# How can two orbitals constructively and destructively interfere simultaneously?

The molecular orbital theory dictates that when two atomic orbitals form molecular orbitals, then two molecular orbitals must form (i.e number of atomic orbitals = number of molecular orbitals). For this to occur there must be bonding molecular orbitals and anti-bonding molecular orbitals. The bonding MO's are caused when the orbitals constructively interfere and the anti-bonding orbitals form as the orbitals destructively interfere. I understand the concept of interference but I do not understand how two atomic orbitals (one from each bonding atom) can simultaneously interfere constructively and destructively. Moreover, I am under the impression that the anti-bonding orbitals exist even if they're not occupied. As a result, is it correct to think that it's the orbitals themselves that are the waves, not the electrons?

In summary:

1. How can two orbitals interfere constructively and destructively?
2. Do anti-bonding orbitals exist when not occupied?
3. Is it the orbitals themselves that are interfering with each other like waves?
• Have a look at chemistry.stackexchange.com/a/15117/4945, maybe it helps a little. – Martin - マーチン Sep 10 '14 at 15:30
• Not so long ago I asked exactly the same question. Dropped two bounties on it. Perhaps you shouldn't until you start learning the real stuff (quantum mechanical explanations). Here is the link (it's now on Physics SE): physics.stackexchange.com/questions/121671/… – Jori Sep 10 '14 at 16:22
• Thanks. Do you know the answers to number 2 and 3 in the summary of the question? – RobChem Sep 10 '14 at 18:31

They do not simultaneously interfere constructively and destructively.

Let me explain. Since the time-independent Schrödinger equation is not separable for molecules due to inter-electronic interaction terms, we can seek solutions to this problem with the variational method.

Let's examine the case of the $\mathrm{H_2}$ molecule. The most natural/physically meaningful/simple choice is to use a trial function that consists of a linear combination of hydrogen 1s atomic orbitals: $$E\left(c_{1},c_{2}\right)=\dfrac{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle }{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}|\psi_{j}\right\rangle }$$

$$H_{ij}\equiv\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle$$

$$S_{ij}\equiv\left\langle \psi_{i}|\psi_{j}\right\rangle$$

$$E\left(c_{1},c_{2}\right)=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}H_{ij}}{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}S_{ij}}$$

After minimizing the energy with respect to $c_1$ and $c_2$, you will end up with two energy levels, each one associated with a wave function: $$E_{+}=\dfrac{H_{11}-H_{21}}{\left(1-S_{12}\right)}\qquad\psi_{+}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}+\psi_{2}\right)$$

$$E_{-}=\dfrac{H_{11}+H_{21}}{\left(1+S_{12}\right)}\qquad\psi_{-}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}-\psi_{2}\right)$$

In other words, by choosing to model the $\mathrm{H_2}$ molecule with two hydrogen $1s$ atomic orbitals, you find two discrete energy levels, each one associated with its wave function (one anti-bonding and one bonding). So this answers your questions:

1. The two atomic orbitals do not simultaneously interfere: The MOs are two different wave functions.

2. Of course the anti-bonding orbital exists when not occupied.

Note that the previous procedure can be generalized for all kind of atomic orbitals.

I want to clarify something about the meaning of the wave function since you're asking: As a result, is it correct to think that it's the orbitals themselves that are the waves, not the electrons?

Let's assume that your system consists of a single particle in a one-dimensional space. The wave function, $\Psi\left(x,t\right)$, is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared, $\left|\Psi\left(x,t\right)\right|^{2}$, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx, gives you the probability for finding the particle between $x$ and $x+dx$ $$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$. Of course, you can integrate the previous equation to find the probability over a certain region of space, $\Omega$ $$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$.