How Do Quantum Numbers Influence Energy Levels in Atomic Systems?

[PreposterousUniverse]

The quantum number n determines the energy, and for each n the allowed values for the angular momentum quantum number are -(n-1),...,(n-1).
This doesn't seem resonable to me. Classically increasing the orbital angular momentum will result in an increase in the energy of the system. But why is it that quantum mechanically the energy doesn't depend on the orbital angular momentum. Could someone explain this?

 

[Dale]

Classically increasing the orbital angular momentum will result in an increase in the energy of the system.

This is not correct. For a fixed energy changing the angular momentum changes the eccentricity of the orbit. You can indeed have different classical orbits with the same energy but different angular momentum.

 

[PreposterousUniverse]

This is not correct. For a fixed energy changing the angular momentum changes the eccentricity of the orbit. You can indeed have different classical orbits with the same energy but different angular momentum.

Could you give an example of this?

 

[Dale]

Could you give an example of this?

Yes, see section 25.3 here: https://www.lehman.edu/faculty/anchordoqui/chapter25.pdf

Note that section 25.3 repeatedly describes $L$ and $E$ as two constants of motion, not one. And 25.3.14 gives the eccentricity $\epsilon$ as a function of both $L$ and $E$

As a concrete example consider a satellite in a circular orbit at some radial distance, the velocity of the satellite is perpendicular to the radius. Now, consider a second satellite at the same radial distance and with the same speed, but whose velocity is directly towards the planet. The energy is the same in both cases but the angular momentum is zero in the second case and non-zero in the first.

For both quantum and classical mechanics at a fixed energy changing the angular momentum changes the shape of the orbit/orbital. There are lots of differences between classical and quantum mechanics, but this isn't one.

 

[DrClaude]

The quantum number n determines the energy, and for each n the allowed values for the angular momentum quantum number are -(n-1),...,(n-1).

Are you talking about the hydrogen atom? The angular momentum quantum number is never negative.

 

[PreposterousUniverse]

Yes, see section 25.3 here: https://www.lehman.edu/faculty/anchordoqui/chapter25.pdf

Note that section 25.3 repeatedly describes $L$ and $E$ as two constants of motion, not one. And 25.3.14 gives the eccentricity $\epsilon$ as a function of both $L$ and $E$

As a concrete example consider a satellite in a circular orbit at some radial distance, the velocity of the satellite is perpendicular to the radius. Now, consider a second satellite at the same radial distance and with the same speed, but whose velocity is directly towards the planet. The energy is the same in both cases but the angular momentum is zero in the second case and non-zero in the first.

For both quantum and classical mechanics at a fixed energy changing the angular momentum changes the shape of the orbit/orbital. There are lots of differences between classical and quantum mechanics, but this isn't one.

But what happens if we consider only circular orbits? Then we have the relation E=L^2/(2m^2)

 

[DrClaude]

The degeneracy of the energy levels of the hydrogen atom is due to a hidden symmetry of the Coulomb potential. It exists only for single-electron atoms (and only when spin-orbit coupling is ignored). For atoms with more than one electron, the energy of the orbitals depends on both $n$ and $l$.

 

[Nugatory]

But what happens if we consider only circular orbits? Then we have the relation E=L^2/(2m^2)

Considering only circular orbits is tantamount to considering only orbits with zero eccentricity; these are the $l=0$ ones.

 

[vanhees71]

The quantum number n determines the energy, and for each n the allowed values for the angular momentum quantum number are -(n-1),...,(n-1).
This doesn't seem resonable to me. Classically increasing the orbital angular momentum will result in an increase in the energy of the system. But why is it that quantum mechanically the energy doesn't depend on the orbital angular momentum. Could someone explain this?

This is a special symmetry of the (non-relativistic) hydrogen atom. The motion of two particles with a $1/r$ interaction potential has not only the usual space-time symmetries with its "10 conservation laws" from Galilei symmetry, but an additional "dynamical symmetry".

In the classical realm this is reflected in the fact that all bound motions are closed trajectories, i.e., "Kepler ellipses". That means that in addition to the 10 conservation laws what's also conserved is the Runge-Lenz vector, which points from the focus of the ellipse to the "perihelion" of the orbit. This additional symmetry makes the energy independent of the modulus of angular momentum, i.e., you can have the same energy with different values for $\vec{L}^2$, i.e., the total energy for bound motion ($E<0$, i.e., elliptical/circular orbits) in the center-mass system is given by
$$E=-\frac{G m_1 m_2}{2 a},$$
where $a$ is the major semiaxis of the ellipse.

In quantum theory, the Runge-Lenz vector is an additional set of generators of symmetry transformations (due to Noether's theorem in the Hamiltonian formulation of classical mechanics as well as in quantum mechanics). It turns out that together with the angular-momentum operators it generates the symmetry groups for the different cases of the energy eigenvalues: for $E<0$ (the bound-state solutions) it's a SO(4) symmetry group, for $E=0$ (scattering states, referring to parabolic orbits in the classical theory) the symmetry group is that of the Euclidean plane ISO(3) (the usual semidirect product of rotations and translations in the 3D Euclidean plane), and finally for $E>0$ (scattering states, referring to hyperbolic orbits in the classical theory) it's the Lorentz group $\text{SO}(1,3)^{\uparrow}$.

That's why the discrete energy eigenstates of the bound-state solutions depend only on the "main quantum number", $n$, and not as well separately on the orbital-angular-momentum quantum number $\ell$. Due to rotational symmetry for any central-potential problem there's always a degeneracy with respect to the "magnetic quantum number", $m$ (i.e., the eigenvalue $m \hbar$ with $m \in \{-\ell,-\ell+1,\ldots,\ell-1,\ell \}$).

Because for any given $n \in \mathbb{N}$ the $\ell \in \{0,1,\ldots,n-1\}$ and for each $\ell$ there are $(2 \ell+1)$ possible $m$ -values for the bound-state energies you have thus the degneracy
$$\sum_{\ell=0}^{n-1} (2 \ell+1)=n^2$$
for each energy-eigenvalue $E_n=-1 \text{Ry}/n^2$.

 

[Dale]

But what happens if we consider only circular orbits? Then we have the relation E=L^2/(2m^2)

Yes, but not all orbits are circular, neither classically nor in QM. In QM the spherical orbitals are the $l=0$ orbitals, so the same restriction applied gives the same result in both QM and classical mechanics.

Again, there are many differences between quantum and classical mechanics, but this is not one.

 

[PreposterousUniverse]

This is a special symmetry of the (non-relativistic) hydrogen atom. The motion of two particles with a $1/r$ interaction potential has not only the usual space-time symmetries with its "10 conservation laws" from Galilei symmetry, but an additional "dynamical symmetry".

In the classical realm this is reflected in the fact that all bound motions are closed trajectories, i.e., "Kepler ellipses". That means that in addition to the 10 conservation laws what's also conserved is the Runge-Lenz vector, which points from the focus of the ellipse to the "perihelion" of the orbit. This additional symmetry makes the energy independent of the modulus of angular momentum, i.e., you can have the same energy with different values for $\vec{L}^2$, i.e., the total energy for bound motion ($E<0$, i.e., elliptical/circular orbits) in the center-mass system is given by
$$E=-\frac{G m_1 m_2}{2 a},$$
where $a$ is the major semiaxis of the ellipse.

In quantum theory, the Runge-Lenz vector is an additional set of generators of symmetry transformations (due to Noether's theorem in the Hamiltonian formulation of classical mechanics as well as in quantum mechanics). It turns out that together with the angular-momentum operators it generates the symmetry groups for the different cases of the energy eigenvalues: for $E<0$ (the bound-state solutions) it's a SO(4) symmetry group, for $E=0$ (scattering states, referring to parabolic orbits in the classical theory) the symmetry group is that of the Euclidean plane ISO(3) (the usual semidirect product of rotations and translations in the 3D Euclidean plane), and finally for $E>0$ (scattering states, referring to hyperbolic orbits in the classical theory) it's the Lorentz group $\text{SO}(1,3)^{\uparrow}$.

That's why the discrete energy eigenstates of the bound-state solutions depend only on the "main quantum number", $n$, and not as well separately on the orbital-angular-momentum quantum number $\ell$. Due to rotational symmetry for any central-potential problem there's always a degeneracy with respect to the "magnetic quantum number", $m$ (i.e., the eigenvalue $m \hbar$ with $m \in \{-\ell,-\ell+1,\ldots,\ell-1,\ell \}$).

Because for any given $n \in \mathbb{N}$ the $\ell \in \{0,1,\ldots,n-1\}$ and for each $\ell$ there are $(2 \ell+1)$ possible $m$ -values for the bound-state energies you have thus the degneracy
$$\sum_{\ell=0}^{n-1} (2 \ell+1)=n^2$$
for each energy-eigenvalue $E_n=-1 \text{Ry}/n^2$.

Thank you for your elaborate answer! Will be pondering about this.

 

[PeterDonis]

for each n the allowed values for the angular momentum quantum number are -(n-1),...,(n-1).

There are three quantum numbers related to angular momentum, not one, and none of them have the range you give.

The orbital angular momentum quantum number $\ell$ ranges from $0$ to $n - 1$.

The "magnetic" (or orbital z-component) angular momentum quantum number $m$ for a state with a given $\ell$ ranges from $- \ell$ to $\ell$.

The spin quantum number $s$ ranges from $- 1/2$ to $+ 1/2$ and accounts for the fact that any given orbital can be occupied by two electrons with opposite spins.

 

[Hyperfine]

For completeness, it should be mentioned that the intrinsic nuclear spin angular momentum, I, also contributes to the eigen energies in appropriate atomic and molecular systems,that is systems with
S > 0 and I > 0.

 

[Vanadium 50]

Although nobody likes a good discussion of the hidden SO(4) symmetry in the one-electron atom more than me, I don't think that answers the question in the best possible way.

We usually talk about the hydrogen atom having quantum numbers (n, l, m) and energy only a function of n. However, one can also solve it for (k, l,m) with n = k + l, and energy being a function of k and l. The wavefunctions are the same in these two cases: I'm just labeling them differently.

In this view, energy due to the radial motion of the electron is given by k and energy given by the angular motion is given by l.

 

[vanhees71]

Well, but even in this labeling the energy depends only on $n=k+l$ and not separately on two quantum numbers, as you'd expect without the additional dynamical symmetry of the $1/r$ potential.

 

[Vanadium 50]

Yes, that view provokes the question "why are the k levels the same as the l levels", but most people are OK with the answer ":they have to be something - it could have been higher, could have been lower, but it turns out its the same" to start with.

I think it also makes the "hidden symmetry" argument more explicit - you are now no longer why the energy depends on something and not on something else; you are explaining why two sets of energy levels are the same. "Symmetry" as a one-word answer works, in my view, better for these kinds of questions.

When one grows up to grad-level quantum mechanics, one sees immediately what's going on: SO(4) breaks to SU(2) x SU(2) and one of those SU(2)'s is angular momentum. It also means that the Stark Effect falls right out, provided you know where to start.