Klein-Gordon Approximation Question

div curl F= 0 · May 9, 2008

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:

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

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

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

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

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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.

lbrits · May 9, 2008

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

lbrits · May 9, 2008

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

ironhill · May 9, 2008

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

E^2 -> m^2 + 2m[tex]\epsilon[/tex]
2EV -> 2mV
V^2 -> 0

Stick that in and you get it.

lbrits · May 9, 2008

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 =)

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Klein-Gordon Approximation Question

div curl F= 0

lbrits

lbrits

ironhill

lbrits

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