Dirac equation and clifford algebra

Click For Summary

Discussion Overview

The discussion revolves around the necessity of Clifford algebra in deriving the Dirac equation, exploring different approaches and intuitions behind the derivation. Participants share their perspectives on whether Clifford algebra is essential or if the equation can be derived using alternative methods, including references to historical contexts and other mathematical frameworks.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Historical

Main Points Raised

  • Some participants argue that Clifford algebra is not strictly necessary to derive the Dirac equation, suggesting that it can be approached through other methods, such as using waves moving in opposite directions.
  • Others contend that while one may not need to know Clifford algebra initially, it becomes relevant when incorporating gamma matrices to reproduce the relativistic energy-momentum relationship.
  • A participant expresses difficulty in finding sources that derive the Dirac equation without involving Clifford algebra, indicating a reliance on existing literature that typically includes it.
  • Another participant mentions that introductory texts often start with a linear Hamiltonian operator to derive the Dirac equation, leading to the conclusion that certain relations, including those of Clifford algebra, emerge from the derivation process rather than being used directly.
  • Historical context is provided, noting that alternative mathematical frameworks like quaternions were once popular for formulating physical equations, and that various derivations of the Dirac equation exist, including those using quaternions.
  • It is emphasized that regardless of the method used for derivation, the Dirac formalism is a natural representation of the Lorentz group and is essential for describing certain interactions in physics.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the necessity of Clifford algebra for deriving the Dirac equation, with multiple competing views presented regarding its relevance and the methods of derivation.

Contextual Notes

Some participants highlight the limitations of existing literature and the assumptions made in various derivations, noting that the discussion does not resolve the mathematical steps or definitions involved in the derivation process.

TimeRip496
Messages
249
Reaction score
5
Is it a must to know clifford algebra in order to derive the dirac equation?
I recently watch drphysics video on deriving dirac equation and he use two waves moving in opposite directions to derive it, without touching clifford algebra. If this possible, what is the intuition behind it?
 
Physics news on Phys.org
No you don't "need to" know Clifford Algebra... you derive it in the process in order to get your equation working [and reproducing the relativistic relationship between energy and mass/momenta] . At least by the time one adds the gamma matrices..
 
  • Like
Likes   Reactions: vanhees71
ChrisVer said:
No you don't "need to" know Clifford Algebra... you derive it in the process in order to get your equation working [and reproducing the relativistic relationship between energy and mass/momenta] . At least by the time one adds the gamma matrices..
Do you have any sources that will allow me to derive the dirac equation without clifford algebra? Cause I have been searching on the Internet and all of it involves clifford algebra. Besides the one by drphysics he didnt really derive it fully such as the dirac matrix.
 
I haven't seen what you are talking about, I don't know which is that drphysics book... Maybe this can help? http://arxiv.org/vc/quant-ph/papers/0607/0607001v1.pdf (sec.4)

However, I don't understand this "clifford algebra" thing. In every introductory book, they never used the Clifford algebra to derive the Dirac equation.
Most of them are starting by a linear Hamiltonian operator, which they act to reproduce the result of the relativistic energy-momentum-mass relation E^2 = m^2 +p^2. By doing so, you see that the objects that you used in your operator were not just complex numbers, but they should satisfy certain relations, one of them is the Clifford Algebra. You don't "use" it, you "result" in it...
 
Well, you can call it as you like. The Clifford algebra is basically the algebra realized in the usual treatment by the Dirac matrices in the one or the other representation. At some time ago in history also quaternions and octonions were en vogue. In the 19th century at least; Maxwell formulated his equations first in terms of quaternions; the usual vector notation was introduced into physics by Heaviside and became widely used in the beginning of the 20th century. It was not the least spread by the famous Handbook "Enzykloädie der mathematischen Wissenschaften", edited by Felix Klein around 1900. You can find a derivation of the Dirac equation, using quaternions in the even more famous book by Sommerfeld "Atombau und Spektrallinien" (freely translated: Atomic Structure and Spectral Lines). As stressed before: No matter, how you derive it you end up with somehow with the Dirac formalism. This is no surprise, because it's a very natural representation of the proper orthochronous Lorentz group augmented by spatial reflections to be able to describe the parity-conserving interactions, especially the electromagnetic interaction, which was the first application of the Dirac formalism of course.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
816
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 61 ·
3
Replies
61
Views
10K
  • · Replies 26 ·
Replies
26
Views
6K
  • · Replies 50 ·
2
Replies
50
Views
4K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 30 ·
2
Replies
30
Views
3K