Lancaster&Blundell "QFT Gifted Amateur" Wick theorem on Fermion Ground State

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In Lancaster & Blundell's "QFT for the Gifted Amateur," Chapter 18 introduces Wick's theorem, which typically applies to time-dependent interaction picture operators. However, the exercises require applying Wick's theorem to time-independent operators, creating ambiguity, particularly in problem 18.5, where the ground state is confused with the full vacuum state. The discussion highlights the challenge of using Wick's theorem with normal-ordered operators, as the resulting calculations seem inconsistent due to the non-zero nature of these operators in the ground state. A proposed solution involves reinterpreting fermion operators into particle-antiparticle pairs, which aligns better with the ground state functioning as a true vacuum. Ultimately, clarity on the application of Wick's theorem in this context is essential for accurate results.
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Homework Statement
Use Wick's theorem to simplify [this fermionic operation]
Relevant Equations
##\langle 0|c^\dagger_{p_1-q}c^\dagger_{p_2+q}c_{p_2}c_{p_1}|0\rangle##
In Lancaster&Blundell QFT for the Gifted Amateur, Chaper 18, the authors introduce Wick's theorem. I have already seen it Fetter&Walecka and in here, but my problem with the theorem is that it is usually announced for interaction picture operators with time-dependences. Nevertheless in the exercises they ask us to use Wick's theorem to rewrite different chains of operators that are not time-dependent. This problem (18.5) struck me specfically because it is completely ambiguous, not only one has to understand that ##|0\rangle## is the ground state (GS) and not the full vacuum, but also how does Wick theorem work here? I mean the operators are already normal ordered...

From previous problems I figured out how to use the Wick theorem for time independent operators. One can use that version of the theorem naively and say that
$$\langle 0|c^\dagger_{p_1-q}c^\dagger_{p_2+q}c_{p_2}c_{p_1}|0\rangle=\text{all contracted pairs}$$
Which provides the right result but it is fishy because this operation contains normal orderings like the original one ##c^\dagger_{p_1-q}c^\dagger_{p_2+q}c_{p_2}c_{p_1}## that are not zero by definition of GS. So why does this work? Does the normal ordering has to be redefined here?

A more natural way I have seen this before is to divide the fermion operators into particle-antiparticle operators so that the ground state works like a true vacuum but it seems that there is no need under this more-seemingly-naive use of Wick's theorem.
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...

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