- #1
Spinnor
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Say we consider the time independent Klein–Gordon equation, see:
http://en.wikipedia.org/wiki/Klein–Gordon_equation
Lets impose the following boundary conditions, the function is zero at infinity and on some small ball of radius R centered on some origin the function is some complex number C. Assume we have a solution to the time independent Klein–Gordon equation such that psi(R) = C and psi(r=infinity) = 0.
Clearly a global phase change of psi(r) by exp[i*theta],
psi(r) --> psi(r)*exp[i*theta]
does not change the energy in the field psi(r) provided,
C --> C*exp[i*theta]
Is it easy to show that a local phase change which depends only on r,
psi(r) --> psi(r)*exp[i*theta(r)]
will increase the energy of the field? Assume,
C --> C*exp[i*theta(R)]
Thanks for any help!
http://en.wikipedia.org/wiki/Klein–Gordon_equation
Lets impose the following boundary conditions, the function is zero at infinity and on some small ball of radius R centered on some origin the function is some complex number C. Assume we have a solution to the time independent Klein–Gordon equation such that psi(R) = C and psi(r=infinity) = 0.
Clearly a global phase change of psi(r) by exp[i*theta],
psi(r) --> psi(r)*exp[i*theta]
does not change the energy in the field psi(r) provided,
C --> C*exp[i*theta]
Is it easy to show that a local phase change which depends only on r,
psi(r) --> psi(r)*exp[i*theta(r)]
will increase the energy of the field? Assume,
C --> C*exp[i*theta(R)]
Thanks for any help!