derive the Coulomb potential from quantum field theory. What's very interesting about the field theory model is that unlike non-relativistic quantum mechanics, QFT actually places restrictions on what types of potentials can exist. So it turns out that you can't just throw whatever potential you want into the Schrodinger Equation and solve for the eigenfunctions and eigenenergies. Some potentials are not possible candidates for fundamental interactions. Thus, the potential in the hydrogen electron's Hamiltonian is not some arbitrary ansatz. It's the correct potential that you get from treating electromagnetic interactions with quantum electrodynamics.

But anyway, if somebody knows what I'm talking about in the Peskin book, then perhaps they could supply the page number and you could check there. Peskin and Schroedter explain it a lot more clearly and accurately than I do.

3. ### RedX

969
As arunma said, quantum field theory can explain the Coloumb potential. Zee's book also discusses this. The 1/r^2 is actually something that comes out of spacetime being 4-dimensional (it's been awhile since I read that part, so I can't go into details now). Not only is the Coloumb potential explained by QFT, but also why the exchange of spin-zero particles is attractive (i.e., why is the strong force attractive) and has exponential decay in addition to 1/r^2 (Yukawa force), and why two like charges repel when the exchange is spin 1 particles (E&M), and also why the exchange of spin 2 particles is attractive (graviton). So the pattern is that the exchange of an even spin particle is always an attractive force, and odd spin is always repulsive. All these things, Coloumb's law, Yukawa's potential, attractive or repuslive, come from QFT.

4. ### meopemuk

This is a well-known fact that lowest-order QFT scattering amplitudes can be reproduced in a quantum-mechanics-like theory with instantaneous potentials (like Coulomb or Yukawa). It is less known that this approach can be extended to higher perturbation orders as well. Then QFT radiative corrections can be absorbed into corrections to interparticle potentials and more complex potentials should be added, in particular those that change the number of particles. This kind of reformulation of QFT is known as the "dressed particle" approach. Here are some references:

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

A. V. Shebeko and M. I. Shirokov, "Unitary transformations in quantum field theory and bound
states", Phys. Part. Nucl., 32 (2001), 15. http://www.arxiv.org/abs/nucl-th/0102037

E. V. Stefanovich, "Relativistic quantum dynamics", http://www.arxiv.org/abs/physics/0504062

Eugene.

5. ### ytuab

189
It's a long story. So I looked at only the small part of your references. I'm sorry if I misunderstand.

In the relativistic QFT, the infinite bare charge and mass exist.

About the Dressing transformation. In page 388
This time the rug was the phase $$\Phi$$ of the transformation operator $$e^{i\Phi}$$.
However this operator had no physical meaning, so there is no harm in choosing it infinite

Did you only transfer the infinity to $$e^{i\Phi}$$?

The idea that the infinity is removed by the infinity is the same as that of the traditional methods?

Does it have any more merits in comparison to the traditional methods?

6. ### meopemuk

In traditional QED you cannot use the usual Hamiltonian obtained from naive analogy with classical electrodynamics. High order perturbation contributions to scattering amplitudes come out divergent, i.e., useless. Tomonaga-Schwinger-Feynman renormalization approach "fixes" this problem by adding divergent counterterms to the above Hamiltonian. If the counterterms are chosen correctly, then the original divergences get cancelled by the divergences from counterterms, and the full renormalized S-matrix is finite and agrees with experiment very well.

Unfortunately, this great result is achieved at the expense of screwing up the Hamiltonian completely. The Hamiltonian of the renormalized theory contains divergent counterterms, so this Hamiltonian H is useless for any type of calculation except the S-matrix calculation (where all divergences cancel out, as I said before). For example, you cannot form the time evolution operator $$\exp(iHt)$$ and therefore you cannot study the interacting time evolution of states and observables. Usually, this problem is not regarded as a big deal, because it is almost impossible to measure the time evolution of particles in scattering experiments. However, I believe that without a well-defined finite Hamiltonian and time evolution operator a theory (QFT) cannot be considered successful and complete.

The dressed particle approach suggests to fix the above problem by choosing a new Hamiltonian H', which is obtained from the (divergent) Hamiltonian H of QED by means of a unitary (dressing) transformation

$$H' = e^{i\Phi} H e^{-i \Phi}$$

It can be proven that the Hermitian operator $$\Phi$$ (which generates the unitary transformation) can be chosen in such a way that

1. The S-matrix calculated with H' is the same as the (accurate) S-matrix calculated with H
2. The relativistic invariance of the theory is preserved.
3. All divergences contained in H get "absorbed" in $$\Phi$$, so that new "dressed particle" Hamiltonian H' is divergence-free.

After this dressing transformation is done we can simply forget about divergent quantities H and $$\Phi$$, and perform all calculations (S-matrix, bound states, time evolution, etc.) with our new finite Hamiltonian H'. Remarkably, in these calculations we will never meet divergences, and we will never need to perform renormalization. So, in this approach, we have managed not only sweep divergences under a rug (as before), we also throw away the rug and all divergences under it.

There is a well-defined prescriprion of how to evaluate the dressed particle Hamiltonian H' in each perturbation order. In the lowest (2nd) order the Hamiltonian H' for charged particles coincides with the well-known Darwin-Breit Hamiltonian. For 2 particles in the non-relativistic approximation, it takes the usual form

$$H' = H_0 + e^2/r$$

Eugene.

7. ### qsa

362
Thank you all for the relpies. I do have the book Zee, the three volumes of weinberg and Hartfield among others. I'll get back about this question later.

So now we should be able to calculate the observables(spacial) or can we. Also I noticed in your blog you talk about bare electrons as virtual photons(I know you prefer weinberg), please, do you have any reference.

8. ### ytuab

189

So you mean this new method also uses the idea that the infinity is removed by the infinity in the divergent problems of the relativistic QFT?

(Sorry. this is what I have wanted to know, because it is very strange to me the idea that the infinity is removed by the infinity in the infinite bare charge and mass.)

9. ### meopemuk

Yes, QFT infinities should be cancelled by some other infinities. There is no other way to get rid of them. Traditional renormalization approach cancels S-matrix infinities by adding infinite counterterms to the Hamiltonian. This is not good, because we need a well-defined finite Hamiltonian to do physical calculations beyond the S-matrix. The "dressed particle" approach takes one more step and cancels infinities in the Hamiltonian by infinities in the "unitary dressing operator" $$e^{i\Phi}$$. The latter infinities are completely harmless, because this "unitary dressing operator" does not take part in any calculations.

In an ideal world we would not need to struggle with all these nasty infinities. We could define a finite "dressed particle" Hamiltonian from the beginning, and never have to worry about divergences and renormalization in our calculations. The general form of such a good Hamiltonian is known, however, it is not clear how to choose coefficients in different interaction terms. There is no any guiding principle for choosing appropriate interactions. On the other hand, traditional QFT (in spite of all its renormalization problems) has a very powerful principle of local gauge invariance. This principle allows us to limit the choice of interactions to just a few types. Somewhat miraculously, these are exactly the types of interactions (electromagnetic, electroweak,..) that are present in nature.

Eugene.

10. ### meopemuk

My point was that with the finite "dressed particle" Hamiltonian we can calculate the time evolution of any observable in an interacting system of particles. Such a calculation is impossible with the Hamiltonian of renormalized QFT, due to the presence of divergent counterterms there

I don't think I ever said that "bare electrons" and "virtual photons" are similar. The only similarity between them is that they both are rather useless and confusing theoretical artefacts. A good consistent theory should not use these two notions.

Eugene.

11. ### qsa

362
this is a direct quote from your blog

"I think that the terminology "dressed electron" is very unfortunate. It appeared for historical reasons, because originally QFT was formulated in terms of "bare particles". In the "bare particle" representation physical electrons look like a complicated mess of virtual photons, electron-positron pairs etc. The "dressed particle" approach cuts all this nonsense, but sadly the old terminology remains."

maybe I misunderstood you.

12. ### meopemuk

Let me clarify. In the traditional QED, physical electron (which is an eigenstate of the full interacting Hamiltonian) is represented as a complicated linear combination of bare particle states. Showing just a few simplest terms in this expansion I can write:

(1 physical electron) = (1 bare electron) + (1 bare electron + 1 photon) + (1 bare electron + 1 bare electron-positron pair) + ...

In textbooks you can often find statements about "bare electron" being surrounded by a "cloud" of virtual particles. Nobody has written exactly all the terms in this expansion. Frankly, in calculations nobody really cares about this expansion either.

The idea of the "dressed particle" approach is to ignore "bare particles" and "virtual clouds" and treat "physical particles" as single fundamental entities. Then the Hamiltonian of QED can be rewritten in terms of creation/annihilation operators of these physical particles. Such a reformulation can be achieved by using the unitary dressing transformation that I've mentioned earlier.

Eugene.

13. ### qsa

362
you were right, I am asking where the Coulomb potential comes from. I read Zee. He only derived the expression for the potential by inserting the EM potential A lagrangian. The J(k)[matter] never "produced" the A. Tell me I am wrong.

14. ### RedX

969
Well, on the front page:

there is a thread that has part of the calculation that derives a 1/r^2 potential. That's what the calculation looks like.

You are right in that the source J(x) never produced A (or $$\phi$$, in the thread). J(x) is an external disturbance that merely perturbs A (not produces it). A was always there. A has an existence outside of whether or not there is a source charge.

15. ### qsa

362
Thanks. After head scratching, the calculations check out. My problem is with A (or $$\phi$$, in the thread). But I thought we are talking about two electrons producing A(in our case), since if the two did not exist so no other fields should either.

16. ### qsa

362
anybody cares to repond to my last post. I would appreciate it a lot.