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In the answer of the question below, I have two doubts:

$1.$ why we divide by $2$ in the square bracket? and

$2$. I know that the states I will need are $n=0, 1, 2, ..., N/2-1$, but why in the summation we took only the terms of $n=0$ and $n=N/2-1$?

$\mathbf{1\;\;\; Sakurai\; 6.1}$

$\textit{A. N identical spin 1/2 particles are subjected to a one-dimensional simple har-}$ $\textit{monics oscillator potential. What is the ground-state energy? What is the Fermi}$ $\textit{energy?}$

The particles are spin $1/2$ and only fermions have half-integer spin. Fermions also obey the Pauli exclusion principle, which states that no two fermions can occupy the same state. Thus, not all the particles will occupy the ground state energy of the $1$D harmonic oscillator. They end up piling up on each other until they are pushed up to higher and higher energy states. If you begin counting at $n = 0$ for the $1$st particle, and $n = 1$ for the $2$nd particle, then both occupy the lowest energy$(1/2\hbar\omega)$, one particle having spin up, and one having spin down. Therefore, it is best to say you have a total of $N/2$ particles and sum up only the spin up(or spin down) and multiply the sum by $2.$ Then you will have the total sum of energies for all the particles, where the $n$th energy level is: $$E_{n} = (n +1/2)\hbar\omega$$ First I will assume I have an even total number of particles, $2$ for each energy state. I will start summing at $n = 0$ and the sum will go to $N/2-1.$ So, summing up the energies of the particles for $N/2$ particles and multiplying by $2$ yields: $$ E_{grdEven} = 2\sum^{N/2-1}_{N=0} E (n + 1/2)\hbar\omega= 2\hbar\omega \frac{N}{2} [\frac{[1/2 + (N/2 — 1) + 1/2)}{2}] = \frac{N^{2}}{4}\hbar\omega$$

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This last line makes use of the fact that $$\sum_{n=0}^N n=\frac{N(N+1)}2.$$ This expression is called a triangular number and there are many great proofs for it. It also splits the sum in two parts \begin{align}\sum_{n=0}^N (n+1/2)&=\sum_{n=0}^N(n)+\sum_{n=0}^N(1/2)\\&=\frac 1 2N(N+1)+\frac 1 2(N+1)\\&=\frac 1 2(N+1)^2\end{align} Obviously you would still have to substitute $N/2-1$ for the upper bounds of the sum to get the final answer. Do you get the answer now if you work it out?

Edit: I said that $\sum_{n=0}^N=\frac N 2$, but it is actually $\frac{1}{2}(N+1)$. Forgot to take into account the $n=0$ term. My bad! Checked it and now it should work out.

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    $\begingroup$ Many Many thanks.. you answered my questions very very well. $\endgroup$ – Logic Nov 28 '19 at 10:18
  • $\begingroup$ Still I could not get the answer above. What I aim getting is: (N/2-1)(N/2-1+1)/2 + (N/2-1)/2=hw/2(N^2-1). What is wrong, I substituted 𝑁/2−1 for the upper bounds of the sum to get the final answer. $\endgroup$ – Logic Dec 1 '19 at 12:10
  • $\begingroup$ Many thanks, this helped me a lot!! $\endgroup$ – Logic Dec 2 '19 at 11:00
  • $\begingroup$ You're welcome. $\endgroup$ – AccidentalTaylorExpansion Dec 2 '19 at 17:20

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