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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 ...


6

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 ...


5

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 ...


5

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,...


5

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 ...


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