User:Phinneganthephysicist/sandbox/wick

Proof of Wick's theorem
We use induction. The $$N=2$$ base case is trivial, because there is only one possible contraction:
 * $$\hat{A}\hat{B} = \mathopen{:}\hat{A}\hat{B}\mathclose{:} + (\hat{A}\,\hat{B}\, - \mathopen{:} \hat{A}\,\hat{B} \mathclose{:}) = \mathopen{:}\hat{A}\hat{B}\mathclose{:} + \mathopen{:}\hat{A}^\bullet\hat{B}^\bullet\mathclose{:}$$

In general, the only non-zero contractions are between an annihilation operator on the left and a creation operator on the right. Suppose that Wick's theorem is true for $$N-1$$ operators $$\hat{B} \hat{C} \hat{D} \hat{E} \hat{F}\ldots$$, and consider the effect of adding an Nth operator $$\hat{A}$$ to the left of $$\hat{B} \hat{C} \hat{D} \hat{E} \hat{F}\ldots$$ to form $$\hat{A}\hat{B}\hat{C}\hat{D}\hat{E} \hat{F}\ldots$$. By Wick's theorem applied to $$N-1$$ operators, we have:



\begin{align} \hat{A} \hat{B} \hat{C} \hat{D} \hat{E} \hat{F}\ldots &= \hat{A} \mathopen{:}\hat{B} \hat{C} \hat{D} \hat{E} \hat{F}\ldots \mathclose{:} \\ &\quad + \hat{A} \sum_\text{singles} \mathopen{:} \hat{B}^\bullet \hat{C}^\bullet \hat{D} \hat{E} \hat{F} \ldots \mathclose{:} \\ &\quad + \hat{A} \sum_\text{doubles} \mathopen{:} \hat{B}^\bullet \hat{C}^{\bullet\bullet} \hat{D}^{\bullet\bullet} \hat{E}^\bullet \hat{F} \ldots \mathclose{:} \\ &\quad + \hat{A} \ldots \end{align} $$

$$\hat{A}$$ is either a creation operator or an annihilation operator. If $$\hat{A}$$ is a creation operator, all above products, such as $$\hat{A}\mathopen{:}\hat{B} \hat{C} \hat{D} \hat{E} \hat{F}\ldots \mathclose{:}$$, are already normal ordered and requires no further manipulation. Because $$\hat{A}$$ is to the left of all annihilation operators in $$\hat{A}\hat{B}\hat{C}\hat{D}\hat{E}\hat{F}\ldots$$, any contraction involving it will be zero. Thus, we can add all contractions involving $$\hat{A}$$ to the sums without changing their value. Therefore, if $$\hat{A}$$ is a creation operator, Wick's theorem holds for $$\hat{A}\hat{B}\hat{C}\hat{D}\hat{E}\hat{F}\ldots$$.

Now, suppose that $$\hat {A}$$ is an annihilation operator. To move $$\hat {A}$$ from the left-hand side to the right-hand side of all the products, we repeatedly swap $$\hat{A}$$ with the operator immediately right of it (call it $$\hat{X}$$), each time applying $$\hat{A}\hat{X} = \mathopen{:}\hat{A}\hat{X}\mathclose{:} + \hat{A}^\bullet\hat{X}^\bullet$$ to account for noncommutativity. Once we do this, all terms will be normal ordered. All the terms added to the sum by pushing $$\hat{A}$$ through the products correspond additional to contractions involving $$\hat{A}$$. Therefore, if $$\hat{A}$$ is an annihilation operator, Wick's theorem holds for $$\hat{A}\hat{B}\hat{C}\hat{D}\hat{E}\hat{F}\ldots$$.

We have proved the base case and the induction step, so the theorem is true.