Is a boron atom isotropic in the absence of a field? If so, how can one write its electronic state?

An atom in the absence of any field should obey a spherical symmetry (unless there's spontaneous symmetry breaking somehow?), but any linear combination of $p$ orbitals seem to possess a privileged axis of orientation. How can one write the state of an electron in a $p$ orbital in an isotropic way?

Writing the Slater determinant with the s orbitals doesn't seem to change anything unless I'm mistaken, but it would be strange for the symmetry of the excited state to depend on the ground state occupation anyway.

  • 1
    $\begingroup$ Classically the solar system is governed by a set of isotropic forces, but this symmetry is (almost) always broken by the angular momentum of the initial conditions and results in a preferred axis perpendicular to the plane of planetary motion. The same thing occurs in quantum mechanics. $\endgroup$ – By Symmetry Sep 20 '18 at 12:02

That's basically determined by how the boron atom got there and what its last interaction was. Atomic eigenstates are stationary in the absence of any external interactions, so the state of the atom will retain the anisotropies of the last interaction it had with external systems.


This is (provably) impossible if the electron is in a pure state.

What you can do is assert that the electron is in the $p$ shell but that this is the only information obtainable, in which case the usual practice is to assign to the system the maximally mixed state that's consistent with the known information about it, so that it will not be describable using a wavefunction, and you will need to use a density matrix instead.

For a $p$ electron for which no orientation information is known, this maximally-mixed density matrix reads $$ \hat \rho = \frac13\bigg( |p_-⟩⟨p_-| + |p_0⟩⟨p_0| + |p_+⟩⟨p_+| \bigg); $$ this is the state you take as the starting point if e.g. you're calculating the photoionization electron momentum spectrum of a boron atom in vacuum which you know is in its electronic ground state but which got to that vacuum e.g. on a gas jet that randomized its orientation in ways you cannot calculate. This density matrix is fully isotropic, and it can also be re-expressed as $$ \hat \rho = \frac{1}{4\pi} \int_{\mathbb S^2} |p_{\hat n}⟩⟨p_{\hat n}| \mathrm d\Omega_{\hat n}, $$ where $|p_{\hat n}⟩$ is an eigenstate of ${\hat n}\cdot \hat{\mathbf L}$ (with any of the three eigenvalues), and the integration is evenly weighted over solid angle over the orientation of the unit vector $\hat n$, which then makes the isotropy automatic.


$$\def\hH{\hat H} \def\hR{\hat R} \def\ket#1{|#1\rangle}$$


True, but this does not imply all states have to be isotropic. Let $\ket\xi$ be an eigenstate of $\hH$ to the eigenvalue $E$. All what you can say is that $$\ket{\xi'}=\hR\ket\xi$$ (with $\hR$ a rotation unitary operator) is still an eigenvector of $\hH$, to the same eigenvalue.

Then two possibilities arise:

  1. State $\ket\xi$ is invariant: $\ket{\xi'}=\ket\xi$. You really have an isotropic state.
  2. $\ket{\xi'}$ is a different vector from $\ket\xi$. This means that eigenvalue $E$ is degenerate.

The first situation prevails for $s$ orbitals, the second one for $p$ orbitals (and for higher angular momenta, of course).

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    $\begingroup$ Wish I could accept two answers, both are answering two different aspects of the question. $\endgroup$ – Xelote Sep 21 '18 at 10:36

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