Dirac equation

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The Schrodinger equation solved for the hydrogen atom gave good agreement with spectral lines, except for line doublets.

To account for these electron spin theory was grafted onto the theory, despite the problem of electron being a point particle.

In 1928 Dirac gives his different answer to this doublet problem with the Dirac equation giving the correct spin for the electron. The spin being connected to relativity.

My question is does the Dirac equation when solved for the Hydrogen atom give all the other correct predictions for spectral lines as the Schrodinger solution does ?

Or are there many other problems/difficulties/unknowns introduced by the Dirac solution ?
 

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  • #2
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The Pauli equation (Schroedinger + spin and a few other relativistic corrections) can be derived as the non-relativistic limit of the Dirac equation. As such it correctly gives the same results as the Schroedinger/Pauli equation for the spectral lines of the Hydrogen atom.

The Dirac equation gives 4-component spinors which are usually interpreted as spin-up particle, spin-down particle, spin-up antiparticle and spin-down antiparticle. It kind of allows the creation of particle/antiparticle pairs out of pure energy. In the hydrogen atom, these effect can be neglected., but in a more general context, it requires the introduction of the "second quantization" that systematically treats the creation and annihilation of particles. That is a big can of worms with infinities creeping up that require "renormalization" and other similar candies.

We all know by now that antiparticles exist, so having predicted that is a huge success for the Dirac equation. But mathematically, renormalization theory still seems to be a mess even if many hugely successful predictions have come out.

I am sure someone with better knowledge will have a more qualified opinion on this :-)
 
  • #3
Bill_K
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No need to make a nonrelativisitic approximation, the full Dirac equation for a central Coulomb potential can be solved exactly. (See one of the standard QM references, such as Messiah vol II.) The resulting energy levels include most of the known corrections such as hyperfine splitting (spin-orbit coupling) and the relativistically correct kinetic energy, which becomes important for inner electrons in high-Z atoms. What it does not include are QFT corrections such as the Lamb shift.

More generally, treating the Dirac equation as an equation for a single particle in an external potential is a valid approximation as long as V(x) << mc2. At that point pair production becomes important.
 
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OK thanks for that.

pretty impressive then.
 
  • #5
dextercioby
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The hyperfine splitting is not a consequence of the Dirac equation. The fine splitting is.
 
  • #6
Jano L.
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Some time ago on this forum we had a discussion about the stability of the atom according to the quantum theory, and one contributor made an interesting remark that due to the fact that in Dirac's eq. the energy is linear in momenta, according to this eq. models of many-electron atoms become unstable, in contrary to models based on Schr. eq, where energy is quadratic in momenta. Maybe in quantum field theory they become stable again, but I think the consequences of the equations are so hard to infer that we do not know for sure.

No need to make a nonrelativisitic approximation, the full Dirac equation for a central Coulomb potential can be solved exactly.
Bill, this makes an impression that the model is fully relativistic and can be solved exactly. But "exactly" above means only exactly within the approximate model. The approximation is that the electron moves in central electrostatic field of the proton. In relativistic theory, this is necessarily an approximation to field worse than 1/c^2 (accompanied by total neglect of readiation from the proton and retardation of its field). Better approximation has to be achieved in other way, maybe somehow in quantum field theory or otherwise.

More generally, treating the Dirac equation as an equation for a single particle in an external potential is a valid approximation as long as V(x) << mc2.
Can you please explain this? What is V(x)?
 
  • #7
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What it does not include are QFT corrections such as the Lamb shift.
So the dirac equation isn't a form of QFT? I've been told different things, but that being said, I haven't had a course in QFT yet (I'm now having a course in elementary particles using Griffiths' book, which does discuss the Dirac equation and such)
 
  • #8
dextercioby
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[...]The approximation is that the electron moves in central electrostatic field of the proton. [...]
Assumed of infinite mass, which is an approximation done in this case specifically, but is unjustified in the non-specially-relativistic Schroedinger equation. The electrostatic approximation for the e-m interaction electron/proton in the Schroedinger equation is justified, of course, because the particles' dynamics is non-specially-relativistic as well.
 
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  • #9
dextercioby
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So the dirac equation isn't a form of QFT? I've been told different things, but that being said, I haven't had a course in QFT yet (I'm now having a course in elementary particles using Griffiths' book, which does discuss the Dirac equation and such)
If the Dirac field were a true classical field (like the e-m field or the gravitational field), then it would have a Lagrangian whose Euler-Lagrange equations would be the Dirac equations.
 
  • #10
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So it is a form of QFT? (I'm sorry if I'm completely missing your point, I'm not familiar enough with these topics)
 
  • #11
dextercioby
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By my version of QFT, it's not, it's a form of relativistic quantum mechanics, that is quantum mechanics (like the one of Schroedinger or Heisenberg) + special relativity instead of Galilean relativity.
 
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  • #12
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Interesting, although I'm slightly confused, since Griffiths seems to mention a Lagrangian for the Dirac equation, while you seem to say it doesn't exist?
 
  • #13
dextercioby
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What Griffiths and many others are using is a <fabrication> of a CLASSICAL Lagrangian for a theory which is purely quantum mechanical. It has no physical meaning, for classical physics cannot account for the existence of the Dirac field.

Surely, one can build classical theories of electrons, but not with spinors.
 
  • #14
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What is a good source for enlightenment on these matters?
 
  • #15
dextercioby
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I would reccomend starting with a course on formal QFT, then you can ask yourself questions like I did. A solid background in mathematics is a must, some answers to your present & future questions are impossible to grasp without it.
 
  • #16
Bill_K
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Assumed of infinite mass,
dextercioby, No one will deny you the right to make the assumption that the proton has infinite mass, but it is pointless and counterproductive. The idea in replacing the Schrodinger equation with the Dirac equation is to make a better approximation, not a worse one. Failure to use the reduced mass means you won't even be able to reproduce the Bohr-Sommerfeld energy levels.
 
  • #17
dextercioby
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Replacing me by μ is indeed a useful trick and is reccomended by Messiah (the only one to my knowledge) to account for the error introduced by the assumption that the nucleus mass is infinite, however, due to the difficuly of separating the 2 particle motion into COM motion and relative "dummy" particle motion in the context of SR*, it remains as a trick, not as a result of specially relativistic dynamics.

*Here again I refer to Greiner's text on field equations as in the other recent thread on this subject (H atom and Dirac's eqn.)
 
  • #18
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Replacing me by μ is indeed a useful trick and is reccomended by Messiah (the only one to my knowledge)
Lol, this is a standard procedure in every textbook on Classical Mechanics for a two-body problem interacting with a central force.
 
  • #19
dextercioby
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Yes, because in Newtonian mechanics it is well justified. In special relativity, it's not.
 
  • #20
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Yes, because in Newtonian mechanics it is well justified. In special relativity, it's not.
Of course. But, then again, there is no concept of potential energy in SR.
 
  • #21
dextercioby
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True, but there's the notion of electromagnetic field potential to replace it.
 
  • #22
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True, but there's the notion of electromagnetic field potential to replace it.
If by electromagnetic field potential, you mean the 4-potential [itex](\phi, \vec{A})[/itex], then yes, which is time dependent and depends on all previous locations and velocities of the proton and the electron! The dynamics of these particles, in turn, depends on the field. This problem, to my knowledge has not been solved exactly in non-quantum mechanics, nor in quantum.
 
  • #23
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If by electromagnetic field potential, you mean the 4-potential [itex](\phi, \vec{A})[/itex], then yes, which is time dependent and depends on all previous locations and velocities of the proton and the electron!
Don't Maxwell's equations relate electromagnetic field and potentials to charge and current density in the present (or rather the past a lightspeed-delay ago)? What does the entire history of the particles matter?
This problem, to my knowledge has not been solved exactly in non-quantum mechanics, nor in quantum.
Could you elaborate on this problem? What is it called?
 
  • #24
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Don't Maxwell's equations relate electromagnetic field and potentials to charge and current density in the present (or rather the past a lightspeed-delay ago)? What does the entire history of the particles matter?
No. ME relate the second derivatives (D'Alembertians) to the instantaneous charge-current density at that point. Due to retardation effects, the solution depends on all densities lying on the light cone passing through the space-time point we are considering. You need to find where the world line of each particle intersects this cone to calculate its contribution to the potential at that point.

Then, the acceleration of each particle depends on the derivatives of the potentials (the fields) at a particular space-time point. Finding a derivative of a function is a "nonlocal" operation in the following sense:

Expand the function [itex]f(x) = \int{\frac{dk}{2\pi} \, e^{i k \, x} \tilde{f}(k)}[/itex], where [itex]\tilde{f}(k) = \int{dx' \, f(x') \, e^{-i k \, x'}}[/itex] is the Fourier transform. The derivative is:
[tex]
f'(x) = \int{\frac{dk}{2 \pi} \, i \, k \, e^{i k \, x} \, \int{dx' \, e^{-i k \, x'} \, f(x')}}
[/tex]
[tex]
f'(x) = \int{dx' \, \left( -\int{\frac{d k}{2 \pi \, i} \, k \, e^{i k \, (x - x')}} \right) \, f(x')}
[/tex]
It is expressed as an integral operator that needs the values of the function for all possible values of x'.

From the previous point, we need the intersection of the past world line of the particle with all the world cones starting from every point in space. This basically amounts to knowing the whole past world line of the particle.

Could you elaborate on this problem? What is it called?
I don't know if it even has a special name. Relativistic two-body problem?! (when I type that, it refers to two point particles interacting only gravitationally http://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity. When I wrote electrodynamics relativistic two body problem, I got the following article "10.1103/PhysRevD.4.2956" [Broken])
 
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  • #25
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No. ME relate the second derivatives (D'Alembertians) to the instantaneous charge-current density at that point. Due to retardation effects, the solution depends on all densities lying on the light cone passing through the space-time point we are considering. You need to find where the world line of each particle intersects this cone to calculate its contribution to the potential at that point.
OK, that's true, but you still don't need to know the entire history of the particles. You just need to know about at the time(s) at which their worldlines intersect the light cones of the point.
 

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