Hamiltonian for Fermionic Fields

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Typically the Standard Model uses the Klein-Gordon equation for scalars, the Dirac equation for spin 1/2 fermions, the Yang-Mills equations for vector bosons and the Schwinger equation for spin 5/2 (though none exist in the SM). The relevant Hamiltonians are easy to deduce from the equations.

As for the Jaynes-Cummings equation, I've never used it at any point in any QFT course.

My thoughts exactly. GD is constantly trying to run before he can even stand, never mind walk. No one wants to do the 'boring' stuff like having to learn basic vector calculus or linear algebra but its essential to understand the more 'sexy' stuff like quantum mechanics and general relativity, that's why universities teach the basics first, then get onto the advanced stuff.

You're clearly not understanding the notation you're using so you're hindering yourself by trying to work with stuff because you understand the basics.

That is not your biggest problem....

Of course the equation doesn't equate to the Hamiltonian. To even consider that shows you don't understand what Hamiltonian and Lagrangian mechanics actually involve and what their utility is.

Learn what Lagrangians and Hamiltonians are and how they are used before trying to do something with complicated examples in quantum field theory, which is another area of physics currently beyond your grasp.

What's the matter, don't want to engage in a little discussion because you don't want to admit you don't understand Hamiltonian mechanics? Just because you start the thread doesn't mean people can't ask you questions. It's a discussion forum.

Judging from the replies in this thread, arfa is not the only one who questions whether you have any clue about this stuff. And this isn't the first time. Stand before you try to run or all you'll do is fall on your face.

 

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I do believe the c's are the creation and annihilation operators, and the equation was extracted from a field theory wiki. The Dirac equation mearly represents the spin states of fermion particles, with their respective matrices alpha and beta. I thought the mu was the muon particle, but is that wrong?

 

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So what equation typically is used for the Hamiltonian for fermions?

 

I am under the impression though James, that the equation was for fermions because the annihilation and creation operators are represented by c, whereas Bosons are represented by an a.

 

The vector components to the summation of energy through the random field noted as H. H is a measure of time when we note it between two objects, but by itself it just notes a level of uncertainty. Where K is temperature and has an effect on the frequency of Mu.

 

Related junk from two years ago:

http://www.physicsforums.com/showthread.php?t=269114

I get \(E_0 = N \left( \frac{3}{5} E_F - \mu \right)\) which simplifies in the case that \(\mu\) is zero is the zero-point energy of an ideal fermi gas.

Alternately, if \(\mu = E_F = \frac{\hbar^2}{2m} \left( \frac{3 \pi^2 N}{V} \right)^{\frac{2}{3}}\) then \( N = \frac{V}{3 \pi^2} (\frac{2m \mu}{\hbar^2})^{\frac{3}{2}} \) then \(\frac{\partial E_0}{\partial \mu} = - \frac{2}{5} \frac{\partial (N \mu)}{\partial \mu} = - \frac{2}{5} \frac{V}{3 \pi^2} (\frac{2m}{\hbar^2})^{\frac{3}{2}} \frac{\partial \mu^{\frac{5}{2}}}{\partial \mu} = - \frac{V}{3 \pi^2} (\frac{2m \mu}{\hbar^2})^{\frac{3}{2}} = - N\) which is remarkable or pedestrian depending on your viewpoint.

I will admit that I'm rusty (over 10 years of disuse) at this and the OP has certainly performed a contextomy upon whatever the original source it. What do you think of my attempt at this?

 

I'm quite sure \(cc^{\dagger}\) refer to the creation and annihilation operators, the things James are asking. After reading Rpenners link, I am not so sure the mu actually refers to muons. Maybe they do, but I am not familiar with the equation.

 

Alphanumeric, you say the Dirac Equation does not equation the Hamiltonian.

\(H= \alpha \cdot pc + \beta Mc^2 \)

But this equation more or less is the Dirac Equation. I am confused, are you saying the link is wrong? I admit, I had not heard of a Dirac Hamiltonian. The name doesn't surprise me nonetheless.

 

In the passage on iodentification of observables, there is a Hamiltonian for the Dirac Theory. Is this the one applied to fermions?

http://en.wikipedia.org/wiki/Dirac_equation

 

The matrices of course given by \(\alpha,\beta\) with their usual algebraic qualities \(\alpha \beta+ \beta \alpha= 0\), the M is for mass, c^2 the speed of light squared, p is momentum. I don't see why you would have thought I would have struggled with that. I find the Dirac Equation relatively easy to understand, and is actually a very beautiful proof.

 

In fact, it has just came to my attention that the ZPF-term energy contribution \(\frac{e^2 h}{4Mc\alpha^2}\) would also contribute to the total energy of the Jaynes-Cummings equation for bosonic atoms, the only thing is whether we consider the contribution is negligible. Electrons contribute atleast 0.06% to an atom's total mass. However negligible, they still add to the total energy, so for the atoms described by the Jaynes-Cummings equation, the correction would be:

\(\hat{H}= \hbar \omega \hat{a}^{\dagger}\hat{a}+ \frac{1}{2}\hbar \omega \hat{\sigma}_z+ \hbar \lambda (\hat{\sigma}_{(+)}\hat{a}+ \hat{\sigma}_{(-)}\hat{a}^{\dagger})+ \frac{e^2 h}{4Mc\alpha^2}\)

 

where \(\hbar \omega \hat{a}^{\dagger}\hat{a}\) is the energy contained in the field, \(\frac{1}{2}\hbar \omega \hat{\sigma}_z\) is the energy transitions between levels \( \hbar \lambda (\hat{\sigma}_{(+)}\hat{a}+ \hat{\sigma}_{(-)}\hat{a}^{\dagger})\) is the interaction of the field with the atom and \(\frac{e^2 h}{4Mc\alpha^2}\) would be the energy interaction between electrons and the ZPF.

 

I don't know how mu enters you equations, or what mu actually is.

I have seen one equation which come close to this, the Fermi Energy where it is clear you have based the equations from:

\(E_F= E_{N/2}= \frac{\hbar^2 \pi^2}{2ML^2}(N/2)^2\)

so forgive me if I don't readily understand the notation.

 

Also if mu was zero and was the zero-point energy, this seems like a contradiction since \(\epsilon= \frac{h \nu}{\frac{h \nu}{KT}-1}+ \frac{h \nu}{2}\). So my point is, the zero point field never reaches zero.

 

unless this was a typo?

''I get \(E_0 = N \left( \frac{3}{5} E_F - \mu \right)\) which simplifies in the case that \(\mu\) is zero is the zero-point energy of an ideal fermi gas.''

 

Ah I'm sitting there thinking why you might have said this, and I just realized I never took into account we are talking about an ideal gas. So the system is non-interacting.

So mu is an interacting term to the ZPF? That would mean the interaction between the electron is zero but the ZPF is not zero.

 

So if that is the case, then \(E_F-\mu\) is the energy measured relative to the chemical potential. So finally, \mu is the chemical potential. At last I found what it means.

 

right, so I have more questions. I took this from wiki;

''Therefore the total energy of the Fermi gas at absolute zero is larger than the sum of the single-particle ground states because the Pauli principle acts as a sort of interaction/pressure that keeps the fermions separated and moving. For this reason, the pressure of a Fermi gas is nonzero even at zero temperature''

Now this is getting messier for me. I am finding my concept of zero point energy field being distorted here on the theory. The article says ''even at zero temperatures,'' the gas has an energy larger than the sum of the single particle ground states.

problem

Zero temperatures are never reached, so how is this tested and known? The zero-point field does not allow a complete zero-temperature of the vacuum (which was why I had my little problem in your text).

Edited: Also it seems an ideal gas is by choice of the model you are working with. If ideal gases do not interact as I have always believed, then yes, I can understand why there is no interaction term in the Hamiltonian. However, this seems to be not always the same in every case.

Edit^2: I keep going away and finding more out. So the fermi ideal gas is in fact workable for only low temperature gases, which would mean overall, electrons do not interact with the zero-point field in empty space. Is this right?