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Potential energy in hydrogen atom 
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#1
Dec2609, 07:06 PM

P: 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
Dec2709, 02:26 AM

P: 908

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 nonrelativistic 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 nonrelativistic 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
Dec2709, 02:52 AM

P: 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 4dimensional (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 spinzero 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
Dec2709, 03:27 AM

P: 1,746

Potential energy in hydrogen atom
This is a wellknown fact that lowestorder QFT scattering amplitudes can be reproduced in a quantummechanicslike 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/nuclth/0102037 E. V. Stefanovich, "Relativistic quantum dynamics", http://www.arxiv.org/abs/physics/0504062 Eugene. 


#5
Dec2709, 08:12 AM

P: 189

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 [tex]\Phi[/tex] of the transformation operator [tex]e^{i\Phi}[/tex]. However this operator had no physical meaning, so there is no harm in choosing it infinite Did you only transfer the infinity to [tex]e^{i\Phi}[/tex]? 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
Dec2709, 03:58 PM

P: 1,746

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 Smatrix calculation (where all divergences cancel out, as I said before). For example, you cannot form the time evolution operator [tex]\exp(iHt)[/tex] 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 welldefined 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 [tex] H' = e^{i\Phi} H e^{i \Phi} [/tex] It can be proven that the Hermitian operator [tex]\Phi[/tex] (which generates the unitary transformation) can be chosen in such a way that 1. The Smatrix calculated with H' is the same as the (accurate) Smatrix calculated with H 2. The relativistic invariance of the theory is preserved. 3. All divergences contained in H get "absorbed" in [tex]\Phi[/tex], so that new "dressed particle" Hamiltonian H' is divergencefree. After this dressing transformation is done we can simply forget about divergent quantities H and [tex]\Phi[/tex], and perform all calculations (Smatrix, 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 welldefined 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 wellknown DarwinBreit Hamiltonian. For 2 particles in the nonrelativistic approximation, it takes the usual form [tex]H' = H_0 + e^2/r [/tex] Eugene. 


#7
Dec2709, 06:42 PM

P: 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.



#8
Dec2709, 06:43 PM

P: 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
Dec2709, 07:04 PM

P: 1,746

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
Dec2709, 07:14 PM

P: 1,746

Eugene. 


#11
Dec2709, 07:22 PM

P: 362

"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, electronpositron pairs etc. The "dressed particle" approach cuts all this nonsense, but sadly the old terminology remains." maybe I misunderstood you. 


#12
Dec2709, 07:49 PM

P: 1,746

(1 physical electron) = (1 bare electron) + (1 bare electron + 1 photon) + (1 bare electron + 1 bare electronpositron 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
Dec2809, 08:25 PM

P: 362




#14
Dec2809, 08:50 PM

P: 969

http://www.physicsforums.com/showthread.php?t=365775 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 [tex]\phi [/tex], 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
Dec2809, 09:25 PM

P: 362




#16
Dec2909, 05:36 PM

P: 362

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



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