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+---
+title: "Wick's theorem"
+firstLetter: "W"
+publishDate: 2021-05-29
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-05-29T14:41:55+02:00
+draft: false
+markup: pandoc
+---
+
+# Wick's theorem
+
+In the [second quantization](/know/concept/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:
+
+$$\begin{aligned}
+    \underline{\hat{b}_\alpha \hat{b}_\beta^\dagger}
+    \equiv \hat{b}_\beta^\dagger \hat{b}_\alpha
+\end{aligned}$$
+
+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 $\ket{0}$:
+
+$$\begin{aligned}
+    \expval{A B}_0
+    \equiv \matrixel{0}{A B}{0}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    A B C D
+    = \underline{A B C D}
+    &+ \underline{A B} \expval{C D}_0 \pm \underline{A C} \expval{B D}_0 + \underline{A D} \expval{B C}_0
+    \\
+    &+ \underline{B C} \expval{A D}_0 \pm \underline{B D} \expval{A C}_0 + \underline{C D} \expval{A B}_0
+    \\
+    &+ \expval{A B}_0 \expval{C D}_0 \pm \expval{A C}_0 \expval{B D}_0 + \expval{A D}_0 \expval{B C}_0
+\end{aligned}$$
+
+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
+1.  L.E. Ballentine,
+    *Quantum mechanics: a modern development*, 2nd edition,
+    World Scientific.