- #1

- 274

- 3

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.

- I
- Thread starter carllacan
- Start date

- #1

- 274

- 3

- #2

- 20,102

- 10,840

$$(\Box+m^2) \psi=(\partial_{\mu} \partial^{\mu} + m^2)\psi=0.$$

That's true for the Dirac equation, because you have

$$(\mathrm{i} \not{\partial}-m) \psi=0.$$

This implies of course

$$(\mathrm{i} \not{\partial}+m)(\mathrm{i} \not{\partial}-m) \psi=0$$

Now multiply out the operator. You get

$$(-\not{\partial}^2-m^2)\psi = 0 \qquad (*),$$

but now

$$\not{\partial}^2=\gamma^{\mu} \gamma^{\nu} \partial_{\mu} \partial_{\nu}.$$

Since the partial derivatives commute this gives from the Dirac-matrices' anti-commutation relations (Clifford algebra of Minkowski space!)

$$\not{\partial}^2= \frac{1}{2} [\gamma^{\mu},\gamma^{\nu}]_+ \partial_{\mu} \partial_{\nu} = g^{\mu \nu} \partial_{\mu} \partial_{\nu} = \Box,$$

and thus (*) is just the proof that the Dirac field describes really particles with mass ##m##.

- #3

Vanadium 50

Staff Emeritus

Science Advisor

Education Advisor

2021 Award

- 28,039

- 12,565

- #4

- 274

- 3

Sure, any free-particle mode for a particle with mass ##m## must obey the "on-shell condition" ##p^2=m^2## (in units where \hbar=c=1). Thus the fields must obey the free KG equation.

Does it mean that all free field operators obey the KG on top of their own equation of motion? Is there any significance to the real scalar field only obeying KG?

- #5

George Jones

Staff Emeritus

Science Advisor

Gold Member

- 7,538

- 1,352

- #6

Paul Colby

Gold Member

- 1,405

- 408

- #7

- 20,102

- 10,840

I don't see, what's weird about it. What don't you understand about my derivation?Thank you both. I already see how to go from the Dirac equation to the KG. My question was: how is it possible that spinors, which are the particles of the Dirac field, follow the equation of motion of the real scalar field particles. It just weirds me out.

Does it mean that all free field operators obey the KG on top of their own equation of motion? Is there any significance to the real scalar field only obeying KG?

The key to the full understanding, why relativistic wave equations look the way they look is the representation theory of the Poincare group. See, e.g., my QFT lecture notes:

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

- #8

- 12,569

- 4,931

There is nothing strange about it. Consider some quantity ##f(t)## satisfying first-order differential equationThank you both. I already see how to go from the Dirac equation to the KG. My question was: how is it possible that spinors, which are the particles of the Dirac field, follow the equation of motion of the real scalar field particles. It just weirds me out.

$$\frac{df}{dt}=0$$

Obviously, it follows that this quantity also satisfies the second order differential equation

$$\frac{d^2f}{dt^2}=0$$

- #9

- 274

- 3

But I guess the fact that none of you understood what I meant means that there's actually nothing interesting about it after all.

Thanks for your answers.

Share: