# Klein-Gordon equation and factorization

1. Apr 21, 2012

### parton

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

Furthermore, we assume the high relativistic limes where $$k = \vert \vec{k} \vert \gg m$$

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

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

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

$$\partial_{t}^{2} + \vec{k}^{2} = (i \partial_{t} + k) (-i \partial_{t} + k)$$

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

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

I hope somebody could help me.

2. Apr 21, 2012

### chrispb

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. Apr 21, 2012

### parton

I think you mean the d'Alembert operator (the Klein-Gordan operator applied on a free field should return 0).

The equation
$$m^{2} \psi(t,x) = 0$$
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
$$(-\partial_{t} + k) \psi(t,\vec{x}) = 0$$
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
$$\left \lbrace (-\partial_{t} + k) (2 k) + m^{2} \right \rbrace \psi(t,\vec{x}) = 0$$
should be considered as some kind of heuristic reasoning to find this "Schroedinger"-equation.
If we write it in the form
$$\partial_{t} \psi(t, \vec{x}) = \left( k + \dfrac{m^{2}}{2k} \right) \psi(t,\vec{x})$$
we see that the RHS is just the expansion of the energy in the relativistic limit:
$$E = \sqrt{k^{2} + m^{2}} \simeq k + \dfrac{m^{2}}{2k}$$
But this is confusing me, because above I used another approximation:
$$E \simeq k$$
So if originally E ~ k is used and therefore, higher order terms
$$\mathcal{O} \left(\dfrac{m^{2}}{2k} \right)$$
are neglected why are this higher order terms not neglected in the "Schroedinger" equation too?

4. Apr 21, 2012

Code (Text):

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
$$(-i\partial_{t} + k) = (k - E)$$
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
$$E - k \simeq \dfrac{m^{2}}{2k}$$
then the equation becomes
$$\left \lbrace (i\partial_{t} + k) (\dfrac{-m^{2}}{2k}) + m^{2} \right \rbrace \psi(t,\vec{x}) = 0$$
which (unless I've got something wrong here) reduces to
$$(-i\partial_{t} + k)\ \psi(t,\vec{x}) = 0$$
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. Apr 22, 2012

### parton

Yes, you are right.

I am just confused by the decomposition
$$\partial_{t}^{2} + \vec{k}^{2} = (-i \partial_{t} + k) (i \partial_{t} + k)$$
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
$$\partial_{t}^{2} + \vec{k}^{2} = (-i \partial_{t} + k) (i \partial_{t} + k)$$
and not
$$\partial_{t}^{2} + \vec{k}^{2} = (i \partial_{t} + k) (-i \partial_{t} + k)$$
?

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