# Does the probability function for a molecule to have energy in a certain mode depend on the minimum required quantum energy level of that mode?

I understand that the probability density for a particular velocity v+dv in one dimension for a molecule in a system is: $$f(v) = \sqrt{\frac{m}{2\pi k_B T}} \cdot e^{\frac{-mv^2}{2k_B T}}$$ I'm noticing that this formula does not distinguish the type of freedom in which the concerning velocity is used. So basically this formula would say that a certain velocity in a translational degree of freedom has an equal probability as the same velocity in a rotational or vibrational degree of freedom, which is incorrect from what I know due to the different required minimum quantum energy for each type of mode (depending on the molecular structure) with respect to the energy that the molecule possesses. Is there a formula that takes the type of mode into account for the probability?

I found a source that takes the frequency of a degree of freedom into account and calculates its minimum required energy according to hf = E but I don't know how to incorporate that formula into the probability function.

• This function considers molecules as hard spheres and is used to study the macroscopic behavior of matter. Treatment of movement of molecules taking into consideration their quantum states is something else and as far as I know very complicated. – rbw May 16 '18 at 13:30
• The formula you give is the Maxwell-Boltzmann distribution for the velocity of a gaseous molecule in one dimension. It is derived assuming translational kinetic energy only of the gas and that there is thermal equilibrium. It is also assumed that components in the y and z direction do not affect motion in the x direction. Translational motion is quantised but quanta are so small that we treat the system classically. This is not true of rotations and vibrations of small molecules. Proteins, &colloids could be treated classically for rotational motion. Similar equations may result. – porphyrin May 16 '18 at 16:08

There's a couple of conceptual problems at the base of your question that need to be addressed.

1) Free translation of particles in space is not discrete - the distribution of quantum states is continuous, and therefore so is the distribution of energies or of velocities - but rotation or vibration are quantised, so they have discrete distributions, so the formula definitely can't be applied to them. You could apply it to molecules as classical objects, ignoring quantum behaviour, but what you are describing then is not a molecule; you won't get realistic results that you could compare to experimental results.

2) The equipartition theorem states that rotation or vibration degrees of freedom will have the same energy as translation degrees of freedom, but the same energy will not translate into the same velocity.

For instance, in a classical treatment, the energy distribution function per degree of freedom in equilibrium that we obtain from Maxwell-Boltzmann is

$$f_\epsilon = \sqrt{ \cfrac{1}{\pi \epsilon_i k_B T} } \exp \left[ \cfrac{-\epsilon_i}{k_B T} \right]$$

Taking into account that classical rotational energy is

$$E_{rot} = \frac{1}{2} I \omega^2$$

we can derive the classical angular velocity distribution:

$$f_\omega = \sqrt{\cfrac{I}{2 \pi k_B T}} \exp \left[ \cfrac{-I \omega_i^2}{k_B T} \right]$$

with the moment of inertia $I$ being dependent on the geometry of the molecule. For instance, in the simplest case of a diatomic molecule approximated as a rigid rotor, it is

$$I = \mu d^2$$

where $\mu$ is the reduced mass and $d$ is the interatomic distance.

However, as we said, this is not a correct description of an actual molecule, in which vibration and rotation movements are quantised. In a quantum treatment, the allowed values for rotational energy in a linear molecule are

$$E_{rot} = J \left( J + 1 \right) \cfrac{\hbar^2}{2I} ~ \mathrm{with} ~ J=0,1,2...$$

note that rotational levels are degenerate (with unequal degeneracy $g_j = 2J+1$) and discrete, so we need to use a discrete partition function $Z_{rot}$:

$$Z_{rot} = \sum_{J=0}^{\infty} \left( 2 J + 1 \right) \exp \left[ \cfrac{-J \left( J + 1 \right) \hbar^2 }{ 2 I k_B T} \right]$$

which in principle we can't calculate analytically. For molecules with a very large moment of inertia $I$ and/or at high enough temperatures $T$, the gap in energy between levels becomes very small and we can approximate this sum as a continuous integral over $J$ (the so-called high temperature limit):

$$Z_{rot} \approx \int\limits_{J=0}^{\infty} \left( 2 J + 1 \right) \exp \left[ \cfrac{-J \left( J + 1 \right) \hbar^2 }{ 2 I k_B T} \right] dJ = \cfrac{ 2 I k_B T}{\hbar^2}$$

Under that assumption, the (discrete) angular momentum distribution of a rotational degree of freedom can be derived as

$$F_J = \left( 2 J + 1 \right) \cfrac{ \pi \hbar^3}{ I k_B T} \exp \left[ \cfrac{-2 \pi \hbar^3 \left( J + 1 \right) }{ I k_B T} \right]$$

Note that we're using angular momentum and not angular velocity here because, while the former is a well-defined magnitude in quantum mechanics, it's not clear the later makes physical sense in a quantum treatment of molecules. We could naively calculate a semi-classical "angular velocity" distribution from that angular momentum distribution, but, again, we'd not be actually describing a molecule.

And a final conceptual problem:

3) Rotation and vibration (and electronic) motion is coupled; to give a simple example, in a diatomic molecule, as rotation angular momentum increases, interatomic distance (and therefore moment of inertia) changes; a simple approximation is to introduce a rotational term representing centrifugal distortion, but for higher precision both movements have to be considered at the same time, what makes for much more complex equations.

Keep also in mind that we've restricted ourselves to the simplest cases in rigid diatomic molecules; more complex systems have much more complex mathematics involved and often have to be treated empirically.

• Thank you so much for your detailed and clear explanation. You have answered my question. One extra thing that I was wondering about. Suppose, for simplicity’s sake, a system has a total of 3 possible rotational energy states, the gaps between the energy levels being too large to continuously intergate over J, and you would like to calculate the probability of energy state J = 2. Should one then calculate this probability by E(rot) / Z(rot), using J = 2 and the upper summing limit of Z(rot) being 3 instead of infinity? (the total number of possible rotational states) – Phy May 17 '18 at 13:28
• If there were a limited number of states, yes, that's how you'd do it. In fact, even if you have infinite states and can't do the high temperature limit approximation, you can still approximate the partition function by doing a discrete sum of as many terms as you can afford, as higher states will most likely be very sparsely populated, so while the solution won't be exact, it will be a good approximation. – user41033 May 17 '18 at 16:15
• Thanks for clarifying. Looking back at my previous statement, I realised that Z(rot) is actually the total number of possible states, not the total summed energy of all the possible states. This means that the probability calculation for a state J should be (J(J+1)) / Z(rot) instead of Ej(rot) / Z(rot). Please correct me if I'm wrong on this. – Phy May 17 '18 at 18:01