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The energy of a quantum harmonic oscillator is given as,

\begin{equation} E_{qho} = \left(n + \frac{1}{2}\right)\hbar\omega,\;\;\;\;\;\; n = 0,1,2,3,... \tag{1} \end{equation}

I understand the implications of the Heisenberg's uncertainty principle not permitting atoms to be at a standstill even at $0$ K. This uncertainty is the reason for some energy possessed by atoms at $0$ K - the zero-point energy (ZPE). Most texts introduce ZPE by noting how with $n = 0$ there is still a remnant energy equal to $\frac{1}{2}\hbar\omega$.

Is $n$ only a number? If so, how has $n = 0$ anything to do with temperature?

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The zero-point energy is of no importance here, since you can always choose your reference energy freely you can energy-shift your hamiltonian by $\frac{1}{2}\hbar\omega$ $$ H = \frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2-\frac{1}{2}\hbar\omega, $$ and the physics of the system will stay the same (the wave function will be the same). Since this wavefunction is not a delta-function located at zero (as it is in classical mechanics) but instead more spread out, you can interpret this as, for example, your atoms still vibrating when in this eigenstate of the hamiltonian.

Regarding your question: Yes, $n$ is just a number which is meant to label the energy eigenstates from lowest to highest. Temperature only plays in indirectly. To define a temperature, you have to define a thermal ensemble (you need more than one particle to do it properly) with an associated density matrix $\rho$. A common choice for this is given by $$ \rho = \frac{1}{z}\sum_{i=1}^{\infty}|i\rangle e^{-E_{i}/kT} \langle i|, z = \sum_{i=1}^{\infty}e^{-E_i/kT} $$ where $|i\rangle$ denote the energy eigenstates and $E_i$ the corresponding energy eigenvalues (in this case for the harmonic oscillator). $T$ is temperature, $k$ just a constant. You can interpret (similarly to a wavefunction expansion coefficient) that the factor $e^{-E_{i}/kT}/z$ is a probability of being in the state $|i\rangle$. You can see that when $T\rightarrow 0$, only the coefficient with the lowest energy eigenvalue will remain (any coefficient with higher $E_i$-value will vanish faster). From this can be deducted that for a general system (not just your harmonic oscillator example) the system will be in the lowest-energy state when $T\rightarrow 0$ (as long as you have a thermal ensemble).

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    $\begingroup$ Thank you for your answer and welcome to the forum. $\endgroup$ – Hitanshu Sachania Jan 8 at 20:25
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The quantum number n simply represents the different energy levels given by the harmonic oscillator.

$\mathbf{n=0}$ does not correspond to a given temperature, but its relative occupation to other energy levels does correspond to a given temperature. As a system rises in temperature, the higher energy levels can be occupied at greater numbers. Likewise, at 0 K there is a requirement that only the lowest energy level is occupied.

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  • $\begingroup$ Quantum mechanics can be weird at times. Why couldn't particles just stick to classical mechanics.😅 $\endgroup$ – Hitanshu Sachania Jan 8 at 20:27
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    $\begingroup$ "Likewise, at 0 K there is a requirement that only the lowest energy level is occupied." Not for particles obeying Fermi-Dirac distribution. Fermions can't occupy the same state, so the 0 K distribution is where all lowest energy levels are occupied. $\endgroup$ – JiK Jan 9 at 10:36
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    $\begingroup$ @JiK While this is true, I don't entirely think this is what the poster was asking about...If you can expand that comment into an answer though it could be a good resource all the same though. $\endgroup$ – Tristan Maxson Jan 9 at 15:34
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Is $n$ only a number?

$n$ is indeed a number. Is it only a number? Well it's a quantum number which means it labels the $n^{\textrm{th}}$ excited energy level of the system (i.e. the $(n+1)^{\textrm{th}}$ smallest eigenvalue of the system's Hamiltonian, with $n=0$ corresponding to the smallest eigenvalue, $n=1$ corresponding to the second smallest eigenvalue, etc.

If so then how does $n = 0$ have anything to do with temperature?

The density matrix of a system with the harmonic oscillator potential is often given in terms of the Hamiltonian $H$ by:

\begin{equation} \rho = \frac{e^{-\beta H}}{\textrm{tr}\left(e^{-\beta H}\right)},~~~~~~~~\beta\equiv \frac{1}{k_BT}. \tag{1} \label{eq:boltzmann} \end{equation}

The diagonals of the density matrix from top-left to bottom-right then tell you the probability of finding the system in $n=0,1,2,\ldots$, meaning that if the top-left element of the density matrix is $p$, the system's probability of being found at the energy level corresponding to $n=0$ is $p$. When $T=0$ we have that the probability of the system being in any excited state ($n>0$) is extremely supressed by the decaying exponential, and you can count on finding the system at the $n=0$ level. When $T$ is larger, the excited states will more likely get populated. As $T$ approaches $+\infty$, the exponential becomes close to 1 and we approach a scenario where the probabilities become equal for each state $n$.

Eq. 1 in this answer is also:

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    $\begingroup$ Demystified, thank you. Sometimes finding correlations where there are none can make things confusing. $\endgroup$ – Hitanshu Sachania Jan 8 at 20:24
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Is $𝑛$ only a number?

In short, $n$ is the energy quantum number of the quantum harmonic oscillator.

If so then how does $𝑛$=$0$ have anything to do with temperature?

In particular, $n$=$0$ means that the harmonic oscillator will stay at its ground state. Usually, the ground state of a quantum system is assumed to be lived at zero temperature. Therefore, you can find a connection between $n=0$ and zero-point.

May it helps.

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  • $\begingroup$ Thank you for your answer, Jack. Those are interesting questions, especially the one about system size and thermal equilibrium. $\endgroup$ – Hitanshu Sachania Jan 9 at 12:06
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As has been stated in several other answers already, $n$ is only a number, and the population of the states with different $n$ depends on the temperature.

However, an important point has not yet been mentioned. The quantum harmonic oscillator is often invoked for nuclear motion. It arises from the second-order Taylor expansion of the Born-Oppenheimer nuclear potential energy surface $V({\bf R}) = V({\bf R}_0) + \nabla V|_{{\bf R}={\bf R}_0} \cdot({\bf R}-{\bf R}_0)+\frac 1 2 ({\bf R}-{\bf R}_0)\cdot \nabla\nabla V|_{{\bf R}={\bf R}_0}\cdot ({\bf R}-{\bf R}_0) + \mathcal{O}(|{{\bf R}-{\bf R}_0}|^3)$

where the first-order term vanishes since $\nabla V|_{{\bf R}={\bf R}_0} ={\bf 0}$ at the minimum.

Since the spatial extent of the states increases with $n$, the importance of anharmonic effects also grows with $n$, or with increasing temperature.

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  • $\begingroup$ Thank you for your answer, Dr Susi. $\endgroup$ – Hitanshu Sachania Jan 10 at 16:47

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