Angular orientation of exact solution of the Hydrogen Schrödinger Equation

Question

I am currently trying to get a better understanding of the basis sets used for computational quantum chemistry and as a start read ch.6 on the exact solution of the Schrödinger equation of the hydrogen atom in "Physical Chemistry: A Molecular Approach" by McQuarrie & Simons.

The solution for the radial and the polar angle part are just given and not formally derived, but I accept that the exact solution can be written as $$ \Psi(r\,\theta,\phi) = R_{nl}(r)\cdot Y^l_m(\theta,\phi), $$ where $Y^l_m(\theta,\phi)$ is the spherical harmonics function with angular momentum quantum number $l$ and magnetc momentum number $m$.

I just wonder though about the geometric meaning of the angular solution. Since we are looking at a free atom, and in the beginning choose an arbitrary angular orientation of our coordinate system, I would assume that the probability of finding an electron (in whatever excited state) at a particular point in space should be independent of $\theta$ and $\phi$. However, from what I understand this is only true for $s$ orbitals $(l=0)$ and for all orbitals of higher angular momentum number ($l>0$) $\theta$ and $\phi$ play a role.

Can somebody help me to understand? Thanks a lot!

Answer 1

Yes, you are right. If there is no external interaction that breaks the rotational symmetry, we are committed to think that the probability of finding an electron around the nucleus cannot depend on the angular orientation.

I am assuming here that the only interaction is the spherically symmetric Coulomb interaction due to the nucleus.

However, as you correctly noted, if $\ell \neq 0$ and we assume that the state is stationary (eigenstate of the Hamiltonian), we are forced to choose the quantum number $m$ as well, breaking the rotational symmetry. What is worse is that the axes have been positioned completely arbitrarily and the atom does not know how we chose them!

In my opinion, the only way out is to assume that, if $\ell>0$ and the energy $E_{n,\ell}$ is defined, then the state is a mixed state, and not a pure state, of the form $$\rho_{n, \ell} = \frac{1}{2\ell+1} \sum_{m=-\ell}^{m= +\ell}|\Psi_{n,l,m}\rangle\langle \Psi_{n,l,m}|$$ with $$\Psi_{n,l,m}(\vec{r}) = R_{n\ell}(r) Y^\ell_{m}(\theta, \phi)\:.$$ That state is an (incoherent) superposition of pure states with the same energy $E_{n,\ell}$ and, in this sense, it is stationary: its temporal evolution is trivial. It is also independent of the choice of the initial orientation of the axes as $$U_R \rho_{n, \ell} U_R^\dagger = \rho_{n, \ell} \quad for \: every \: R\in SO(3)$$ as is immediately demonstrated using the action of the unitary representation $U_R$ of rotations on spherical harmonics.

In the real world, the description is much more complex since there are other types of interactions which make the previous discussion quite academic: there is the electron spin and thus the spin-orbit interaction and the further interaction due to other electrons around the same nucleus. A better description should include all these interactions at least at the level of corrections.

Finally, the states other than the ground one are unstable, due to the interaction with the electromagnetic field (also when it stays in the vacuum state).

Answer 2

We can find a nice visual representation of the Spherical Harmonics in this wiki article. You can see that in fact, only $l=0$ is spherically symmetric.

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We know that:

• The ground state is the only true stable state. All the other states are subjected to spontaneous emission of radiation
• if we find a system in some excited state, it got that way through absorption, i.e., some external interaction with an EM field.

The interaction usually does not have spherical symmetry. The dipole interaction, $H_{dip}=-\vec\mu\cdot\vec E$ for example, respects rotational symmetry, but w.r.t. the direction of polarization of the electric field. The spherical harmonics are all symmetric to rotations, and usually they are constructed considering the direction of the dipolar interaction.

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The hydrogen atom eigenstates are built by considering some direction. It is necessary, in order to diagonalize the angular momentum operators $L^2,L_z$. The $z$ direction, despite being arbitrary, is fixed after we choose it. And usually, we choose it as the direction of the dipole interaction of the atom with some external EM field. The interaction breaks the spherical symmetry, and the excited states does not follow this symmetry, despite being rotationally symmetric.

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So if you are travelling in the universe and you find an excited hydrogen atom alone, you could be sure that it interacted with something, and the axis of symmetry of the state is telling you what the direction of the interaction was.

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EDIT After reading the fantastic answer of Valter Moretti, I noticed that I completely forgot about the possibility of mixed states. It makes sense that an interaction with completely polarized light gives rise to a perfectly pure state. If we consider unpolarized light, on the other hand, we could have a mixed state as he showed. It seems that in order to have a pure state (for the atom and for the light), we need to choose a direction, losing symmetry.

Answer 3

"Since we are looking at a free atom, and in the beginning choose an arbitrary angular orientation of our coordinate system, I would assume that the probability of finding an electron (in whatever excited state) at a particular point in space should be independent of $\theta$ and $\phi$."

An arbitrary orientation of the coordinate system doesn't imply that a spatial function is independent of angular coordinates, just that the associated outputs will be arbitrarily rotated.

More intuitively, if you are in the center of a spherically symmetric object, like a spherically symmetric cloud, wherever you look (whichever $(\theta, \phi)$ you choose), you see the same thing, regardless of how you are oriented yourself. But that's a special case; if the object has some other shape, there will be directions that look different compared to other directions. In other words, what you see is angle dependent in a well-defined way, it's just that the coordinate-to-value mapping will be determined by how you are oriented.

It's like tilting the coordinate grid of the world map in an arbitrary way - the continents still have the same outlines (the non–spherically symmetric features of the object are still represented in the image), but the new latitudes & longitudes are different. Same principle when the mapping is three dimensional, extending also in the radial direction.