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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.

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# Klein-Gordon equation and factorization

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