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Expectation value of \bgroup\color{col1}$ \hat{\mathcal H}_0$\egroup

Let us consider a \bgroup\color{col1}$ N$\egroup-Fermion state (Slater determinant), cf. Eq. (III.1.24),
$\displaystyle \vert\Psi\rangle$ $\displaystyle =$ $\displaystyle \vert\nu_1\nu_2...\nu_N\rangle_A =
\frac{1}{\sqrt{N!}}\sum_p \hat{\Pi}_p {\rm sign}(p)
\vert\nu_{p(1)}\nu_{p(2)}...\nu_{p(N)}\rangle.$ (2.2)

We wish to calculate the expectation value \bgroup\color{col1}$ \langle \Psi \vert\hat{\mathcal H}_0\vert \Psi\rangle$\egroup with \bgroup\color{col1}$ \hat{\mathcal H}_0$\egroup from Eq. (IV.2.1). Consider for example the free Hamiltonian \bgroup\color{col1}$ \hat{H}_0^{(1)}$\egroup for the first particle,
$\displaystyle \langle \Psi \vert\hat{H}_0^{(1)}\vert \Psi\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{N!}\sum_{pp'} {\rm sign}(p){\rm sign}(p')
\langle \nu_{p...
...\nu_{p(1)}\vert\hat{H}_0^{(1)} \vert\nu_{p'(1)}\nu_{p'(2)}...\nu_{p'(N)}\rangle$  
  $\displaystyle =$ $\displaystyle \frac{1}{N!}\sum_{pp'} {\rm sign}(p){\rm sign}(p')
\langle \nu_{p...
..._{p'(N)}\rangle
\langle\nu_{p(1)}\vert\hat{H}_0^{(1)} \vert\nu_{p'(1)}\rangle
.$  

For \bgroup\color{col1}$ N-1$\egroup numbers we must have \bgroup\color{col1}$ p(2)=p'(2)$\egroup,..., \bgroup\color{col1}$ p(N)=p'(N)$\egroup (otherwise the term is zero), but if you have a permutation with \bgroup\color{col1}$ N-1$\egroup terms fixed, the last term ist automatically fixed and we have \bgroup\color{col1}$ p=p'$\egroup, thus (note \bgroup\color{col1}$ {\rm sign}(p)^2=1$\egroup)
$\displaystyle \langle \Psi \vert\hat{H}_0^{(1)}\vert \Psi\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{N!}\sum_{p}
\langle\nu_{p(1)}\vert\hat{H}_0^{(1)} \vert\nu_{p(1)}\rangle.$ (2.3)

The sum of the single-particle Hamiltonians yields
    $\displaystyle \langle \Psi \vert\hat{\mathcal H}_0\vert \Psi\rangle=$ (2.4)
  $\displaystyle =$ $\displaystyle \frac{1}{N!}\sum_{p}
\langle\nu_{p(1)}\vert\hat{H}_0^{(1)}\vert\n...
...2)}\rangle +... +
\langle\nu_{p(N)}\vert\hat{H}_0^{(N)} \vert\nu_{p(N)}\rangle,$  

but all the Hamiltonians \bgroup\color{col1}$ \hat{H}_0^{(i)}$\egroup have the same form, the sum \bgroup\color{col1}$ \sum_p$\egroup just gives \bgroup\color{col1}$ N!$\egroup identical terms, and therefore
$\displaystyle \langle \Psi \vert\hat{\mathcal H}_0\vert \Psi\rangle= \sum_{i=1}^{N}\langle\nu_{i}\vert\hat{H}_0\vert\nu_{i}\rangle,$     (2.5)

where we can omit the index \bgroup\color{col1}$ ^{(i)}$\egroup in \bgroup\color{col1}$ \hat{H}_0^{(i)}$\egroup and write \bgroup\color{col1}$ \hat{H}_0$\egroup for the free Hamiltonian of a single particle (note that \bgroup\color{col1}$ \hat{\mathcal H}_0$\egroup in Eq. (IV.2.1) is the total free Hamiltonian; some books use \bgroup\color{col1}$ \hat{h_0}$\egroup instead of \bgroup\color{col1}$ \hat{H}_0$\egroup to make this distinction clearer, but small letters are not nice as a notation for a Hamiltonian).


next up previous contents index
Next: Expectation value of Up: Hamiltonian for Fermions Previous: Hamiltonian for Fermions   Contents   Index
Tobias Brandes 2005-04-26