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So I am pretty new to quantum mechanics (mostly related to atomic structures), and have some difficulties in the following setup:

Consider a particle of mass $m$ confined to a one-dimensional box of length $L$. The particle moves in the box with momentum $p$ colliding elastically with the walls. We consider the quantum mechanics of this system.

At each energy state, the particle may be represented by a standing wave given by the de Broglie hypothesis. Express its wavelengths $λ_{dB}$ in terms of $L$ in the $n^{\text{th}}$ energy state.

First of all, (it may seem silly), what is energy state here? In atomic structure, I know it corresponds to electrons permissible energy in shells, as per Bohr's quantization. But here in this setup, I couldn't grasp the meaning of 'energy state'.

Secondly, question asked to express particle as a standing wave. I know the standing wave concept but from a classical mechanics point of view, not quantum mechanical. I couldn't figure which concept should I apply here.enter image description here

Interestingly, the answer is $\lambda_{dB}=\frac{2L}{n}$, which is the same as the general formula of standing wave (classical) wavelength as shown in the figure. I don't know how the $n$ in the figure of standing waves and the $n$ corresponding to energy state in our state are so related, so this is my major doubt. I may be confused between these formulae, but I can't help myself here.

Please help clear my doubts/confusion.

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Well the energy for a particle in a box is as follows: $$E_n = \frac{n^2h^2}{8mL^2}$$ Where each n represent the state, so $E_n$ represents the energy per state.

The n in the lambda represents the excited state of the wavelength, so for n=1 you have the ground state, n=2 is the first excited state etc. It represents the amount of knots etc in a wave. Hope this cleared things up! If not, let me know.

To come back to your question of where this formula came from: The momentum is known to be $p=m v \longrightarrow v = \frac{p}{m}$ Inside a one-dimensional box there's kinetic energy: $E= \frac{1}{2}mv^2$. We can rewrite this as $E = \frac{p^2}{2m}$. since $p = \frac{h}{\lambda} = \frac{hn}{2L}$ So we get $$E_n = \frac{n^2h^2}{8mL^2}$$ when we plug all of this in.

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  • $\begingroup$ Can you please, tell that how you wrote $E_n$? I am pretty novice in quantum mechanics. I know nothing beyond basic quantum mechanical model of atom. $\endgroup$ Nov 2, 2021 at 18:13
  • $\begingroup$ I included that now, take a look. @KshitijKumar $\endgroup$ Nov 2, 2021 at 18:22
  • $\begingroup$ By the way, $p = \frac{h}{\lambda}$ because $p = \frac{E}{c} = \frac{hf}{c} = \frac{h}{\lambda}$ $\endgroup$ Nov 2, 2021 at 18:30
  • $\begingroup$ One more doubt is that wavelength gets shorter for increasing value of $n$, so is it sort of frequency like factor? Also how are we exciting particle, through external means? $\endgroup$ Nov 2, 2021 at 18:30
  • $\begingroup$ Yeah the wavelength gets shorter, because for a higher n, the energy increases. I'm not sure what you mean with the last question. @KshitijKumar $\endgroup$ Nov 2, 2021 at 18:33

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