I've been working through the following set of slides related to Wick's theorem: http://www.euroschoolonexoticbeams.be/site/files/2008_JDobaczewski_lecture.pdf

From slide 19 onwards the following is stated:

Given a multi-particle state $\lvert \Psi \rangle$, any operator $A$ can be decomposed as $A=A_0+A_++A_-$, with

\begin{equation} A_0 = \langle\Psi\lvert A \lvert\Psi\rangle,\\ A_+ = \big(1-\lvert\Psi\rangle \langle\Psi\lvert \big)A \lvert\Psi\rangle \langle\Psi\lvert,\\ A_- = \big(A-\langle\Psi\lvert A \lvert\Psi\rangle \big) \big(1-\lvert\Psi\rangle \langle\Psi\lvert \big). \end{equation}

A contraction of two operators $A$ and $B$ is then defined as

\begin{equation} A^{\circ} B^{\circ} = A_-B-cBA_-, \end{equation}

where c is either $+1$ or $-1$ depending on the type of multi-particle system you're dealing with (bosons or fermions). It's then said, if $\lvert\Psi\rangle$ is a multi-particle product state, the contraction of two operators is a number and given as

\begin{equation} A^{\circ} B^{\circ} = \langle\Psi\lvert AB \lvert\Psi\rangle-\langle\Psi\lvert A \lvert\Psi\rangle\langle\Psi\lvert B \lvert\Psi\rangle. \end{equation}

This last statement is the one I can't grasp. I've specifically been trying to show it's true for the case where $A$ and $B$ are fermionic creation or annihilation operators but with no luck. The contraction of fermionic operators is worked from slide 26 onward, where it's stated, if $a_\mu$ is an annihilation operator and $\lvert\Psi\rangle$ a slater determinant, then

\begin{equation} {a_\mu}_- = \begin{cases} a_\mu,& \text{if } \mu \text{ is not an occupied state of }\lvert\Psi\rangle,\\ 0, & \text{otherwise} \end{cases} \end{equation}

which I also can't show. I keep getting ${a_\mu}_- = {a_\mu}\big(1-\lvert\Psi\rangle \langle\Psi\lvert \big)$. I'm not sure if I've missed something or if I just can't see why these last two expressions are the same and I'm hoping someone could help make it a bit clearer.

  • $\begingroup$ It could just be that the decomposition isn't unique, and the decomposition at the start is just an example that it can always be done, rather than the unique way it can be done. $\endgroup$ – snulty Mar 17 '17 at 18:16

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