1 / 73

The World

The World. Particle content. All the particles are spin ½ fermions!. Schrodinger Wave Equation. He started with the energy-momentum relation for a particle. Expecting them to act on plane waves. he made the quantum mechanical replacement:. How about a relativistic particle ？.

jenna-holloway
Download Presentation

The World

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The World Particle content All the particles are spin ½ fermions!

  2. Schrodinger Wave Equation He started with the energy-momentum relation for a particle Expecting them to act on plane waves he made the quantum mechanical replacement: How about a relativistic particle？

  3. The Quantum mechanical replacement can be made in a covariant form. Just remember the plane wave can be written in a covariant form: As a wave equation, it does not work. It doesn’t have a conserved probability density. It has negative energy solutions.

  4. Plane wave solutions for KG Eq. Time dependence can be determined. There are two solutions for each 3 momentum p (one for +E and one for –E ) It has negative energy solutions.

  5. The proper way to interpret KG equation is it is not a Wavefunction Equation but actually a Field equation just like Maxwell’s Equations. Plane wave solutions just corresponds to Plane Waves. It’s natural for plane waves to contain negative frequency components. Expansion of the KG Field by plane： If Φ is a real function, the coefficients are related:

  6. Dirac This result is too beautiful to be false; it is more important to have beauty in one's equations than to have them fit experiment.

  7. Blaming the negative energy problem on the second time derivative of KG Eq., Dirac set out to find a first order differential equation. This Eq. still needs to give the proper energy momentum relation. So Dirac propose to factor the relation! For example, in the rest frame: Made the replacement First order diff. Eq.

  8. Now put in 3-momenta: Suppose the momentum relation can be factored into linear combinations of p’s: Expand the right hand side: We get and we need:

  9. It’s easier to see by writing out explicitly: Oops! What! or No numbers can accomplish this! Dirac propose it could be true for matrices.

  10. 2 by 2 Pauli Matrices come very close

  11. Dirac find it’s possible for 4 by 4 matrices We need: that is He found a set of solutions: Dirac Matrices

  12. Dirac find it’s possible for 4 by 4 matrices Pick the first order factor: Make the replacement and put in the wave function: If γ’s are 4 by 4 matrices, Ψ must be a 4 component column: It consists of 4 Equations.

  13. The above could be done for 2 by 2 matrices if there is no mass. Massless fermion contains only half the degrees of freedom. A pure left-handed or a pure right-handed must be massless.

  14. Now put in 3-momenta: Suppose the momentum relation can be factored into linear combinations of p’s: Expand the right hand side: Now β, γdo not need to be the same. We can choose: 交叉項抵銷

  15. Massless fermions contain only half the degrees of freedom.

  16. Expansion of a solution by plane wave solutions for KG Eq. If Φ is a real function, the coefficients are related:

  17. Plane wave solutions for Dirac Eq. Multiply on the left with There are again two sets of solutions for each 3 momentum p (one for +E and one for –E ) Negative Energy again! First order equation doesn’t escape it!

  18. How about u?

  19. We need: You may think these are two conditions, but no. Multiply the first by So one of the above is not independent if

  20. We need: or How many solutions for every p? Go to the rest frame!

  21. Go to the rest frame! uA is arbitrary u has two solutions corresponding to spin up and spin down in the rest frame.

  22. uB is arbitrary Two solution (spin down and spin up antiparticle)

  23. There are four solutions for each 3 momentum p (two for particle and two for antiparticle) or It’s not hard to find four independent solutions. - We got two positive and two negative energy solutions! Negative energy is still here! In fact, they are antiparticles.

  24. Electron solutions:

  25. Positron solutions: u to v

  26. Expansion of a solution by plane wave solutions for KG Eq. Expansion of a solution by plane wave solutions for Dirac Eq.

  27. 4-columns 4-rows

  28. How about internal lines? Find the Green Function of Dirac Eq. Now the Green Function G is a 4 ˣ 4 matrix

  29. Using the Fourier Transformation Fermion Propagator

  30. 4 ˣ 4 matrices

  31. Bilinear Covariants Ψ transforms under Lorentz Transformation: Interaction vertices must be Lorentz invariant.

  32. W W -ig -ig νe e νμ μ The weak vertices of leptons coupling with W

  33. Bilinear Covariants Ψ transforms under Lorentz Transformation: Interaction vertices must be Lorentz invariant. How do we build invariants fromtwoΨ’s ? A first guess:

  34. Maybe you need to change some of the signs: It turns out to be right! We can define a new adjoint spinor: is invariant!

  35. In fact all bilinears can be classified according to their behavior under Lorentz Transformation:

  36. Feynman Rules for external lines

  37. Photons: It’s easier using potentials: forms a four vector.

  38. 4-vector again Charge conservation

  39. Now the deep part: E and B are observable, but A’s are not! A can be changed by a gauge transformation without changing E and B the observable: So we can use this freedom to choose a gauge, a condition for A:

  40. For free photons: Almost like 4 KG Eq. Energy-Momentum Relation

  41. Polarization needs to satisfy Lorentz Condition: Lorentz Condition does not kill all the freedom: We can further choose then Coulomb Guage The photon is transversely polarized. For p in the z direction: For every p that satisfy there are two solutions! Massless spin 1 particle has two degrees of freedom.

  42. Feynman Rules for external photon lines

  43. Gauge Invariance Classically, E and B are observable, but A’s are not! A can be changed by a gauge transformation without changing E and B the observable: Transformation parameter λ is a function of spacetime. But in Qunatum Mechanics, it is A that appear in wave equation: In a EM field, charged particle couple directly with A.

  44. Classically it’s force that affects particles. EM force is written in E, B. But in Hamiltonian formalism, H is written in terms of A. Quantum Mechanics or wave equation is written by quantizing the Hamiltonian formalism: Is there still gauge invariance?

  45. B does not exist outside.

  46. Gauge invariance in Quantum Mechanics: In QM, there is an additional Phase factor invariance: It is quite a surprise this phase invariance is linked to EM gauge invariance when the phase is time dependent. This space-time dependent phase transformation is not an invariance of QM unless it’s coupled with EM gauge transformation!

  47. Derivatives of wave function doesn’t transform like wave function itself. Wave Equation is not invariant! But if we put in A and link the two transformations: This “derivative” transforms like wave function.

  48. In space and time components: The wave equation: can be written as It is invariant!

  49. This combination will be called “Gauge Transformation” It’s a localized phase transformation. Write your theory with this “Covariant Derivative”. Your theory would be easily invariant.

  50. There is a duality between E and B. Without charge, Maxwell is invariant under: Maybe there exist magnetic charges: monopole

More Related