Proof of Wick's Theorem for 3 fields

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SUMMARY

The discussion centers on the proof of Wick's Theorem for three fields, specifically addressing the time-ordered product of field operators. The initial equation used is T(φ(x) φ(y)) = :φ(x) φ(y): + G(x-y), where G(x-y) represents the commutation relation between the creation and annihilation operators. The user identifies a common mistake in their understanding of normal ordering, realizing that all creation operators must be on the left and all annihilation operators on the right to satisfy the theorem's requirements. This clarification is crucial for correctly applying Wick's Theorem in quantum field theory.

PREREQUISITES
  • Understanding of quantum field theory concepts, particularly field operators.
  • Familiarity with the notation and properties of creation and annihilation operators.
  • Knowledge of time-ordering and normal ordering of operators.
  • Basic grasp of commutation relations in quantum mechanics.
NEXT STEPS
  • Study the formal proof of Wick's Theorem for multiple fields.
  • Learn about the implications of normal ordering in quantum field theory.
  • Explore examples of time-ordered products in quantum field calculations.
  • Investigate the role of propagators in quantum field theory and their derivations.
USEFUL FOR

Students and researchers in quantum field theory, particularly those working with time-ordered products and Wick's Theorem, will benefit from this discussion. It is also valuable for anyone looking to deepen their understanding of operator algebra in quantum mechanics.

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Homework Statement



Question attached:

wick3question.png

Homework Equations


[/B]
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.
 
ok so I've figured out what i was doign wrong, and it's a pretty quick spot, and I assume perhaps a common sort of mistake , so I'm surprised no one replied but hey..

basically my definition of 'normal-ordered' was as soon as you had any permutation of the operators that would cause either the bra or ket of the vacuum to vanish ,so either to a creation ladder on the lhs or a annihilating on the rhs , thee job was done. when instead you needed ALL creation operators on the left and all annihilating operators on the rhs. I'm actually not to sure why this is, the use of wicks theorem I've seen is when a bunch of ladder operators are sandwhiched between the vacuum bra and ket, and so this would suffice to cause it to vanish. However if I think about a proof by induction, adding more fields to it, it makes sense that you'd want it 'fully normal-ordered'...

ta
 

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