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Klein-Gordon equation and factorization

  1. Apr 21, 2012 #1

    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.
  2. jcsd
  3. Apr 21, 2012 #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.
  4. Apr 21, 2012 #3
    Thanks for your answer.
    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?
  5. Apr 21, 2012 #4
    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
    [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?)
  6. Apr 22, 2012 #5
    Yes, you are right.

    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]
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