From aab299218975a8e775cda26ce256ffb1fe36c863 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Sun, 30 May 2021 15:54:40 +0200 Subject: Expand knowledge base --- content/know/concept/wicks-theorem/index.pdc | 194 +++++++++++++++++++++++++++ 1 file changed, 194 insertions(+) create mode 100644 content/know/concept/wicks-theorem/index.pdc (limited to 'content/know/concept/wicks-theorem/index.pdc') diff --git a/content/know/concept/wicks-theorem/index.pdc b/content/know/concept/wicks-theorem/index.pdc new file mode 100644 index 0000000..824885a --- /dev/null +++ b/content/know/concept/wicks-theorem/index.pdc @@ -0,0 +1,194 @@ +--- +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. -- cgit v1.2.3