Klein-Gordon equation and factorization

In summary, the author reads a text where a "Schroedinger-equation" for a neutrino field is being derived, but does not understand a particular step. Using a Dirac field, the Klein-Gordon equation is satisfied: \left( \partial_{t}^{2} + \vec{k}^{2} + m^{2} \right) \psi_{\vec{k}}(t) = 0 where the field was expanded in plane waves and the spatial derivatives were already computed. Furthermore, we assume the high relativistic limes where k = \vert \vec{k} \vert \gg m. Linearziation of the
  • #1
parton
83
1
Hi!

I read a text were some kind of "Schroedinger-equation" for a neutrino field is being derived. But there is a particular step I do not understand.

Consider a Dirac field [tex] \psi(t, \vec{x}) [/tex] of a neutrino, satisfying the Klein-Gordon equation:
[tex] \left( \partial_{t}^{2} + \vec{k}^{2} + m^{2} \right) \psi_{\vec{k}}(t) = 0 [/tex]
where the field was expanded in plane waves [tex] \psi(t,\vec{x}) = \psi_{0} e^{i (\vec{k} \cdot \vec{x} - E \cdot t)} [/tex] and the spatial derivatives were already computed.

Furthermore, we assume the high relativistic limes where [tex] k = \vert \vec{k} \vert \gg m [/tex]

Now we can linearzie the equation above using
[tex] \partial_{t}^{2} + \vec{k}^{2} = (-i \partial_{t} + k) (i \partial_{t} + k) [/tex]
Applying only the 2nd factor onto the field [tex] \psi(t,\vec{x}) [/tex]
will lead to
[tex] \left \lbrace (-\partial_{t} + k) (E + k) + m^{2} \right \rbrace \psi(t,\vec{x}) = 0 [/tex]

Now the following approximaton of the energy-momentum relation is used:
[tex] E = \sqrt{\vec{k}^{2} + m^{2}} \simeq k [/tex]
If we use this approximation in the equation above, we have
[tex] \left \lbrace (-\partial_{t} + k) (2 k) + m^{2} \right \rbrace \psi(t,\vec{x}) = 0 [/tex]
which is actually the equation I need.

But my problem is, what happens if I exchange the order in the factorization above:

[tex] \partial_{t}^{2} + \vec{k}^{2} = (i \partial_{t} + k) (-i \partial_{t} + k) [/tex]

Going through the same steps as above I end up with
[tex] \left \lbrace (\partial_{t} + k) (k - k) + m^{2} \right \rbrace \psi(t,\vec{x}) = 0 [/tex]
so finally I have
[tex] m^{2} \psi(t,\vec{x}) = 0 [/tex]
which does not make sense.

So my question is why is it possible to make the linarization
[tex] \partial_{t}^{2} + \vec{k}^{2} = (-i \partial_{t} + k) (i \partial_{t} + k) [/tex]
and why do I have to pay attention to the order of the factors?

I hope somebody could help me.
 
Physics news on Phys.org
  • #2
I didn't read carefully enough to spot where you made the mistake, but I'll tell you the answer; the Klein-Gordan operator acting on a plane wave returns m^2 times the plane wave, even in the relativistic limit. Therefore, no matter what order you do things, you should expect a cancellation between two big numbers (k^2 + m^2) and (-k^2) such that you get the small number m^2.
 
  • #3
Thanks for your answer.
the Klein-Gordan operator acting on a plane wave returns m^2 times the plane wave, even in the relativistic limit.
I think you mean the d'Alembert operator (the Klein-Gordan operator applied on a free field should return 0).

The equation
[tex] m^{2} \psi(t,x) = 0 [/tex]
makes no sense. But I just recognized that if the relativistic limit corresponds to the massless limit, this equation would make sense if we set m=0.

And the other equation (if we exchange the order of the operators) would be
[tex] (-\partial_{t} + k) \psi(t,\vec{x}) = 0 [/tex]
So if the mass is vanishing in the high relativistic limit there is no contradiction anymore and the order of the operators does not matter.

But nevertheless I am confused. As long as I do not neglect the mass the order is important and depending on it, I find two different equations.

Maybe the "derivation" of the equation
[tex] \left \lbrace (-\partial_{t} + k) (2 k) + m^{2} \right \rbrace \psi(t,\vec{x}) = 0 [/tex]
should be considered as some kind of heuristic reasoning to find this "Schroedinger"-equation.
If we write it in the form
[tex] \partial_{t} \psi(t, \vec{x}) = \left( k + \dfrac{m^{2}}{2k} \right) \psi(t,\vec{x}) [/tex]
we see that the RHS is just the expansion of the energy in the relativistic limit:
[tex] E = \sqrt{k^{2} + m^{2}} \simeq k + \dfrac{m^{2}}{2k} [/tex]
But this is confusing me, because above I used another approximation:
[tex] E \simeq k [/tex]
So if originally E ~ k is used and therefore, higher order terms
[tex] \mathcal{O} \left(\dfrac{m^{2}}{2k} \right) [/tex]
are neglected why are this higher order terms not neglected in the "Schroedinger" equation too?
 
  • #4
Code:
Isn't the basic issue here that, whilst it may be reasonable to approximate k ≈ E when calculating the factor (E + k), doing this is much less so when calculating
[tex] (-i\partial_{t} + k) = (k - E) [/tex]
For example, it is often reasonable to say that 1,000,000 + 1,000,001 ≈ 2,000,000, but much less so to suggest that 1,000,001 - 1,000,000 ≈ 0.

If, instead, we use the approximation
[tex] E - k \simeq \dfrac{m^{2}}{2k} [/tex]
then the equation becomes
[tex] \left \lbrace (i\partial_{t} + k) (\dfrac{-m^{2}}{2k}) + m^{2} \right \rbrace \psi(t,\vec{x}) = 0 [/tex]
which (unless I've got something wrong here) reduces to
[tex] (-i\partial_{t} + k)\ \psi(t,\vec{x}) = 0 [/tex]
and doesn't really help here. Actually, I don't really see why the same result should follow regardless of which factor we decide to operate with on the wavefunction first.

(Also, haven't you left out an i that should be before the ∂t in some of your equations?)
 
  • #5
(Also, haven't you left out an i that should be before the ∂t in some of your equations?)

Yes, you are right.

Actually, I don't really see why the same result should follow regardless of which factor we decide to operate with on the wavefunction first.

I am just confused by the decomposition
[tex] \partial_{t}^{2} + \vec{k}^{2} = (-i \partial_{t} + k) (i \partial_{t} + k) [/tex]
If the order is important then how do I know how to decompose the d'Alembert operator in the right way?

How do I know that i should write
[tex] \partial_{t}^{2} + \vec{k}^{2} = (-i \partial_{t} + k) (i \partial_{t} + k) [/tex]
and not
[tex] \partial_{t}^{2} + \vec{k}^{2} = (i \partial_{t} + k) (-i \partial_{t} + k) [/tex]
?

Or should I argue that I am intersted in the "difference" between energy and momentum and therefore, need to consider
[tex] \partial_{t}^{2} + \vec{k}^{2} = (-i \partial_{t} + k) (i \partial_{t} + k) [/tex]
 

1. What is the Klein-Gordon equation?

The Klein-Gordon equation is a relativistic wave equation that describes the behavior of spinless particles, such as mesons. It combines the concepts of special relativity and quantum mechanics, and is used to describe the motion and interactions of these particles.

2. What is the factorization method used for in relation to the Klein-Gordon equation?

The factorization method is a mathematical technique used to solve the Klein-Gordon equation. It involves rewriting the equation in terms of creation and annihilation operators, which simplifies the equation and allows for easier solutions to be found.

3. What are the key features of the Klein-Gordon equation?

The Klein-Gordon equation is a second-order partial differential equation with four independent variables: time and three spatial coordinates. It is Lorentz invariant, meaning that its form remains the same under transformations between inertial reference frames. It also has both positive and negative energy solutions, which led to the prediction of antimatter.

4. How does the Klein-Gordon equation relate to the Schrödinger equation?

The Klein-Gordon equation is a relativistic version of the Schrödinger equation, which describes the behavior of non-relativistic particles. Both equations are used to describe the quantum mechanical behavior of particles, but the Klein-Gordon equation also takes into account special relativity effects, such as time dilation and length contraction.

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

The Klein-Gordon equation has applications in various fields of physics, including particle physics, field theory, and quantum mechanics. It is used to study the behavior of fundamental particles and their interactions, as well as to describe the behavior of waves in quantum mechanical systems. It has also been used in the development of quantum field theories, which are essential for understanding the behavior of particles at the subatomic level.

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
1
Views
989
  • High Energy, Nuclear, Particle Physics
Replies
3
Views
845
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
229
  • High Energy, Nuclear, Particle Physics
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
540
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
1K
  • Classical Physics
Replies
3
Views
1K
Back
Top