# Fetter & Walecka's derivation of second-quantised kinetic term in many-particle systems

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?

• More from same book: physics.stackexchange.com/q/234816/2451, physics.stackexchange.com/q/232680/2451 Nov 25, 2016 at 16:59
• 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 $E_W$ Nov 27, 2016 at 12:01

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 $$\bar{C}(n_1, n_2, \dots, n_{E_k}-1, \dots, n_W+1, \dots, n_{\infty}, t)$$.