# Symmetry of an in-phase and an out-of-phase wavefunction “approaching” each other in an MO

Hiho.

I have a question concerning this image (Chemistry, Catherine E. Housecroft et al., Pearson Education, 2010).

In the text, it states the following:

"Each of the new molecular wavefunctions, $\Psi$(in-phase) and $\Psi$(out-of-phase), has an energy associated with it wich is dependent of the distance apart of the two hydrogen nuclei. As Figure 4.13 [provided image] shows, the energy of $\Psi$(in-phase) is always lower than $\Psi$(out-of-phase)."

The in-phase-function has a stabilized minimum, but the out-of-phase-function seems stable only at an infinite distance. For my intuition, the two graphs should be symmetric, since the two wavefunctions are alike and approach each other equally. Why do they behave like this?

Why are those functions not symmetric?

A following question would be, when I combine those two functions, I'll get a minimum at the x-value I marked with "minimum". Would this be the distance at which a nucleus in a stable diatomic molecule would be found like in this figure?

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Thank you for your help and have a nice day :)

• Yes. As per the axis label on the chart, 'x' is the optimal inter-nuclear distance for a wave function in-phase. Note that when considering a diatomic, the wave function is represented by a linear combination of atomic orbitals to form molecular orbitals, not just two 1s orbitals. – LordStryker Mar 16 '14 at 15:28
• OK, my second question is answered. But then again, how does it come, the two functions are not symmetric? PS: Yes, thanks for the hint; in this particular example, it's H2. – ste Mar 16 '14 at 17:29

## 1 Answer

Considering this diatomic molecule, the total wavefunction $\Psi$ is formed via a linear combination of the atomic orbitals $\psi_1$ and $\psi_2$ (which is an approximation, but not a too bad one). $$\Psi^+ = c_1 \psi_1 + c_2 \psi_2$$ $$\Psi^- = c_1 \psi_1 - c_2 \psi_2$$

The (electronic) energy of the system can be described using the following (stationary) Schrödinger equation: $$H_\text{el} \Psi_\text{el} = E_\text{el} \Psi_\text{el}$$

So somehow this all factors into each other and in the end gives you the curves (for the potential energy of the system vs. internuclear distance) that you posted. Where is the symmetry? Well, it is still there, with respect to the center of one atom. If you imagine rotating the functions around the $y$ axis, you will get a symmetric image. And you're allowed to do this because space expands in all three directions. So in theory (if you follow through and expand the function to everywhere) you get a 3D image of the energy at every point around one atom.

But: The wavefunction in itself is not all that interesting to us, because there is no "real" physical information that can be directly extracted from it. But fret not, we can at least get out the electron probability density by complex-squaring it: $$\rho = |\Psi|^2 = \Psi \Psi^*$$

And this is where those nice small bubbles come into play that we see in your second figure. And we have again found the symmetry that you seek, because both linear combinations have $D_{\infty h}$ symmetry (I think... I'm not all too good with point groups and the such).