- #1
binbagsss
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- 11
Homework Statement
Question attached:
Homework Equations
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Using the result from two fields that
## T(\phi(x) \phi(y))= : \phi(x) \phi(y) : + G(x-y)##
Where ##G(x-y) = [\phi(x)^+,\phi(y)^-] ##
## : ## denotes normal ordered
and ##\phi(x)^+ ## is the annihilation operator part , and ## \phi(x)^- ## is the creation operator part.
The Attempt at a Solution
Assume non-trivially that ## z^0 > x^0 > y^0 ##
Then ## T(\phi(z),\phi(x),\phi(y)) = \phi(z) T(\phi(x) \phi(y)) ##
##=(\phi(z)^+ + \phi(z)^-) T (\phi(x),\phi(y)) ##
Since ##\phi(z)^-## is already normal ordered, look at the term multiplied by ##\phi(z)^+##:
##=\phi(z)G(x-y) + \phi(z)^+:\phi(x)\phi(y): ## (1)
The term to be concerned with from
##\phi(z)^+:\phi(x)\phi(y):## is ##\phi(z)^+\phi(x)^-\phi(y)^-=\phi(x)^-\phi(z)^+\phi(y)^- +[\phi(z)^+,\phi(x)^-]\phi(y)^-= \phi(x)^-\phi(y)^-\phi(z)^+ +\phi(x)^-[\phi(z)^+,\phi(y)^-] + [ \phi(z)^+,\phi(x)^-]\phi(y)^-##
So putting this with (1) I have
## T(\phi(z),\phi(x),\phi(y)) = : \phi(z) (\phi(x) \phi(y)): + [ \phi(z)^+,\phi(x)^-]\phi(y)^- +\phi(x)^-[\phi(z)^+,\phi(y)^-] +\phi(z)(G(x-y)) ##
So looking at the solution the last term is right, but the other propagator terms , should have a factor of both the creation and annihilation parts of the field, ##\phi(y)^- + \phi(y)^+ ## and ## \phi(x)^+ + \phi(x)^- ## , multiplying the propagator? and should be multiplying the RHS of the propagator rather than the LHS ? I'm not sure what I have done wrong...
Many thanks in advance.