Why do Dirac spinors obey the Klein-Gordon equation?

  • I
  • Thread starter carllacan
  • Start date
  • #1
274
3
The solutions to the Dirac equation are also solutions of the Klein-Gordon equation, which is the equation of motion for the real scalar field. I can see that the converse is not true, but why do spinors follow the equation for real-field particles? Is there any physical meaning to it?
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
17,077
8,180
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
$$(\Box+m^2) \psi=(\partial_{\mu} \partial^{\mu} + m^2)\psi=0.$$
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
26,437
9,958
If you square the Dirac Equation, you get the Klein-Gordon equation. So any solution of the Dirac Equation is a solution of the Klein-Gordon equation.
 
  • #4
274
3
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.

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
7,414
1,054
Moving beyond single-particle states, quantum fields are operators on mutli-particle states (Fock spaces), and quantum fields for spin 0 and spin 1/2 satisfy different commutation/anti-commutation relations.
 
  • #6
1,283
345
I might also point out the components of E and B fields of electromagnetism obey a wave equation ##(m = 0)##. The reasons are relativity and the fact that space time is homogeneous and isotropic.
 
  • #7
vanhees71
Science Advisor
Insights Author
Gold Member
17,077
8,180
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?
I don't see, what's weird about it. What don't you understand about my derivation?

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
Demystifier
Science Advisor
Insights Author
Gold Member
11,360
3,975
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.
There is nothing strange about it. Consider some quantity ##f(t)## satisfying first-order differential equation
$$\frac{df}{dt}=0$$
Obviously, it follows that this quantity also satisfies the second order differential equation
$$\frac{d^2f}{dt^2}=0$$
 
  • Like
Likes vanhees71
  • #9
274
3
No, I don't see anything wrong or weird with the math. It is just that I found it curious that the Dirac field obeyed the equation for the scalar real field, and I though there might be some interesting physics behind the fact, like fermions and scalar real particles having some common property, or fermions being scalar real field particles with an additional characteristic, or somethin.

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.
 

Related Threads on Why do Dirac spinors obey the Klein-Gordon equation?

Replies
14
Views
5K
Replies
4
Views
2K
Replies
3
Views
1K
Replies
5
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
6K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
3K
Top