Quantum Chemistry 6.10 - Rigid Rotor Wavefunctions

[@ianlee5812]

I noticed that the wavefunction for the rigid rotor be described as identical to the particle in a sphere, except l is replaced with J and ml is replaced with ImI? I also noticed that the eigenenergy is identical to the particle in a sphere, except l is replaced with J. Is all this a coincidence or are the 2 models connected in some way?

[@nkyu3035]

at 8:00, you mentioned each spehrical harmonics mean 1s, 2s, 2p, 3s, ... in comment below, you also have written that "The quantum numbers J and m (Psi_J,m) from the rigid rotor are the same as the quantum number l and m (Psi_n,l,m) from the hydrogen atom." then there's no quantum number in spherical harmonics that matches 'n' of (Psi_n,l,m). so, shouldn't they be ns, np_z(?), np_x(?) and np_y(?), nd_(??)0, and so on?

[@TMPChem]

Good question. The principle quantum number n is equal to the first number in the orbital. This determines the radial shape of the orbital. It determines the energy of the orbital. Rigid rotor does not have an analog for this quantum number, because r is fixed. The orbital angular momentum quantum number l determines s, p, d, f, etc. It determines the shape of the orbital. This is equivalent to the quantum number J from rigid rotor (sometimes also called l). The z-axis orbital angular momentum quantum number m is equivalent to the m from rigid rotor. The value of m determines things like p_z vs p_x vs p_y, etc. It determines the angular orientation of the orbital (what axis it points towards).

[@nkyu3035]

is 'the Center of Mass in the drawing' the orign of Spherical polar coordinate to represent Psi(Theta,Phi)? I know the motion of system and the shape of Psi don't always match, but somehow it seems wrong for the shape Psi(with r=l) to have the larger radiance than two circles in the drawing, I mean l > l_1 or l_2 .

[@TMPChem]

There are two different ways we can represent this system in angular coordinates. The first would be two have two particles of mass m1 and m2, each a fixed distance r1 and r2 from the origin, where the center of mass is the origin of the spherical polar coordinate system and each particle has their own values of theta and phi, though they are not independent, because the center of mass must remain at the origin. This is a "two-particle" representation. Instead, since the coordinates are co-dependent, we prefer a "one-particle" representation. Here there is one particle of mass [ mu = (m1 * m2) / (m1 + m2) ], and distance r from the origin. Effectively one atom of infinite mass is fixed at the origin, and another of mass mu is free to rotate to any value in theta or phi. This makes it much easier to interpret a single value of theta and phi for the rigid rotor wavefunctions.

[@nkyu3035]

I got it. thank you!

[@minkway522]

Should the Hamiltonian Operator be positive L^2/2I ? I don't know, but it is positive in your other two videos: "Rotation Operator" and "Angular Momentum Eigenvalues".

[@TMPChem]

You're correct. The Hamiltonian operator should be positive L^2/(2 I). I probably made the typo thinking about the fact that L^2 starts with a negative sign (- hbar^2 ...). This is why it is wise to take are of sign conventions, as negative signs are easily lost during algebra if we're not careful.

[@nkyu3035]

Why did you mention Capital-Phi(Phi+2pi)=Capital-Phi(phi)? and Why is the coefficient of Capital-Phi_m equal to 1/sqrt(2pi)? and at the right top eqn, it looks like the position of m and J of spherical harmonics has to be changed.

[@TMPChem]

I mention that just to note that phi is periodic, and every 2pi radians (360 degrees) of rotation results in starting over again at the same point, thus the restriction that 0

[@nkyu3035]

uhoh.. I feel like i'm being fooled. realized again I shouldn't study in bad condition. Thank you anyway!

[@minkway522]

Does the quantum number "m" have any physical meaning? I saw previous videos so I know where did j come from, but not very clear about "m".

[@TMPChem]

The quantum numbers J and m (Psi_J,m) from the rigid rotor are the same as the quantum number l and m (Psi_n,l,m) from the hydrogen atom. J is related to the total angular momentum of the molecule, and m is related to its orientation. J tells you how fast the molecule is rotating, and m tells you what axis it's rotating around (x, y, z, something else).

[@minkway522]

Got it! Thank you so much!

[@shraddhaverma4111]

I didn't got it.... 😕 that 5th line is confusing with peach color one

[@TMPChem]

Hi Ezra. We took the Schrodinger equation for the rigid rotor (in terms of the angular momentum squared operator) and applied the differential equations technique "separation of variables" to it. In separation of variables, we assume that a function of multiple variables can be expressed as a product of one-dimensional functions of each of its variables. We then substitute this expression, and try to separate the equation such that each side of the equals sign only depends on one variable. If this can be done, then each side must be equal to a constant, and we can solve each side of the equation individually. This particular differential equation is still quite difficult, so we're not showing the solution, but just stating what the result is. It's presented here to give a little bit more of an idea of where the rigid rotor solutions come from, up to the point where the math becomes more advanced than we're interested in presenting here.

[@nikitashlapakov8425]

I think there are mistakes in the naming of harmonics with the hydrogen orbitals. There should not be any n-quantum numbers. The harmonics just show the shape of the orbitals. The correct sequence is: s, p(z), p(x) and p(y), d(z2), d(yz) and d(xz), d(xy) and d(x2-y2). But anyway, thank You so much for Your labor!

[@yousefmohsen6287]

Exactly!