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Classical Harmonic oscillator's energy depends on temperature as it equals $k_B$$T/2$. However, quantum harmonic oscillator energy is $(n+1/2)hf$. So, when T=0, quantum predicts motion.

I have been wondering what would happen if there is change in pressure? If the pressure is changed, does the quantum and classical prediction differ?

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  • $\begingroup$ sorry - not sure what you mean by pressure - if we think about mass on a spring simple harmonic motion then if 'force constant' of the spring then nothing much changes, but the Quantum levels get further apart as the spring gets stronger. Also I think you will find that the harmonic oscillator might get $k_B T$ rather than $k_B T/2$ energy. Certainly for a diatomic molecule it has $3 k_B T/2$ translational energy, $k_B T$ rotational energy (for two degrees of freedom in rotation), but $k_B T$ for the one vibrational mode - typically explanation $k_B T/2$ of KE + $k_B T/2$ of PE for vibration. $\endgroup$ – tom Nov 16 '14 at 22:15
  • $\begingroup$ For the molecules in one dimension, what will the effect of change in pressure result? I mean like will the classical as well as quantum both be affected by change in pressure as in temperature? $\endgroup$ – Harry Watson Nov 16 '14 at 22:44
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Pressure and volume are thermodynamic conjugates. So, to truly depend on pressure a harmonic oscillator would have to have a volume. In that sense a harmonic oscillator is independent of pressure---volume and pressure simply don't appear as variables! On the other hand we can control the oscillator's "size" by changing its spring constant (changing frequency).

In all mechanical systems we have control parameters (generalizing the idea of volume) with associated forces (generalizing pressure). For any parameter $x$ that influences a mechanical system, there is an associated force $X = -\partial E/\partial x$ applied by the system. Note that this force depends on the state of the system and likely fluctuates over time.

In statistical thermodynamics the state of the system is random but with a particular distribution and so we can talk about a well-defined mean value of the force. This turns out to be given (equivalently) by either of these expressions: $$\langle X \rangle = -(\partial \langle E\rangle / \partial x)_S$$ $$\langle X \rangle = -(\partial A / \partial x)_T$$ The first is the derivative of enery with respect to the parameter, holding entropy constant. The second is the derivative of Helmholtz free energy with respect to the parameter, holding temperature constant.

For the quantum harmonic oscillator, we have a free energy $$ A = E_0 + \tfrac{1}{2}hf + kT\ln(1 - e^{-hf/kT}) $$ where $E_0$ is the minimum of the potential energy, $h$ is Planck constant, and $f$ is the oscillation frequency. Note that the classical oscillator is a special case for low frequency, where $A \approx E_0 + kT\ln(hf/kT)$.

Now, we can think of $E_0$ and $f$ as control parameters. Changing $E_0$ is boring -- it just adds or removes that change in energy directly to the system. But if we change $f$ (by changing spring constant) while keeping $E_0$ constant, then there is a "force" of $$ \langle F \rangle = -(\partial A/\partial f)_T = -\tfrac{1}{2}h - \frac{1}{e^{hf/kT} - 1}h $$

What is the meaning of this? First, it is negative meaning that we must supply energy to the system when we increase frequency. For very high frequencies where the oscillator is frozen into its ground state, the force is $-h/2$. This is is a zero-point-motion force analogous to Casimir force. For very low frequencies the oscillator is basically behaving classically and has an $f$-independent mean energy of $E_0+kT$, however there is still a frequency-changing force of $-kT/f$. This is analogous to the pressure $kT/V$ from a gas molecule in a box. When we increase frequency we decrease the "volume" of the harmonic oscillator.

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It's a little unclear what you are asking, but one influence that pressure does have is that it can effectively broaden energy levels and result in "pressure broadened" or "collisionally broadened" radiative transitions.

Essentially what is going on is that at high pressures (which for a given temperature, means high densities), collisions between atoms/molecules become more frequent and close encounters with another atom/molecule can perturb the energy levels. Thus if you were to think about atoms/molecules as harmonic oscillators (they aren't, but that's not important here) It turns out that the energy levels are perturbed in such a way that the radiative transitions between them have shortened lifetimes and are broader by a factor that depends on the number density and the square root of temperature (for collisional broadening; i.e. it increases with pressure). Similarly, the level-perturbing effects of near encounters with other atoms, sometimes known as the Stark effect, also increases with pressure (essentially as the number density).

These effects are commonly encountered at the kinds of pressures and temperatures present in the photospheres of stars.

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Ah - do you mean pressure as external collisions onto the oscillator?

I think that pressure would have to be very very high to disrupt the oscillation and note that atoms in a crystal have zero point motion (the motion you get at T=0) and zero point energy.

The pressure that you would need to disrupt this zero point motion in a solid would be very high- the sort of pressure that you get in white dwarf/neutron stars I expect.

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