The Klein-Gordon equation has the Schrodinger equation as a nonrelativistic limit, in the following sense:(adsbygoogle = window.adsbygoogle || []).push({});

Start with the Klein-Gordon equation (for a complex function ##\phi##)

## \partial_\mu \partial^\mu \phi + m^2 \phi = 0##

Now, define a new function ##\psi## via: ##\psi = e^{i m t} \phi##. Then the equation for ##\psi## is:

## \ddot{\psi} -2 i m \dot{\psi} -\nabla^2 \psi = 0##

Now, if we assume that ##\ddot{\psi}## is small compared with the other terms, then we have approximately:

## -2 i m \dot{\psi} -\nabla^2 \psi = 0 \Rightarrow i \dot{\psi} = - \frac{1}{2m} \nabla^2 \psi##

That's Schrodinger's equation. So that seems to work. But now, instead of looking at equations of motion, let's look at the field theory.

The Schrodinger equation follows from a field-theoretic lagrangian density:

##\mathcal{L} = - i (\psi^* \dot{\psi} - \dot{\psi^*} \psi) - \frac{1}{2m} (\nabla \psi^*) \cdot (\nabla \psi)##

which corresponds to the hamilton density:

##\mathcal{H} = \frac{1}{2m} (\nabla \psi^*) \cdot (\nabla \psi)##

Now, what I would expect is that just as the Klein Gordon equation has the Schrodinger equation as a nonrelativistic limit, the relativistic hamiltonian density should have the appropriate nonrelativistic limit, as well. But it doesn't quite work.

A relativistic lagrangian density that yields the Klein-Gordon equation is:

##\mathcal{L} = \frac{1}{2} \dot{\phi^*} \dot{\phi} - \frac{1}{2} (\nabla \phi^*) \cdot (\nabla \phi) - \frac{m^2}{2} \phi^* \phi##

This corresponds to a Hamiltonian density (I think).

##\mathcal{H} = \frac{1}{2} \dot{\phi^*} \dot{\phi} + \frac{1}{2} (\nabla \phi^*) \cdot (\nabla \phi) + \frac{m^2}{2} \phi^* \phi##

Now, let's try the same trick: Let ##\phi = e^{-imt}\psi##. Then in terms of ##\psi##:

##\mathcal{H} = \frac{1}{2} \dot{\psi^*}\dot{\psi} + \frac{im}{2} (\psi^* \dot{\psi} - \dot{\psi^*} \dot{\psi}) + \frac{1}{2} (\nabla \psi^*) \cdot (\nabla \psi) + m^2 \psi^* \psi##

Assuming once again that the first term is negligible compared to the others gives you:

##\mathcal{H} = \frac{im}{2} (\psi^* \dot{\psi} - \dot{\psi^*} \psi) + \frac{1}{2} (\nabla \psi^*) \cdot (\nabla \psi) + m^2 \psi^* \psi##

That doesn't look anything like the nonrelativistic Hamiltonian density. It doesn't even have the right units (although I guess you could fix that by rescaling ##\psi##).

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# I Nonrelativistic limit of scalar field theory

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