0
$\begingroup$

I'm studying junior physics where I learned the particle in a box and also Bohr's atom. Where the energy gape was give by

  1. $\Delta E_{n+1\rightarrow n}=((n+1)^2-n^2)E_1$ and
  2. $\Delta E_{n+1\rightarrow n}=(1/(n+1)^2-1/n^2)E_1)$ where I define
  3. $R(\lambda)=\Delta E_{n+1\rightarrow n}/mean(E_n+E_{n+1})=(E_{n+1}-E_{n})/((E_n+E_{n+1})/2) $.

For a particle in a box $\lim_{n\rightarrow \infty}R(\lambda)=2(2n+1)/(2n^2+2n+1)\approx2/n=0$ for Bohr's atom $\lim_{n\rightarrow \infty}R(\lambda)=-2(2n+1)/(2n^2+2n+1)\approx -2/n=0$. Which meant, as the increase of the energy states, the system would be hard to control since the portion of the energy gap approach 0.

My question was that: is there any way to create a quantum system at high energy state such that $\Delta E_{n+1\rightarrow n}$ and $R(\lambda)$ was comparable rather large.

$\endgroup$
2
  • $\begingroup$ What do you mean by "quantum system"? Is any operator good as a hamiltonian? Or are you only looking for hamiltonians of the form $H=T+V$? $\endgroup$ Mar 15 '18 at 23:09
  • $\begingroup$ @EmilioPisanty As long as it's quantum because of the quantization. $E_n$ was energy(usually total energy but the most important thing was such that it could be extracted from a single form of interaction or not, i.e. "only" be excited by lazer and was able to ignore the rest of interactions, where in the case of total energy, any interaction was good). operator was good and most of the case I saw by far was $H=T+V$. But as long as its energy eigenvalue, any Hamiltonian was all right. $\endgroup$
    – J C
    Mar 15 '18 at 23:20
-1
$\begingroup$

In principle, the simple quantum harmonic oscillator does. If you read through that derivation, bound states are separated by a constant $hw$, where $w$ is the oscillator strength.

$\endgroup$
1
  • $\begingroup$ If $E_n=n+1/2$ then $\lim_{n\rightarrow \infty }R(\lambda)≈1/(n+1/2)$ approached $0$ as well. $\endgroup$
    – J C
    Mar 15 '18 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.