# Shape of atomic orbitals

• I

## Summary:

The non spherical orbitals of spherical Hamiltonian

## Main Question or Discussion Point

Oritals, other than s-orbitals, don't have spherical symmetry while the atomic Hamiltonian does have spherical symmetry. How is this possible?

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hilbert2
Gold Member
Summary:: The non spherical orbitals of spherical Hamiltonian

Oritals, other than s-orbitals, don't have spherical symmetry while the atomic Hamiltonian does have spherical symmetry. How is this possible?
You can compare this to the case of a free particle in 1D. The Hamiltonian operator

##\displaystyle\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}##

is invariant in any coordinate translation ##x\mapsto x+\Delta x##, but the energy eigenstates are of form

##\displaystyle\psi (x) = Ae^{ikx} + Be^{-ikx}##

which are not translation invariant unless ##k=0## and ##\psi## is a constant function.

aaroman
PeterDonis
Mentor
2019 Award
How is this possible?
Because the symmetry of the Hamiltonian shows up in the symmetry of the entire set of solutions, not in every individual solution. The entire set of solutions has spherical symmetry, but not all individual solutions do.

aaroman
Nugatory
Mentor
Orbitals, other than s-orbitals, don't have spherical symmetry while the atomic Hamiltonian does have spherical symmetry. How is this possible?
That's true of any central force problem, even classically. For example the Sun's gravitational field is (too good approximation) spherically symmetrical but the solar system obviously is not.

A. Neumaier
2019 Award
Orbitals, other than s-orbitals, don't have spherical symmetry while the atomic Hamiltonian does have spherical symmetry. How is this possible?
The reason is that the named orbitals commonly drawn are basis-dependent. It is like drawings of a circle from different (basis-dependent) perspectives, which produces unsymmetric ellipses.

The general orbital is quite arbitrary, and the set of all orbitals has spherical symmetry.

aaroman, vanhees71 and TeethWhitener
DrClaude
Mentor
Exercise: Show that the sum of the norm of all orbitals of a given angular momentum ##l##, i.e.,
$$\sum_{m=-l}^l \left| Y_{l,m} (\theta, \phi) \right|^2$$
is spherically symmetric.

aaroman, vanhees71 and TeethWhitener
Suppose that an electron is in a d-orbital, say ##d_{z^2}## . The probability of existence of electron in one direction may be different from that in another direction! I think this discrepancy can be explained as follows:
There is no preferred z-direction and the shape of orbitals helps to determine the dynamics of electrons.

DrClaude
Mentor
Suppose that an electron is in a d-orbital, say ##d_{z^2}## . The probability of existence of electron in one direction may be different from that in another direction! I think this discrepancy can be explained as follows:
There is no preferred z-direction and the shape of orbitals helps to determine the dynamics of electrons.
What is the source of atoms?

If the source is a thermal one, then there will be an equal probability of the atom being in all states of the same energy, as is the case for states that only differ in the magnetic quantum number. In that case, the electronic distribution is isotropic, see my post #6 above.

If the atom is in a specific state, as in your example above, then there was some state preparation (or selection) and there is no requirement for an isotropic distribution, since the preparation process can break the symmetry. In that case, the atom is said to be polarized, and there is a preferred direction for the electron.

hokhani, vanhees71, PeroK and 1 other person
Eigenfunctions of spherically symmetric Hamiltonians actually do have spherical symmetry, just not in the sense that you're thinking of. Say you have a Hamiltonian ##H## with spherical symmetry, i.e. ##[H,U(\theta)]=0## , then eigenstates ##|\omega\rangle## of ##H## will remain eigenstates with the same eigenvalue under rotation.

aaroman, hokhani and vanhees71