Relativistic quantum field theory:antiparticles!

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  • #1
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When Heisenberg, Schrodinger and Dirac proposed the first quantum field theories (QFT) they did not obey Einstein's special theory of relativity (SR) :they were not invariant under the transform laws of special relativty.

So Dirac develops a new equation for the wave function nature of the electon which did obey special relativity... But now he had to posit the existence of another particle, an antiparticle, the positron! So a QFT consistent with SR predicts a doubling of the number of particles: thanks a lot Dirac!!

(The above perspective is derived from Charles Seife, THE STATE OF THE UNIVERSE, pages 158-162. (Seife says "Dirac found an unexpected result")

What happens in a QFT which obeys SR to cause more particles to emerge? Do we have a modern interpretaton? Why does SR invariance, a geometric constraint, cause antiparticles to emerge in QFT?
 
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  • #2
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I suspect some insights can be gleaned from

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

under "DIRAC'S COUP". But the description there is too obscure for me. Is "Physical Interpretation" in Wiki the answer...can somebody interpret the math gobbledegook??
Thanks
 
  • #3
Vanadium 50
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r...can somebody interpret the math gobbledegook??

I'm afraid not. The answer is in "the math gobbledegook" that you don't like. Essentially, you are doing something very similar to replacing a linear equation with a quadratic equation, and a quadratic equation has two solutions.
 
  • #4
malawi_glenn
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SR is not merley a geometric constraint. It is a constraint in the 4-dim Minkowski space (so in that sense it is a geometric "constraint" but not just in 3 space)

So the antiparticle interpretation arose due to the fact, as Vanadium mentioned, that a quadratic equation has two solutions. And E^2 = p^2 +m^2, I wouldnt call that geometrical constraint :P
 
  • #5
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In the Wikipedia article, under HISTORY they explain:

Two of the four solutions of the equation correspond to the two spin states of the electron. The other two solutions make the peculiar prediction that there exist an infinite set of quantum states in which the electron possesses negative energy. This strange result led Dirac to predict, via a remarkable hypothesis known as "hole theory," the existence of particles behaving like positively-charged electrons..... When asked later why he hadn't actually boldly predicted the yet unfound positron with its correct mass, Dirac answered "Pure cowardice!"

Vanadium, its not that I don't like but rather I have not studied the advanced math involved here. I decided a few years ago to focus on experienced scientists interpretations of the math rather than embark on extensive math studies on my own....If Dirac was "surprised" and you guys tell me there are more variables involved, that suggests nobody really knows the physical answer to my question....and that may also be implied by some Wikipedia comments. Maybe there just is not one understood today...in which case I'm not the only "dummy"...then thats ok.

In your posted explanations, I had already assumed there were extra variables as posted....conjugate solutions, perhaps....I studied electron and hole semiconductor theory many years ago but did not know... or forgot ....it came from Dirac....way back then it seemed like a "stop gap explanation" to account for twice the experimentally observed current flow due soley to electrons.

and I understand the time/space Minkowski constraint...but my question is how such a spacetime constraint manifests itself to cause new particles to emerge! That sounds more like string theory where geometrical constraints of additional dimensions are hypothesized to constrain string vibrations and as a result effect particle (string) properties.
 
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  • #6
malawi_glenn
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SR gives you E^2 = p^2 + m^2, hence Quadratic!

Classically you only have E = kinE = mv^2/2

The particle hole theory in semicondcutor is not exactly what the Dirac Sea is (which is remnisent today where we have better knowledge in Quantum Field theory). in the semiconductor you only have a "hole" in the electron sea, no antiparticiles there...

And hehe you say you don't math enough to do "what"? But you are adressing things to String Theory? :-p
 
  • #7
atyy
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What happens in a QFT which obeys SR to cause more particles to emerge? Do we have a modern interpretaton? Why does SR invariance, a geometric constraint, cause antiparticles to emerge in QFT?

This has nothing to do with relativity. There are QFTs in non-relativistic condensed matter physics. There you write the Schroedinger equation for lots of identical particles which are not created or destroyed. You can write the same equation as a QFT for "fake particles", such as phonons, which can be created or destroyed. The analogy in classical physics is a classical string consisting of many particles. The classical string can have many "modes" or "excitations" or "fake particles" which are created and destroyed. Historically it wasn't discovered this way, but QFT is essentially a quantum formalism in which "fake particles" can be created and destroyed.
 
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  • #8
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This has nothing to do with relativity.

well, Charles Seife in his book says it DOES...but maybe you are right in which case that would address my question....That's the same reaction I had when when I read Seife's explanation!!!! It still sounds "crazy" but so does time dilation.

But Seife is no "crackpot" ....I am able to correlate a lot of his explanations with Kiku, Greene, Smolin and others....and there were sometimes different but accurate perspectives..... maybe he got this wrong....


Malawai...yes, but your "example" equation has nothing to do with my question unless the additional quadratic terms somehow posits new particles....
 
  • #9
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Naty, if you are familiar with the non-relativistic Schroedinger equation around page 19/20 of the draft (downloadable) version of Mark Srednicki's QFT text available here:

http://www.physics.ucsb.edu/~mark/qft.html

has a pretty clear motivation/discussion of how to get from the non-relativistic equation to the Klein-Gordon equation (not much more handywavy than what appears to be the usual way of motivating Schroedinger's equation itself) :

[tex]
- \Box \psi = \frac{m^2 c^2}{\hbar^2} \psi
[/tex]

It appears to me, that roughly speaking, the Dirac equation results from taking roots of the operators on both sides. This "undoes" the squaring that was done in the first place (in the text above) arriving at the Klien Gordon equation, and still leaves you with an differential operator on functions to be determined that essentially still expresses an energy conservation relationship (like the normal Schr. eqn and the Klien Gordon equation).

This leaves you with something of the form:

[tex]
i \hbar \sqrt{\Box} \psi = \pm m c \psi
[/tex]

Now, there is a lot of math in the wiki Dirac equation page (and as an amature part time self study physics student I personally admit to NOT understanding lots of it;) That said, a meaning can be given to the root of the wave equation operator (there's an infinite number of such roots, you can describe them as spacetime "four-vector" gradients, and they all vary by a Lorentz boost or rotation).

My guess here is that the positron vs. electron postulate comes from attempting to assign or pick between the two possible signs of these roots. One isn't any better than the other mathematically, so why shouldn't both be allowed.

2c from somebody who first saw these equations last night, and was trying to get some intuitive feeling for the motivation and meaning for them before looking at my old school QM text that almost exclusively deals with how to solve the equation.

Hopefully I'm not too far off base.
 
  • #10
malawi_glenn
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NO the dirac equation is NOT obtained by "taking the root of the operators on both sides!
 
  • #11
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NO the dirac equation is NOT obtained by "taking the root of the operators on both sides!

If the Klien Gordon equation is written in clifford algebra terms (as in the Doran/Lasenby Geometric Algebra text)

[tex]
\Box = (\gamma^\mu \partial_\mu)^2
[/tex]
[tex]
\gamma^\mu \cdot \gamma_\nu = \delta^\mu_\nu
[/tex]
[tex]
{\gamma_0}^2 = 1
[/tex]
[tex]
{\gamma_i}^2 = -1
[/tex]

Then one _can_ take roots in the way described to arrive at the equation:

[tex]
i \hbar \gamma^\mu \partial_\mu \psi - m c \psi = 0
[/tex]

which is given in the wikipedia page as the covariant form of the Dirac equation. As just an algebra exersize this is possible. Whether actually doing that is justifiable is another story.
 
  • #12
malawi_glenn
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I have heard many times that taking the square root of Klein Gordon eq is not correct, altough one could do it naively.

BUT can you show that [tex]\Box = (\gamma^\mu \partial_\mu)^2[/tex] ? I can't do it.


I get:

[tex]\Box = \eta ^{\mu\nu}\partial _{\nu} \partial _{\mu} = \frac{1}{2}\gamma ^{\mu}\gamma^{\nu} \partial _{\nu} \partial _{\mu} +\frac{1}{2}\gamma ^{\nu}\gamma^{\mu} \partial _{\nu} \partial _{\mu} [/tex]

Dirac equation are for spin 1/2 particles ans KG for scalars... so the wavefuntions are not "equal" either.
 
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  • #13
malawi_glenn
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Malawai...yes, but your "example" equation has nothing to do with my question unless the additional quadratic terms somehow posits new particles....

The equations solution suggest that particles with equal properties but opposite energy can exist, that was the suggestion/postulate of Dirac.
 
  • #14
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I have heard many times that taking the square root of Klein Gordon eq is not correct, altough one could do it naively.

BUT can you show that [tex]\Box = (\gamma^\mu \partial_\mu)^2[/tex] ? I can't do it.


[tex]\begin{align*}
(\gamma^\mu \partial_\mu)^2
&=
(\gamma^\mu \partial_\mu) \cdot (\gamma^\nu \partial_\nu)
+ (\gamma^\mu \partial_\mu) \wedge (\gamma^\nu \partial_\nu) \\
&= (\gamma^\mu \partial_\mu) \cdot (\gamma_\nu \partial^\nu) \\
&= \gamma^\mu \cdot \gamma_\nu \partial_\mu \partial^\nu \\
&= {\delta^\mu}_\nu \partial_\mu \partial^\nu \\
&= \partial_\mu \partial^\mu \\
\end{align*}
[/tex]
 
  • #15
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Malawi..My post #8 was referring to your post #6...Is that what your are referring to in your post # 13...??


In another active thread, THE PHYSICAL MEANING OF PAULI SPIN MATRICES, someone just posted:
Then, I recommend that you read up on the Dirac equation as presented from a relativistic quantum mechanics point of view (rather than quantum field theory).

And maybe my question would have been better expressed if I had asked what considerations go into such a change in perspective IF that is what causes positrons to pop into existence...
 
  • #16
malawi_glenn
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[tex]\begin{align*}
(\gamma^\mu \partial_\mu)^2
&=
(\gamma^\mu \partial_\mu) \cdot (\gamma^\nu \partial_\nu)
+ (\gamma^\mu \partial_\mu) \wedge (\gamma^\nu \partial_\nu) \\
&= (\gamma^\mu \partial_\mu) \cdot (\gamma_\nu \partial^\nu) \\
&= \gamma^\mu \cdot \gamma_\nu \partial_\mu \partial^\nu \\
&= {\delta^\mu}_\nu \partial_\mu \partial^\nu \\
&= \partial_\mu \partial^\mu \\
\end{align*}
[/tex]

show how you did the first identity, how can you obtain the wedgeproduct?

Can you try to show why it works with my way of doing it?
 
  • #17
malawi_glenn
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Malawi..My post #8 was referring to your post #6...Is that what your are referring to in your post # 13...??

Yes that is correct.
 
  • #18
malawi_glenn
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Peeter, is that derivation given in a textbook or similar, I can't even get:

[tex]\gamma^\mu \cdot \gamma_\nu = \delta^\mu_\nu[/tex]

working, i get:

[tex]\gamma^\mu \cdot \gamma_\nu = 2\eta ^\nu_\mu - \gamma_\mu \gamma^\nu[/tex]
 
  • #19
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show how you did the first identity, how can you obtain the wedgeproduct?

Can you try to show why it works with my way of doing it?

The definition of the product of two vectors in a clifford algebra is:

[tex]
a b = a \cdot b + a \wedge b
[/tex]

With a symmetric and antisymmetric split one has:

[tex]
a \cdot b = \frac{1}{2}( a b + b a )
[/tex]

[tex]
a \wedge b = \frac{1}{2}( a b - b a )
[/tex]

(this is definition in clifford algebra introductions, but a consequence in the axiomatic treatment when dot and wedge are defined in terms of grade selection)

I'm not exactly sure what your way of doing it is, but think that this may be closer to it:

[tex]
\begin{align*}
\Box
&= \eta ^{\mu\nu}\partial_{\nu} \partial_{\mu} \\
&= \frac{1}{2}\gamma^{\mu}\gamma^{\nu} \partial_{\nu} \partial_{\mu} +\frac{1}{2}\gamma^{\nu}\gamma^{\mu} \partial_{\nu} \partial_{\mu} \\
&= \frac{1}{2}\left(\gamma^{\mu}\gamma_{\nu} + \gamma_{\nu}\gamma^{\mu} \right) \partial^{\nu} \partial_{\mu} \\
&= \gamma^\mu \cdot \gamma_\nu \partial^{\nu} \partial_{\mu} \\
&= \partial^{\mu} \partial_{\mu} \\
\end{align*}
[/tex]
 
  • #20
malawi_glenn
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Hmm must try that algebra thing.

When you tried help me, can you motivate:
[tex]\frac{1}{2}\left(\gamma^{\mu}\gamma_{\nu} + \gamma_{\nu}\gamma^{\mu} \right) \partial^{\nu} \partial_{\mu} \\&= \gamma^\mu \cdot \gamma_\nu \partial^{\nu} \partial_{\mu} [/tex]

?
 
  • #21
Hurkyl
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If the Klien Gordon equation is written in clifford algebra terms
...
Then one _can_ take roots in the way described to arrive at the equation:
But there are more than 2 square roots....
 
  • #22
Avodyne
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Here's how the Dirac square-root thing works in a more pedestrian notation. First we write [itex]\gamma^\mu\gamma^\nu[/itex] as the sum of a commutator and an anticommutator:

[tex]\gamma^\mu\gamma^\nu = {\textstyle{\frac12}}[\gamma^\mu,\gamma^\nu]+{\textstyle{\frac12}} \{\gamma^\mu,\gamma^\nu\}.[/tex]

The first term is antisymmetric on exchange of [itex]\mu[/itex] and [itex]\nu[/itex], while the seoncd term is symmetric. Also, partial derivatives commute, so [itex]\partial_\mu\partial_\nu[/itex] is symmetric. Then, in general, if you contract both indices of an antisymmetric tensor [itex]A^{\mu\nu}[/itex] with those of a symmetric tensor [itex]S_{\mu\nu}[/itex], you get zero: [itex]A^{\mu\nu}S_{\mu\nu}=0[/itex].
So, the commutator term vanishes when contracted. And the anticommutator is

[tex] {\textstyle{\frac12}}\{\gamma^\mu,\gamma^\nu\}=\eta^{\mu\nu}.[/tex]
 
  • #23
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But there are more than 2 square roots....

Given a particular choice of basis (ie: observer frame), are there more than two roots? Other root selections rotate or lorentz rotate the frame I think.
 
  • #24
Hurkyl
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Given a particular choice of basis (ie: observer frame), are there more than two roots? Other root selections rotate or lorentz rotate the frame I think.
Multiplication is independent of basis. If there are more than two square roots, then it doesn't matter what basis you analyze the problem in.

I don't know whether or not you can apply additional constraints so that the resulting system of equations has only two solutions. It's that simply proving x²=a² is definitely not enough to infer that [itex]x = \pm a[/itex].
 
  • #25
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Hmm must try that algebra thing.

When you tried help me, can you motivate:
[tex]\frac{1}{2}\left(\gamma^{\mu}\gamma_{\nu} + \gamma_{\nu}\gamma^{\mu} \right) \partial^{\nu} \partial_{\mu} \\&= \gamma^\mu \cdot \gamma_\nu \partial^{\nu} \partial_{\mu} [/tex]

?

That's the dot product again. If I'm to attempt to use the "pedestrian" notation, as Avodyne puts it, I'll have to learn it first.

Can you tell me how in the that notation you would define your reciprocal frame vectors [itex]\gamma^\mu[/tex] in terms of the [itex]\{\gamma_\mu\}[/itex] frame?
 

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