# A Atomic Orbitals and Symmetry

1. Jul 20, 2016

### hokhani

The Hamiltonian of an atomic electron is spherically symmetric so we expect to have symmetric distribution of electrons around the nucleus. However, as an example, p-orbitals don't have spherical symmetry and $p_x$-orbitals imply that electrons may be found in the x-direction with higher probability compared with another direction between x and y axis! What is the reason of this discrepancy?

2. Jul 20, 2016

### kq6up

In my mind, it has to do with the wave nature of the probability amplitude. They form standing waves. If one things about the way standing waves can form on a drum head. At higher resonances above the fundamental these standing waves do not have rotational symmetry in the 2-D drum head.

Hopefully that helps.

Regards,
KQ6UP

3. Jul 21, 2016

### Truecrimson

A wave function does not have to be invariant under a symmetry of the Hamiltonian. Take for example a free particle. the Hamiltonian is translationally invariant, but it has plane waves as solutions. In fact, any linear combination of plane waves, which mean any function (loosely speaking, ignoring mathematical subtlety for a moment) will also be a solution.

4. Jul 21, 2016

### Truecrimson

The s orbital is special because it is a unique (non-degenerate) ground state, which we can prove must have the same symmetry as the Hamiltonian. Let $H$ be a symmetric Hamiltonian, $U$ be the symmetry transformation $U^{\dagger}HU=H$, and $|\psi \rangle$ be the unique ground state $H|\psi \rangle = E_0|\psi \rangle$. Then $$H (U|\psi \rangle ) = UH|\psi \rangle = E_0(U|\psi \rangle).$$ That is, $$U|\psi \rangle = |\psi \rangle .$$

5. Jul 22, 2016

### Khashishi

The orbitals are just particular solutions to the Schrodinger equation. The actual orbit will be some linear combination of them. The linear combination will obey whatever symmetries are imposed by the environment.
You should know that the angular part of the orbital is just given by spherical harmonics. Spherical harmonics obey various summation rules:
$\sum\limits_{m=-l}^l |Y_l^m(\theta, \psi)|^2 = \frac{2l+1}{4\pi}$
Depending on the convention, the coefficient may vary, but the point is that there is no angular dependence in the sum.
An atom in equal superposition of px, py, pz will be spherically symmetric.

6. Jul 22, 2016

### hokhani

Thanks all. But I don't get convinced. Although the wave function of electron with energy $E_p$ is a linear combination of the three p orbitals but when we measure experimentally, the electron would be only in $p_x$or $p_y$ or $p_z$. In other words, we can never detect the electron somewhere between $p_x$ and $p_y$.

7. Jul 23, 2016

### Truecrimson

If I understand correctly, you accept that the wave function could be a linear combination of the orbitals. Then the fact that the electron will be found in one of these orbitals must be true, because these orbitals is a basis set of the wave functions! That "the electron would be only in $p_x$ or $p_y$ or $p_z$" does not mean that the wave function was $p_x$ or $p_y$ or $p_z$.

8. Jul 25, 2016

### Khashishi

No, we don't measure the electron in these states. We can measure the atom in a general p state, but not in the $p_x$ or $p_y$ state. In the case of hydrogen, we measure it in various states which we label with term symbols which describes an energy eigenstate under the effects of spin orbit coupling.
For example,
2p $^2\mathrm{P}^o_{1/2}$
That describes an electron in the 2p state, which is coupled to the spin such that the total angular momentum quantum number is j=1/2.
Nowhere do we claim that the electron is in a $p_x$ or $p_y$ state.

Now, if there is an external field, like a magnetic field in the z direction, then we can put the atom in a $p_z$ state, since the energy eigenvalues are separated. To measure the $p_x$ state, we have to put an external field in the x direction to separate the energies.