Klein-Gordon Equation

  • Thread starter ehrenfest
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  • #1
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Homework Statement


I do not understand how the Klein-Gordin equation can hold when you have a del operator on one side and a partiall derivative on the other. Doesn't the del operator give a vector and the partial derivative operator yield a scalar?





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The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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Depends on what 'del' you are talking about. The operator in the KG equation is a laplacian - which yeilds a scalar.
 
  • #3
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How would you show that [tex] \psi(r,t) = exp\left(i(k*r - \omega t \right)[/tex] is a solution to the K-G equation?

What do you get to cancel with the m in the the mu term?

Even after I apply the derivative operators and use the Schrodinger equation, I get an m one side and an m^2 on the other from the mu^2 term:

[tex]2 m i \omega \psi = \frac{-\omega^2}{c^2} \psi + \mu^2 \psi [/tex]
 
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  • #4
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Can you show your KG equation here?
 
  • #5
mjsd
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How would you show that [tex] \psi(r,t) = exp\left(i(k*r - \omega t \right)[/tex] is a solution to the K-G equation?

What do you get to cancel with the m in the the mu term?

Even after I apply the derivative operators and use the Schrodinger equation, I get an m one side and an m^2 on the other from the mu^2 term:

[tex]2 m i \omega \psi = \frac{-\omega^2}{c^2} \psi + \mu^2 \psi [/tex]
when working out the d'Alembertian, note that this is now in polar coordinates..so your del^2 should be written in the appropriate coord system too... not sure whether this is the cause of the problem but without seeing more of your workings, it is hard to pin point
 
  • #6
olgranpappy
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How would you show that [tex] \psi(r,t) = exp\left(i(k*r - \omega t \right)[/tex] is a solution to the K-G equation?
I would take two derivatives w.r.t. time.
And two derivative w.r.t. space
and subtract them.
This will be equal to:
[tex]
(\omega^2-k^2)\psi
[/tex]

And then I would use the fact that (presumably, since there's not much in your post to go on):
[tex]
\omega^2=k^2+m^2
[/tex]

What do you get to cancel with the m in the the mu term?
what's mu? what's m? you need to supply more information.

Even after I apply the derivative operators and use the Schrodinger equation...
why are we now introducing the Schrodinder equation? You're solving the KG equation...
 
  • #7
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I am using the following version of the Klein-Gordon equation:

[tex](\Box^2 + \mu^2) \psi = 0,[/tex]

where

[tex]\mu = \frac{mc}{\hbar} \,[/tex]

and

[tex]\Box^2 = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2.\, [/tex]

After I take the two time derivatives and the Laplacian, I get

[tex]w^2/c^2-k^2- \frac{m^2*c^2}{ \hbar^2} = 0[/tex]

I do not understand how the RHS simplifies
 
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  • #8
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Psi is only a solution for certain [tex]\omega(k)[/tex]. As a matter of fact, [tex]\omega(k)[/tex] is exactly the energy-momentum relation relativity demands for a free particle (E² - p² = m², c=1).

The RHS=0 is as simple as it gets :biggrin:

EDIT: Btw, you can use \cdot for a multiplication dot in tex.
 
  • #9
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Psi is only a solution for certain [tex]\omega(k)[/tex]. As a matter of fact, [tex]\omega(k)[/tex] is exactly the energy-momentum relation relativity demands for a free particle (E² - p² = m², c=1).

The RHS=0 is as simple as it gets :biggrin:

EDIT: Btw, you can use \cdot for a multiplication dot in tex.
Firstly, how can we just let c = 1? c should be about 300,000,000 m/s.

Second, where did the relation omega^2 = k^2 + m^2 come from?

Third, we are still left with an h-bar^2, even if we use that relation, right?
 
  • #10
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Firstly, how can we just let c = 1? c should be about 300,000,000 m/s.
300,000,000 m/s = 1 light year/year. It's just a unit conversion.
 
  • #11
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Yes, you can see chosing c=1 as a unit conversion. It is a common choice in relativistic physics. hbar=1 would also be a common choice in particle physics, but that would not be needed for my post since I didn't write any equation that involves an hbar. You're probably missing that [tex] E = \hbar \omega, \ \vec p = \hbar \vec k [/tex].
 
  • #12
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1) The choose of c=1, implies that your measures ( equipaments) is calibrated in another system of coordinates, not meter, second... You can make this change if you keep in mind that the dimension of your quantities was change.

2) This relation come from the relation of energy and momentum of relativity theory( c=1), is the same relation used to demonstrated the KG equation to a free particle.

3) Remember that energy and momentum (expectation values) is hbar*omega and hbar*k ( k is a vector) , so with this you have E^2=p^2+m^2.
 
  • #13
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1
Yes, you can see chosing c=1 as a unit conversion. It is a common choice in relativistic physics. hbar=1 would also be a common choice in particle physics, but that would not be needed for my post since I didn't write any equation that involves an hbar. You're probably missing that [tex] E = \hbar \omega, \ \vec p = \hbar \vec k [/tex].
Everything makes sense now, except I thought that those last two equations were only true for photons. How did you know only photons were being considered in the problem?
 
  • #14
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You can make an association with mass and wave lenght, or wave lenght and momentum. Holds true for photons ( carrier momentum,through radiation, but dont have mass) and to particles with mass.
 
  • #15
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36
Everything makes sense now, except I thought that those last two equations were only true for photons. How did you know only photons were being considered in the problem?
[tex] E = \hbar \omega, \vec p = \hbar \vec k, E^2 - (pc)^2 = \left( mc^2 \right)^2 [/tex] are true for all relativistic free particles with definite momentum (plane waves in quantum theories, classical particles in classical relativity). You just get/have different values m for different particle types.
 

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