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I have read that if in the quantum harmonic oscillator, $n$ is very large, then the probability density is similar to the classical one. In the case of a simple harmonic oscillator:

$$P_{clas}=\frac{1}{\pi \sqrt{A^2-x^2}}$$ $$P_{qho}\propto\frac{1}{\sqrt{2n-1-x^2}}.$$

But what about the energies? I know that if i take the limit with n tends to infinite: $\frac{\Delta E_n}{E_n}\rightarrow 0$. So the energy spectrum becomes continuous as in the classical case. Is that the only thing you can know about energy?

If $A^2=2n-1$, then: $$E_{clas}=\frac{1}{2}\omega^2mA^2$$ $$E_{qho}=(2n+1)\hbar\omega/2$$ So in order to become equal: $\omega m =\hbar$. Which means that the mass and the frequency must be very small and this would lead us to have to use quantum mechanics instead of classical mechanics. Is this correct?

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    $\begingroup$ Hello. You should define the quantities $P_{\text{clas}}$ and $P_{\text{qho}}$. Also, what is $n$? Is it the number of photons in a fock state? The classical limit of the quantum harmonic oscillator is usually understood for a quantum state being the coherent state (in the large average photon number). $\endgroup$
    – StarBucK
    Commented Dec 15, 2023 at 12:30
  • $\begingroup$ Linked. $\endgroup$ Commented Dec 15, 2023 at 14:13

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Yes, it is correct. Instead of thinking of a spectrum, it's easier to bridge the two in terms of density of states. In quantum mechanics, the density of states is just: $$ D(E) = \sum_n \delta(E-E_n) $$ with the sum over an energy eigenbasis.

In the case of the harmonic oscillator, you have for $n\in\mathbb N$: $$ E_n = \hbar\omega(n+1/2) $$ So as $\hbar\to 0$, you as a weak limit: $$ D(E)\to \frac{1}{\hbar\omega} $$

This is consistent with classical mechanics. The density of states in this case is (using the semiclassical limit, or using the heuristic that a patch of phase space has $\frac{dxdp}{h}$ number of states): $$ D(E) = \frac{1}{h}\int\delta(H-E)dxdp $$ with $H$ the Hamiltonian. In the case of the harmonic oscillator, the isolines are ellipses and you do get: $$ D(E) = \frac{2\pi}{h\omega} $$

These results are typically most useful in statistical physics.

Hope this helps.

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$$P_{clas}=\frac{1}{\pi \sqrt{A^2-x^2}}$$ $$P_{qho}\propto\frac{1}{\sqrt{2n-1-x^2}}$$

Your formula for the classical probability (with $A$ being the maximum amplitude of $x$) is correct. But in the formula for the quantum probability density someone has deliberately omitted all the constants. The correct formula including the constants would be (because $\sqrt{\frac{\hbar}{m\omega}}$ has the dimension of a length): $$P_{qho}\propto\frac{1}{\sqrt{2n-1-\frac{m\omega}{\hbar}x^2}}$$ or $$P_{qho}\propto\frac{1}{\sqrt{\frac{\hbar}{m\omega}(2n-1)-x^2}}$$

If $A^2=2n-1$, then ...

Therefore you cannot simply equate $A^2$ and $2n-1$. Instead, you can equate $$A^2=\frac{\hbar}{m\omega}(2n-1)$$

Then, using $$E_{clas}=\frac{1}{2}\omega^2mA^2$$ $$E_{qho}=(2n+1)\hbar\omega/2$$ you find $$E_{qho}=E_{clas}+\hbar\omega$$

So, in the limit $\hbar\to 0$, you have confirmed $E_{qho}\approx E_{clas}$, as it should be due to the correspondence principle.

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