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On page 9 of Quantum theory of many-particle systems by Alexander L. Fetter and John Dirk Walecka, during the derivation of the second-quantised kinetic term, there is an equality equation below:

\begin{align} \sum_{k=1}^{N} \sum_{W} & \langle E_k|T|W\rangle C(E_1, ..., E_{k-1}, W, E_{k+1},...,E_N, t) \\&= \sum_{k=1}^{N}\sum_{W}\langle E_k|T|W\rangle\times \bar{C}(n_1, n_2,...,n_{E_k}-1, ..., n_{W}+1,...,n_\infty, t) \end{align}

Why is the number of particles with quantum numbers $n_{E_k}$ decreased by 1 whereas the number of particles with quantum numbers $n_W$ increased by 1?

Anybody know how to get this equality?

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  • $\begingroup$ More from same book: physics.stackexchange.com/q/234816/2451, physics.stackexchange.com/q/232680/2451 $\endgroup$
    – Qmechanic
    Commented Nov 25, 2016 at 16:59
  • $\begingroup$ Seems I understand the coefficient on the right-hand side. What we want to do is get the coefficient with particle 1 in given state $E_1$,..., particle N in given state $E_N$, i.e. $C(E_1, E_2, ..., E_N, t)$, which corresponds to $\bar{C}(n_1, n_2, ..., n_{E_k}, ..., n_W,..., n_{\infty}, t)$. Actually $\bar{C}(n_1, n_2, ..., n_{E_k}, ..., n_W,..., n_{\infty}, t)$ defaults the $k$th particle in state $E_k$, while what really happens on the right-hand side is that the $k$th particle is in state W running over all possible quantum states. So $n_{E_k}-1$ in state $E_k$, $n_W+1$ in state $E_W$ $\endgroup$
    – bitsoal
    Commented Nov 27, 2016 at 12:01

1 Answer 1

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The coefficient $C(E_1, E_2, \dots, E_k, \dots, E_N, t)$ corresponds to the configuration:

$1^{st}$ particle in state $E_1$, $2^{nd}$ particle in state $E_2$, $\dots,$ $k^{th}$ particle in state $E_k$, $\dots,$ $N^{th}$ particle in state $E_N$.

Say, this corresponds to the following occupation numbers:

$n_1$ particles in state $1$, $n_2$ particles in state $2$, $\dots,$ $n_l$ particles in state $l$, $\dots,$ $n_{\infty}$ particles in state ${\infty}$.

where $n_1+n_2+\dots+n_{\infty}=N$. Then, we rewrite the coefficient for this configuration in terms of occupation numbers as $\bar{C}(n_1, n_2, \dots, n_l, \dots, n_{\infty}, t)$.

Now, the new coefficient after the action of the kinetic energy operator is $C(E_1, E_2, \dots, E_{k-1}, W, E_{k+1}, \dots, E_N, t)$, which corresponds to the configuration:

$1^{st}$ particle in state $E_1$, $2^{nd}$ particle in state $E_2$, $\dots,$ $(k-1)^{th}$ particle in state $E_{k-1}$, $k^{th}$ particle in state $W$, $(k+1)^{th}$ particle in state $E_{k+1}$, $\dots,$ $N^{th}$ particle in state $E_N$.

$i.e.,$ the $k^{th}$ particle, which was originally occupying the state $E_k$, is currently in state $W$. Therefore, the number of particles occupying the state corresponding to $E_k$ should be decreased by one, and the number of particles occupying the state corresponding to $W$ should be increased by one. (In other words, the kinetic energy operator has annihilated a particle in the state corresponding to $E_k$, and created a particle in the state corresponding to $W$.) So, the coefficient for this new configuration in terms of occupation numbers is $\bar{C}(n_1, n_2, \dots, n_{E_k}-1, \dots, n_W+1, \dots, n_{\infty}, t)$.

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