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In the usual Fourier expansion of Schrodinger fields \begin{align} \Psi(\vec{x}) = \frac{1}{(2\pi)^{\frac{3}{2}}} \int d^3 k \hat{a}_k e^{-i (wt-\vec{k}\cdot \vec{x})}, \quad \Psi^{*}(\vec{x}) = \frac{1}{(2\pi)^{\frac{3}{2}}} \int d^3 k \hat{a}^{\dagger}_k e^{i (wt-\vec{k}\cdot \vec{x})} \end{align} while Foruier expansion of Klein-Gordon fields are followings

\begin{align} \Psi(\vec{x}) = \int \frac{d^3 k}{(2\pi)^{3}2w_k} \left[\hat{a}_k e^{-i (wt-\vec{k}\cdot \vec{x})}+\hat{a}^{\dagger}_k e^{i(wt-\vec{k}\cdot \vec{x})}\right] \end{align}

Here i wonder why the expansion of schrodinger field and klein gordon fields are different. First in schrodinger field $\Psi$ is wrttien in one variable($a$), while klein gordon field is written in two variables. ($a, a^{\dagger}$). Second the factor of $2w_k$ comes out. I guess this comes from the Lorentz invariance of measure $\frac{d^3k}{(2\pi)^3 2w_k}$. hmm, in this sense, i guess Lorentz invariance is key difference between two theories.

Can anyone gives some reliable explanation about the difference between Schrodinger fields and Klein-Gordon fields?

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2 Answers 2

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Klein-Gordon equation describes massive relativistic particles (scalar boson) which satisfy the equation: $$E_p^2 = p^2 + m^2$$ Thus the energy (or equivalently $\omega$) has 2 solutions for a given momentum $E_p = \pm \sqrt{p^2 + m^2}$. Therefore, a general solution is necessarily a linear superposition of the solution with $E>0$ and the other with $E<0$ justifying the 2 terms in the Fourier development. The normalization with $\frac{1}{2\omega_k}$ is the relativistic normalization valid even when you change the frame with a boost (as you suggested in your question).

For Schrodinger, the energy is non-relativistic satisfying $E = \frac{p^2}{2m}$ and thus admitting only 1 solution for a given momentum. The Fourier transform is then simpler.

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  • $\begingroup$ @phy_math: Your expansion is in values $\pm p^\mu = \pm (E, \vec{p})$ and not in $\pm E$. Also, there are complex field solutions of the KGE, for which the usuall Schrödinger-Ansatz is correct $\endgroup$
    – image357
    Commented Apr 3, 2015 at 13:47
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I'd say the second equation is just a way of writing the field as a hermitian field-operator. This is usually not required for Schrödinger-fields, since the field-equations incorperate complex numbers, while the KGE does not. This allows the KG-fields to be "real-valued"/hermitian (though "complex-valued"/non-hermitian fields are in princpile fine, too).

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