Categories: Physics, Quantum mechanics.

# Wick’s theorem

In the second quantization formalism,
**Wick’s theorem** helps to evaluate products
of creation and annihilation operators by
breaking them down into smaller products.

Firstly, let us define the **normal product** or **normal order** as
a product of second quantization operators
reordered such that
all creation operators are on the left of
all annihilation operators.
For two operators this is written as follows,
at least in the case of bosons:

For fermions, the result must be negated for each swapping of adjacent operators (and every reordering of operators can be treated as a sequence of such swaps):

$\begin{aligned} \underline{\hat{f}_\alpha \hat{f}_\beta^\dagger} \equiv - \hat{f}_\beta^\dagger \hat{f}_\alpha \end{aligned}$The normal product of three or more operators works in the same way, but might not be unique depending, on how many of each type there are.

Next, the **contraction** of the operators $A$ and $B$
is defined as the vacuum matrix element,
i.e. the expectation value of $\Ket{0}$:

Unsurprisingly, a contraction can only be nonzero if $A = \hat{c}_\alpha$ is an annihilation and $B = \hat{c}_\alpha^\dagger$ a creation for the same state $\alpha$.

Wick’s theorem states:
**any product of second quantization operators can be
rewritten as a sum of normal products,
from which 0, 1, 2, etc. contractions have been removed
in every possible way.**
For fermions, the sign of a term must also be swapped
every time two adjacent operators are swapped.
As an example, for four operators:

Where the negative signs apply to fermions only. We take the normal product with 0 contractions removed ($\underline{ABCD}$), then with 1 contraction removed in every possible way (first two lines), then with 2 contractions removed in every possible way (last line), and so on.

## Proof

We will prove this by induction, with the base case being two operators, where Wick’s theorem becomes as follows:

$\begin{aligned} A B = \underline{AB} + \Expval{A B}_0 \end{aligned}$This must be proven separately for fermions and bosons. For fermions, a general consequence of the definition of the anticommutator is:

$\begin{aligned} \hat{f}_\alpha \hat{f}_\beta^\dagger = - \hat{f}_\beta^\dagger \hat{f}_\alpha + \{\hat{f}_\alpha, \hat{f}_\beta^\dagger\} \end{aligned}$This anticommutator is known to be $\delta_{\alpha\beta}$, so we can inconsequentially take its inner product with the vacuum state $\Ket{0}$:

$\begin{aligned} \hat{f}_\alpha \hat{f}_\beta^\dagger &= - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\{\hat{f}_\alpha, \hat{f}_\beta^\dagger\}}{0} = - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\hat{f}_\alpha \hat{f}_\beta^\dagger + \hat{f}_\beta^\dagger \hat{f}_\alpha}{0} \\ &= - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\hat{f}_\alpha \hat{f}_\beta^\dagger}{0} = \underline{\hat{f}_\alpha \hat{f}_\beta^\dagger} + \expval{\hat{f}_\alpha \hat{f}_\beta^\dagger}_0 \end{aligned}$Which agrees with Wick’s theorem. For bosons, we use the commutator:

$\begin{aligned} \hat{b}_\alpha \hat{b}_\beta^\dagger = \hat{b}_\beta^\dagger \hat{b}_\alpha + [\hat{b}_\alpha, \hat{b}_\beta^\dagger] \end{aligned}$This commutator is known to be $\delta_{\alpha\beta}$, so we take the inner product with $\Ket{0}$, like before:

$\begin{aligned} \hat{b}_\alpha \hat{b}_\beta^\dagger &= \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{[\hat{b}_\alpha, \hat{b}_\beta^\dagger]}{0} = \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{\hat{b}_\alpha \hat{b}_\beta^\dagger - \hat{b}_\beta^\dagger \hat{b}_\alpha}{0} \\ &= \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{\hat{b}_\alpha \hat{b}_\beta^\dagger}{0} = \underline{\hat{b}_\alpha \hat{b}_\beta^\dagger} + \expval{\hat{b}_\alpha \hat{b}_\beta^\dagger}_0 \end{aligned}$Which again agrees with Wick’s theorem. Next, we prove that if it holds for $N$ operators, then it also holds for $N + 1$. To begin with, consider the following statement about right-multiplying by an extra $A_{N+1}$, with $s = 1$ for bosons and $s = -1$ for fermions:

$\begin{aligned} \underline{A_1 ... A_N} A_{N+1} = \underline{A_1 ... A_N A_{N+1}} + \sum_{n = 1}^N s^{n + N} \Expval{A_n A_{N+1}}_0 \underline{A_1 ... A_{n-1} A_{n+1} ... A_N} \end{aligned}$If $A_{N + 1}$ is an annihilation operator, then this is trivial: appending it does not break the existing normal order, and $\Expval{A_n A_{N+1}}_0 = 0$ for all $A_n$.

However, if $A_{N + 1}$ is a creation operator, then to restore the normal order, we move it to the front by swapping, which introduces a bunch of (anti)commutators:

$\begin{aligned} \underline{A_1 ... A_N} A_{N+1} &= s^N A_{N+1} \underline{A_1 ... A_N} + \sum_{n} s^{n + N} \{[A_n, A_{N+1}]\} \underline{A_1 ... A_{n-1} A_{n+1} ... A_N} \\ &= \underline{A_1 ... A_N A_{N+1}} + \sum_{n} s^{n + N} \Expval{A_n A_{N+1}}_0 \underline{A_1 ... A_{n-1} A_{n+1} ... A_N} \end{aligned}$Where $\{[]\}$ is the anticommutator or commutator, respectively for fermions or bosons.

If we take Wick’s theorem for $N$ operators $A_1 ... A_N$, and right-multiply it by $A_{N + 1}$, then each term will contain a product of the form $\underline{A_{v} ... A_{w}} A_{N+1}$. Using the relation that we just proved, each such product can be rewritten as follows:

$\begin{aligned} \underline{A_v ... A_w} A_{N+1} &= \underline{A_v ... A_w A_{N+1}} + \sum_{n} s^{n + N} \Expval{A_n A_{N+1}}_0 \underline{A_v ... A_{n-1} A_{n+1} ... A_w} \end{aligned}$Inserting this back into Wick’s theorem, we get new terms with contractions of $A_{N+1}$. After a lot of rearranging, the result turns out to just be Wick’s theorem for $N\!+\!1$ operators. Therefore, if Wick’s theorem holds for $N$ operators, it also holds for $N\!+\!1$.

We showed that Wick’s theorem holds for $N = 2$, so, by induction, it holds for all $N \ge 2$.

## References

- L.E. Ballentine,
*Quantum mechanics: a modern development*, 2nd edition, World Scientific.