# Proof of Wick's theorem

I'm tackling proof of Wick's theorem. By induction.
Let us suppose we have already proved $$C_2 \cdots C_n = N(C_2 \cdots C_n + (\text{all possible contractions}) ) \quad (C_i=a\,\, \text{(annihilation) or }a^{\dagger}\text{(creation)} \,\, ) ,$$ and multiply $C_1$ from the left. Then we move $C_1$ inside $N()$.
Though the real computation manipulates all the terms inside $N()$, let us consider the term with no contractions.
If $C_1=a$,
\begin{alignat}{2} C_1 N(C_2 \cdots C_n) &=&& a (a^{\dagger} \cdots a^{\dagger} a \cdots a) \\ &=&& N(:C_1 C_2:C_3 \cdots C_n \, +\, C_2:C_1 C_3: \cdots C_n\,+ \cdots\cdots\,+\, C_2\cdots C_{n-1} :C_1 C_n: ) \\ &&&+ (a^{\dagger} \cdots a^{\dagger} a \cdots a)a \\ &=&& N( :C_1 C_2:C_3 \cdots C_n\, +\, C_2:C_1 C_3: \cdots C_n \,+\cdots\cdots\,+ C_2\cdots C_{n-1} :C_1 C_n:\,+\,C_1 \cdots C_n ) \end{alignat} there's no problem.
But if $C_1=a^{\dagger}$, \begin{alignat}{2} C_1 N(C_2 \cdots C_n) &=&& a^{\dagger} (a^{\dagger} \cdots a^{\dagger} a \cdots a) \\ &=&& N(:C_1 C_2:C_3 \cdots C_n \,+ \,C_2:C_1 C_3: \cdots C_n \,+\cdots\cdots\,+\, C_2\cdots C_{n-1} :C_1 C_n: ) \\ &&&+ (a^{\dagger} \cdots a^{\dagger} a \cdots a)a^{\dagger} \end{alignat} then the last term of the last line isn't equal to $N( C_1 \cdots C_n)$. Why?
To get better insight, I calculated a example. From $$a_2 a_3^{\dagger} = N(a_2 a_3^{\dagger} + :a_2 a_3^{\dagger}: ) ,$$ (I added subscripts just for comprehensibility. So for instance $a_2^{\dagger}=a_3^{\dagger}$.) I got \begin{alignat}{2} a_1^{\dagger} a_2 a_3^{\dagger} &=&& a_1^{\dagger} a_3^{\dagger}a_2 + a_1^{\dagger}:a_2 a_3^{\dagger}: \\ &=&& N(a_1^{\dagger} a_2 a_3^{\dagger} + a_2 : a_1^{\dagger} a_3^{\dagger}: + a_1^{\dagger}:a_2 a_3^{\dagger}: ) . \end{alignat} But according to Wick's theorem, the relation $$a_1^{\dagger} a_2 a_3^{\dagger} = N(a_1^{\dagger} a_2 a_3^{\dagger} + : a_1^{\dagger} a_2: a_3^{\dagger} + a_2 : a_1^{\dagger} a_3^{\dagger}: + a_1^{\dagger}:a_2 a_3^{\dagger}: )$$ should hold. How does the term $: a_1^{\dagger} a_2: a_3^{\dagger}$ emerge?
Thanks.

• I might be wrong, but AFAIK Wicks theorem is about fields $\phi$ and not about their ladder operators. In other words, you must manipulate $\phi_1\phi_2\phi_3$, not $a_1a_2a_3$, etc. Commented Dec 6, 2016 at 13:30
• @AccidentalFourierTransform, Qmechanic : Sorry. I misunderstood two subjects. $: :$ indicates the normal ordering as well as $N()$. I thought it as contraction and thought contraction as commutation relation. I'm now summarizing and will post as an answer. Commented Dec 7, 2016 at 8:29
• @AccidentalFourierTransform : Wick's theorem already appears in Quantum Mechanics. See here. Commented Dec 7, 2016 at 8:30
• @Qmechanic : In QFT contraction is invariant under changing the order, but in Quantum Mechanics it's not invariant. the definition of contraction is $C_2C_3\equiv <0|C_2C_3|0>$. If $C_2=a,\,C_3=a^{\dagger}$, then $<0|C_2C_3|0>=1,\,<0|C_3C_2|0>=0$. So RHS is not invariant. Commented Dec 7, 2016 at 8:30

Let's consider the case in which there exist $h$ pairs of $d_i, d_i^{\dagger}$. $$d_1, \cdots , d_h ,d_1^{\dagger} \cdots d_h^{\dagger}$$ $$[d_i, d_j^{\dagger}] = \delta_{ij}$$ In this formula and induction, subscripts denote the class of operators. In the example below($a_i$'s), they doesn't. The definition of contraction is $$<0|d_i d_j|0> =0,\, <0|d_i d_j^{\dagger}|0> = \delta_{ij}\, <0|d_i^{\dagger} d_j|0> =0\, <0|d_i^{\dagger} d_j^{\dagger}|0> =0.$$ From $$C_2 \cdots C_n = N(C_2 \cdots C_n + (\text{all possible contractions}) ) \quad (C_i=d_j\,\, \text{(annihilation) or }d_j^{\dagger}\text{(creation)} \,\, ) ,$$ If $C_1=d_i$, using $$d_i d_j^{\dagger} = [d_i, d_j^{\dagger}] + d_j^{\dagger}d_i = <0|d_i d_j^{\dagger}]|0> + d_j^{\dagger}d_i$$ $$d_i d_j = <0|d_i d_j|0> + d_j d_i$$ we move $C_1$ inside $N()$.
\begin{alignat}{2} C_1 N(C_2 \cdots C_n) &=&& d_i (d_j^{\dagger} \cdots d_k^{\dagger} d_l \cdots d_m) \\ &=&& N(<0|C_1 C_2|0>C_3 \cdots C_n \, +\, C_2<0|C_1 C_3|0> \cdots C_n\,+ \cdots\cdots\,+\, C_2\cdots C_{n-1} <0|C_1 C_n|0> ) \\ &&&+ (d_j^{\dagger} \cdots d_k^{\dagger} d_l \cdots d_m)d_i \\ &=&& N( <0|C_1 C_2|0>C_3 \cdots C_n\, +\, C_2<0|C_1 C_3|0> \cdots C_n \,+\cdots\cdots\,+ C_2\cdots C_{n-1} <0|C_1 C_n|0>\,\,\\ &&& +C_1 \cdots C_n ) \end{alignat} The number of $d\,\text{or}\,d^{\dagger}$ may be zero.
If $C_1=d_i^{\dagger}$, by the fact $$<0|d_i^{\dagger} d_j^{\dagger}|0>=0 , \quad <0|d_i^{\dagger} d_j|0>=0 ,$$ no need to move $C_1$ to the right.
\begin{alignat}{2} C_1 N(C_2 \cdots C_n) &=&& d_i^{\dagger} (d_j^{\dagger} \cdots d_k^{\dagger} d_l \cdots d_m) \\ &=&& N(<0|C_1 C_2|0>C_3 \cdots C_n \,+ \,C_2<0|C_1 C_3|0> \cdots C_n \,+\cdots\cdots\,+\, C_2\cdots C_{n-1} <0|C_1 C_n|0> ) \\ &&&+ d_i^{\dagger} (d_j^{\dagger} \cdots d_k^{\dagger} d_l \cdots d_m) \\ &=&& N( <0|C_1 C_2|0>C_3 \cdots C_n\, +\, C_2<0|C_1 C_3|0> \cdots C_n \,+\cdots\cdots\,+ C_2\cdots C_{n-1} <0|C_1 C_n|0>\, \\ &&&+\,C_1 \cdots C_n ) \end{alignat} The number of $d\,\text{or}\,d^{\dagger}$ may be zero.
For the terms in which contracted $C_j$'s exist , the calculation is just in the same way since a contraction is just a multiplicative constant.
When $C_i$'s are linear superpositions of $d_j,\,d_j^{\dagger}$, i.e. $$C_1=\sum_{j_1} \alpha_{j_1}C_{1j_1},\, C_2=\sum_{j_2} \alpha_{j_2}C_{2j_2}, \cdots C_n=\sum_{j_n} \alpha_{j_n}C_{nj_n}$$ Where $C_{ij_i}$ is $d_k\,\text{or}\,d_k^{\dagger}$.
Replace $C_1\, ,\to \, C_{1j_1} \quad C_i\, \to \, \alpha_{j_i}C_{ij_i}(i\ge2)$ in the above proof.
The proof of \begin{alignat}{2} C_{1j_1} N(C_2 \cdots C_n) &=&& N( <0|C_{1 j_1} C_2|0>C_3 \cdots C_n\, +\, C_2<0|C_{1 j_1} C_3|0> \cdots C_n \,+\cdots\cdots\,+ C_2\cdots C_{n-1} <0|C_{1j_1} C_n|0>\, \\ &&&+\,C_{1j_1} \cdots C_n ). \end{alignat} holds as it stands for each $C_{1j_1}=d_i,\, C_{1j_1}=d_i^{\dagger}$. Then multiply by $\alpha_{i_1}$, and operate $\displaystyle \sum_{i_1,i_2,\cdots,i_n}$. We'll also get \begin{alignat}{2} C_1 N(C_2 \cdots C_n) &=&& N( <0|C_1 C_2|0>C_3 \cdots C_n\, +\, C_2<0|C_1 C_3|0> \cdots C_n \,+\cdots\cdots\,+ C_2\cdots C_{n-1} <0|C_1 C_n|0>\, \\ &&&+\,C_1 \cdots C_n ). \end{alignat}
And in the example, a contraction is not a commutation relation. $$a_2 a_3^{\dagger} = N(a_2 a_3^{\dagger} + <0|a_2 a_3^{\dagger}|0> )$$ is obviously correct. And since $<0|a_1^{\dagger} a_2|0> = <0| a_1^{\dagger} a_3^{\dagger}|0>=0,$ $$a_1^{\dagger} a_2 a_3^{\dagger} = N(a_1^{\dagger} a_2 a_3^{\dagger} + <0|a_1^{\dagger} a_2|0> a_3^{\dagger} + a_2 <0| a_1^{\dagger} a_3^{\dagger}|0> + a_1^{\dagger} <0|a_2 a_3^{\dagger}|0> )$$ holds as Wick's theorem states.