- #1

kof9595995

- 679

- 2

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter kof9595995
- Start date

- #1

kof9595995

- 679

- 2

- #2

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,875

- 420

- #3

kof9595995

- 679

- 2

So the saying "K-G equation describes spinless particles" is not accurate?

- #4

- 2,335

- 313

[EDIT] Actually saying "The K-G equation describes spinless particles" is totally accurate, it just isn't complete. Being described by the K-G equation doesn't imply anything about spin.

- #5

- 13,242

- 1,026

Historically, in 1926 about 6 people found the same equation which described a free relativistic particle with mass 'm'. At that time spin of the particle was not an issue. Even as we speak, no matter the spin, the equation has the same structural form: d'Alembertian, mass squared and <wavefunction>.

Later, when the theory of representations of the restricted Lorentz and Poincare groups was built, some spinor indices came on the <wavefunction> [itex] \Psi [/itex] to account for the spin, but both the d'Alembertian and the mass squared remained there.

So the equation has remained since 1926 essentially the same.

P.S. As is assumed in field theory and relativistic QM [itex] c=1 [/itex] and [itex] \hbar =1 [/itex].

Later, when the theory of representations of the restricted Lorentz and Poincare groups was built, some spinor indices came on the <wavefunction> [itex] \Psi [/itex] to account for the spin, but both the d'Alembertian and the mass squared remained there.

So the equation has remained since 1926 essentially the same.

P.S. As is assumed in field theory and relativistic QM [itex] c=1 [/itex] and [itex] \hbar =1 [/itex].

Last edited:

- #6

kof9595995

- 679

- 2

- #7

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,875

- 420

Also, I think the Hilbert space of the free theory can be taken to be the set of positive-frequency solutions of the classical field equation. (I probably just left out a bunch of technical details, because I don't have time to think about it now). There's an unpublished set of notes by Robert Geroch titled "Special topics in particle physics" that has some of the details. This is a direct link to the pdf.

- #8

- 13,242

- 1,026

The field equation fully characterizes the field. It encompasses both the spin and the mass, which are the 2 Casimir invariants of the Poincare algebra.

The field equation's solutions provide a start of the so-called <canonical quantization> approach. Likewise, the field equations give us the propagator used in the path-integral approach.

- #9

kof9595995

- 679

- 2

- #10

- 12,943

- 5,259

The correct statement is: The ONE-COMPONENT KG equation describes spinless particles. Higher-spin wave functions have more than one component, even if each of them satisfies the KG equation.

- #11

kof9595995

- 679

- 2

Yeah I agree, but now it's a bit fuzzy to me "what's the role of field equation in QFT"

- #12

RedX

- 970

- 3

For example, using the Klein-Gordan Hamiltonian for your partition function, you can derive say the pressure of massless spinless particles, and you'll get that it is half that of the pressure from blackbody radiation. This is because massless spin 1 particles have two degrees of freedom due to spin, while spin 0 particles only have 1.

- #13

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,875

- 420

"The field equation" is what you get from the Lagrangian. For the Dirac field, that's the Dirac equation. The field satisfies the field equation. Its components satisfy the Klein-Gordon equation, but since the field components aren't scalar fields, the method I described before won't give you a spin-0 representation.Yeah I agree, but now it's a bit fuzzy to me "what's the role of field equation in QFT"

- #14

kof9595995

- 679

- 2

Hmm, if that's all, it seems field equations do not really play a big role in QFT

- #15

RedX

- 970

- 3

Hmm, if that's all, it seems field equations do not really play a big role in QFT

Well you solve the field equation to find the Green's function (i.e., free-field propagator) by setting the RHS equal to a delta function.

These form the stems of your Feynman diagrams, and in principle knowledge of these stems, along with an interaction term, gives you scattering. The trick is that using perturbation theory along with Wick's theorem reduces solving interactions into solving for the free-field propagator and multiplying a bunch of them and adding them together.

What's amazing about these Green's functions for the Klein-Gordan propagator is that they propagate backwards and forwards in time, because no matter which direction you close the contour integral (dictated by the sign of t-t' as required by Jordan's lemma), there'll be a pole, so you get:

[tex]

\Delta(x-x')=i\theta(t-t') \int \frac{d^3k}{2(2\pi)^3E_k}e^{ik(x-x')}

+i\theta(t'-t) \int \frac{d^3k}{2(2\pi)^3E_k}e^{-ik(x-x')}

[/tex]

And this shows up in the field equation, and is not just something that comes from QFT.

- #16

- 2,335

- 313

The field equation

In QFT (and QM in general) we can resolve the dynamics in terms of free propagation and interactions. Free propagation is dictated by the appropriate wave equation and interactions by the gauge field coupling. These are succinctly expressed respectively by Feynman diagram lines and vertices.

- #17

kof9595995

- 679

- 2

- #18

- 12,943

- 5,259

They do, in the Heisenberg picture.Hmm, if that's all, it seems field equations do not really play a big role in QFT

Share:

- Last Post

- Replies
- 12

- Views
- 588

- Replies
- 1

- Views
- 240

- Last Post

- Replies
- 3

- Views
- 294

- Last Post

- Replies
- 21

- Views
- 411

- Last Post

- Replies
- 3

- Views
- 363

- Replies
- 45

- Views
- 2K

- Replies
- 3

- Views
- 261

- Last Post

- Replies
- 1

- Views
- 238

- Replies
- 5

- Views
- 666

- Last Post

- Replies
- 3

- Views
- 539