Klein-Gordon Equation: Understanding Scalar & Vector

In summary, the homework statement is trying to solve a problem that depends on what 'del' you are talking about. The operator in the KG equation is a laplacian - which yeilds a scalar. How would you show that \psi(r,t) = exp\left(i(k*r - \omega t \right) is a solution to the K-G equation?
  • #1
ehrenfest
2,020
1

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?





Homework Equations





The Attempt at a Solution

 
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  • #2
Depends on what 'del' you are talking about. The operator in the KG equation is a laplacian - which yeilds a scalar.
 
  • #3
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
Can you show your KG equation here?
 
  • #5
ehrenfest said:
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
ehrenfest said:
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
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
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
Timo said:
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
ehrenfest said:
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
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
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
Timo said:
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
You can make an association with mass and wave lenght, or wave length and momentum. Holds true for photons ( carrier momentum,through radiation, but don't have mass) and to particles with mass.
 
  • #15
ehrenfest said:
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.
 

1. What is the Klein-Gordon equation?

The Klein-Gordon equation is a relativistic wave equation that describes the behavior of a scalar or vector particle. It was first proposed by physicists Walter Gordon and Oskar Klein in the 1920s.

2. What is the significance of the Klein-Gordon equation?

The Klein-Gordon equation is significant because it was the first relativistic wave equation to be proposed and it has been used to describe a variety of particles, including the Higgs boson. It also led to the development of other important equations, such as the Dirac equation.

3. How does the Klein-Gordon equation differ from the Schrödinger equation?

The Schrödinger equation describes the behavior of non-relativistic particles, while the Klein-Gordon equation is a relativistic equation. Additionally, the Klein-Gordon equation is second order in time and space, while the Schrödinger equation is first order in time.

4. What is the physical interpretation of the Klein-Gordon equation?

The Klein-Gordon equation describes the evolution of a particle with spin zero or spin one. It can be interpreted as a wave equation for a particle moving in a field, such as an electromagnetic field. The solutions to the equation represent the probability amplitudes for finding the particle at a given position and time.

5. What are some applications of the Klein-Gordon equation?

The Klein-Gordon equation has been used in a variety of fields, including quantum field theory, particle physics, and cosmology. It has also been applied to study the behavior of particles in strong magnetic fields and in the early universe. Additionally, the Klein-Gordon equation has been used to develop models of bosonic particles, such as the Higgs boson.

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