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For the complex Klein-Gordon Lagrangian density in the non-relativistic limit, we can decompose the complex scalar field into the form $$\phi=\frac{1}{\sqrt{2m}}e^{-imt}\psi.$$ When substituting the explicit form of $\phi$ into the lagrangian density, you end up with this term:

$$L=\frac{i}{2}\psi^*\dot{\psi}- \frac{i}{2}\dot{\psi^*}\psi +\frac{1}{2m}\dot{\psi^*}\dot{\psi} -\frac{1}{2m}\partial_i\psi^*\partial^i\psi -\frac{i}{2}\psi^*\partial^i\psi -\frac{1}{2m}\dot{\psi^*}\partial^i\psi +\frac{1}{2}\partial_i\psi^*\psi -\frac{1}{2m}\partial_i\psi^*\dot{\psi}.$$

The dot means the time derivative. In the solutions, this was simplified so that only the last four terms remained. What I do not understand is how exactly the first four terms were cancelled out? Any tips on how to do this would be greatly appreciated! I understand this question has probably been asked before, but this exact specific issue is really confusing me.

Qmechanic
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Chris G
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You need to keep the $c^2$'s --- particularly in $e^{-imc^2 t}$. Let's set $\hbar=1$ for convenience, and start from the KG equation: $$ \frac 1{c^2} \frac {\partial^2 \phi}{\partial t^2}- \frac {\partial^2 \phi}{\partial x^2}+m^2c^2 \phi=0. $$ Write $$ \phi(x,t) = \psi(x,t) e^{-imc^2t} $$ where $\psi$ is slowly varying in $t$ compared to the $e^{-imc^2t}$ factor. Then you get $$ \left(- m^2c^2 \psi - 2i m \dot \psi + \frac 1 {c^2}\ddot \psi- \frac {\partial^2 \psi}{\partial x^2}+m^2c^2 \psi\right)e^{-imc^2 t}=0. $$
Which is equivalent to $$ i\dot \psi - \frac 1 {2m c^2}\ddot \psi= -\frac 1 {2m} \frac{\partial^2 \psi}{\partial x^2}. $$ The non-relativistic limit ignores the $(1/2m c^2)\ddot \psi$ as being very small compared to the other terms.

mike stone
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