Quantum Mechanics – Relativistic Correction to the Hydrogen Atom and Spherically Symmetric Operators

atomic-physicsperturbation-theoryquantum mechanicsspecial-relativity

We know, the lowest order relativistic correction to the Hamiltonian for the Hydrogen atom is
$$H^{'}=-\frac{p^{4}}{8m^{3}c^{2}}$$
Where,
$$p=-\frac{\hbar}{i}\nabla$$
So, is $p^{2}=-\hbar^{2}\nabla^{2}$ and $p^{4}=\hbar^{4}\nabla^{2}(\nabla^{2})$?

In various sources, for calculating the first order correction to the energy, a fact has been utilized that the perturbation is spherically symmetric.

a) First of all, what should a spherically symmetric operator look like?

b) Secondly, does this mean that the operator $\nabla^{2}(\nabla^{2})$ is spherically symmetric? If so, how to prove that? Or is there any easy way to intuitively understand that? Any help is appreciated.

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An operator is called spherically symmetric when it is invariant under rotations. The Laplace operator is invariant under rotations and translation (the Euclidean transformations).

In fact, the algebra of all scalar linear differential operators, with constant coefficients, that commute with all Euclidean transformations, is the polynomial algebra generated by the Laplace operator. 1

For a hint to how to show that indeed the Laplace operator is invariant under rotations see e.g. this math.SE post. For some intuitive hints on why the Laplace operator is spherically symmetric see this & this posts on math.SE and physics.SE.

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[Physics] Relativistic correction to Hydrogen atom – Perturbation theory

Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $ \psi_{a} $ and $ \psi_{b} $, both of which correspond to energy $ E_{0} $. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent.

Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks into two distinct energy levels, $ E_{1} $ and $ E_{2}$.

The subtlety is that these two distinct energy levels do not necessarily correspond to the two degenerate states (it is not necessarily the case that $ H'\psi_{a} = E_{1}\psi_{a} $ and $ H'\psi_{b} = E_{2}\psi_{b} $). It is possible that the two distinct perturbed energies correspond to linear combinations of the two degenerate states, i.e. $ H'(\alpha \psi_{a} + \beta \psi_{b}) = E_{1}(\alpha \psi_{a} + \beta \psi_{b}) $, and similarly for some other linear combination (orthogonal to the first).

This subtlety is the reason that in degenerate perturbation theory, the first-order corrections to the energy require calculation of off-the-diagonal elements, i.e. things that look like $ \langle \psi_{a} | H' | \psi_{b} \rangle $. This is annoying, because in first order perturbation theory, we only needed to calculate one inner product. In degenerate perturbation theory, we have to compute a whole matrix of inner products to calculate the correction to the energy.

Since it is tedious to compute inner products, it'd be nice to know a trick to find out if the off-the-diagonal elements of the perturbing Hamiltonian are 0. The trick is given and proven on pg. 259-260 of Griffiths.

In the case of your question, the original Hamiltonian is spherically symmetric (Coulomb potential has no angular dependence). Also, the the perturbing Hamiltonian is spherically symmetric. If you look at the form of the angular momentum operator, you will note that it only involves things with $ \theta $ and $ \phi$ (derivatives and cosines and stuff). The nice thing about that, is that if I have a Hamiltonian that is purely radial (depends only on r), than it definitely commutes with L.

All Griffiths is doing is showing that = the conditions of the theorem on pg. 259 are satisfied, which shows that degenerate perturbation theory collapses to non-degenerate perturbation theory.

I'm not familiar with the idea of states being "connected" by a Hamiltonian. I would speculate that two states are connected by a Hamiltonian if the expected value of one state is nonzero given that it is initially in the other state.

Hope this helps!

Quantum Mechanics – Are All Atoms Spherically Symmetric and Why Are Half-Filled Sub-Shells Especially Symmetric

In general, atoms need not be spherically symmetric.

The source you've given is flat-out wrong. The wavefunction it mentions, $\varphi=\frac{1}{\sqrt3}[2p_x+2p_y+2p_z]$, is in no way spherically symmetric. This is easy to check: the wavefunction for the $2p_z$ orbital is $\psi_{2p_z}(\mathbf r)=\frac {1}{\sqrt {32\pi a_0^5}}\:z \:e^{-r/2a_{0}}$ (and similarly for $2p_x$ and $2p_y$), so the wavefunction of the combination is $$\varphi(\mathbf r)=\frac {1}{\sqrt {32\pi a_0^5}}\:\frac{x+y+z}{\sqrt 3} \:e^{-r/2a_{0}},$$ i.e., a $2p$ orbital oriented along the $(\hat{x}+\hat y+\hat z)/\sqrt3$ axis.

This is an elementary fact and it can be verified at the level of an undergraduate text in quantum mechanics (and it was also obviously wrong in the 1960s). It is extremely alarming to see it published in an otherwise-reputable journal.
On the other hand, there are some states of the hydrogen atom in the $2p$ shell which are spherically symmetric, if you allow for mixed states, i.e., a classical probabilistic mixture $\rho$ of hydrogen atoms prepared in the $2p_x$, $2p_y$ and $2p_z$ states with equal probabilities. It is important to emphasize that it is essential that the mixture be incoherent (i.e. classical and probabilistic, as opposed to a quantum superposition) for the state to be spherically symmetric.
As a general rule, if all you know is that you have "hydrogen in the $2p$ shell", then you do not have sufficient information to know whether it is in a spherically-symmetric or an anisotropic state. If that's all the information available, the initial presumption is to take a mixed state, but the next step is to look at how the state was prepared:
- The $2p$ shell can be prepared through isotropic processes, such as by excitation through collisions with a non-directional beam of electrons of the correct kinetic energy. In this case, the atom will be in a spherically-symmetric mixed state.
- On the other hand, it can also be prepared via anisotropic processes, such as photo-excitation with polarized light. In that case, the atom will be in an anisotropic state, and the direction of this anisotropy will be dictated by the process that produced it.
It is extremely tempting to think (as discussed previously e.g. here, here and here, and links therein) that the spherical symmetry of the dynamics (of the nucleus-electron interactions) must imply spherical symmetry of the solutions, but this is obviously wrong $-$ to start with, it would apply equally well to the classical problem! The spherical symmetry implies that, for any anisotropic solution, there exist other, equivalent solutions with complementary anisotropies, but that's it.
The hydrogen case is a bit special because the $2p$ shell is an excited state, and the ground state is symmetric. So, in that regard, it is valid to ask: what about the ground states of, say, atomic boron? If all you know is that you have atomic boron in gas phase in its ground state, then indeed you expect a spherically-symmetric mixed state, but this can still be polarized to align all the atoms into the same orientation.

As a short quip: atoms can have nontrivial shapes, but the fact that we don't know which way those shapes are oriented does not make them spherically symmetric.
So, given an atom (perhaps in a fixed excited state), what determines its shape? In short: its term symbol, which tells us its angular momentum characteristics, or, in other words, how it interacts with rotations.
- The only states with spherical symmetry are those with vanishing total angular momentum, $J=0$. If this is not the case, then there will be two or more states that are physically distinct and which can be related to each other by a rotation.
- It's important to note that this anisotropy could be in the spin state, such as with the $1s$ ground state of hydrogen. If you want to distinguish the states with isotropic vs anisotropic charge distributions, then you need to look at the total orbital angular momentum, $L$. The charge distribution will be spherically symmetric if and only if $L=0$.
A good comprehensive source for term symbols of excited states is the Levels section of the NIST ASD.

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[Physics] Relativistic correction to Hydrogen atom – Perturbation theory

Quantum Mechanics – Are All Atoms Spherically Symmetric and Why Are Half-Filled Sub-Shells Especially Symmetric

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