# Klein-Gordon Approximation Question

1. May 9, 2008

### div curl F= 0

I'd be greatful for a bit of help on this question, can't seem to get the answer to pop out:

A particle moving in a potential V is described by the Klein-Gordon equation:

$$\left[-(E-V)^2 -\nabla^2 + m^2 \right] \psi = 0$$

Consider the limit where the potential is weak and the energy is low:
$$|V| << m \;;\; |\epsilon| << m \;;\; \epsilon = E - m$$

Show that in this limit the KG equation can be approximated by the Schrodinger equation:

$$\left[-\nabla^2 + 2mV \right] \psi = 2m\epsilon \psi$$

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It seems that a taylor expansion is required or other approximation is needed to get the factors of 2 but I've tried several now and can't seem to make the answer.

Any help much appreciated.

2. May 9, 2008

### lbrits

Divide the original equation by $$m^2$$, and Taylor expand. Of course $$E$$ is of order $$m$$, so keep that in mind when you divide by "large" things.

Last edited: May 9, 2008
3. May 9, 2008

### lbrits

Sorry, divide everything through by $$E$$, then $$V/E$$ is small and you can Taylor expand the square and drop the term $$V^2/E^2$$. Now multiply through by E again and divide by m?

4. May 9, 2008

### ironhill

No need for Taylor expansions, just algebra, keep stuff of 1st order of smallness, expand the square;

E^2 -> m^2 + 2m$$\epsilon$$
2EV -> 2mV
V^2 -> 0

Stick that in and you get it.

Last edited: May 9, 2008
5. May 9, 2008

### lbrits

Uhm, heh, about the Taylor expand thing... "If all you have is a hammer then everything looks like a nail", it doesn't bother me that it's already a polynomial =)