This question is mostly about the BE distribution as presented in the Einstein model lecture. In the lecture notes there’s a plot of the occupation number n_B as a function of \hbar\omega the oscillators frequency. My question is why is the occupation number high for lower \omega? ( I understand that if we plug in a small \omega that n_B becomes small ) Is it because a lower \omega means the phonons in the oscillator have less energy so we expect them all to be around the ground state (n = 0) which means the occupation number for those states are high? Isn’t the \omega supposed to be a fixed quantity depending on the type of atom that we’re modelling as a QHO? What is the physical reasoning behind the high occupation number?
Moreover, what does the red dashed line represent in that plot (left one under Einstein Model)?
The n_B as a function of \hbar \omega plot demonstrates a simple law: the stiffer the spring, the harder it is to excite (stretch) it. In this case, \hbar \omega controls the stiffness of the Quantum Harmonic Oscillator (QHO) because larger frequency increases the energy spacing \Delta E = \hbar \omega of the QHO levels. Since the energy spacing \Delta E is larger, fewer QHO levels become thermally excited and thus n_B is smaller.
The red dashed line indicates thermal energy: T k_B = \hbar \omega.
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