# Trying to understand the statement: 'Resonance is not a flickering between the contributing states. '

So, the famous resonance definition: it is the weighted average of different Lewis structures....

Well, this is purely wrong.

Resonance is nothing but quantum 'superposition of the wavefunctions representing different electron distributions in the same nuclear framework.'-Peter Atkins.

For-instance, to describe the wavefunction of $$\ce{HCl}$$, quantum superposition is applied viz. $$\psi_{\ce{HCl}}= \psi_{\ce{H^+ Cl^-}}C_1 + \psi_{\ce{H-Cl}}C_2$$ where $$C_1$$ is the probability-amplitude to find the electron-pair at $$|\psi_{\ce{H^+ Cl^-}}\rangle$$ provided it was prepared in $$|\psi_{\ce{HCl}}\rangle$$ that is, $$C_1 =\langle \psi_{\ce{H^+ Cl^-}}|\psi_{\ce{HCl}}\rangle;$$ also $$C_2 = \langle \psi_{\ce{H-Cl}}|\psi_{\ce{HCl}}\rangle$$ and $$C_1^2 + C_2^2= 1.$$

As written by Peter Atkins in his book Elements of Physical Chemistry:

A better description of the wavefunction for the molecule is a superposition of the covalent & ionic descriptions, & we write (with a slightly simplified notation) $$\psi= \psi_{\ce{H_Cl}} + \lambda\psi_{\ce{H^+ Cl^-}}$$ with $$\lambda$$ some numerical coefficient. In general, we write $$\psi = \psi_\text{covalent} + \lambda \psi_\text{ionic}$$ [...]According to the general rules of quantum mechanics, in which probabilities are related to squares of wavefunctions, we interpret the square of $$\lambda$$ as the relative proportion of the ionic contribution. If $$\lambda^2$$ is very small, the covalent description is dominant. If $$\lambda^2$$ is very large, the ionic description is dominant.

[...] The interpretation of the wavefunction, which is called resonance hybrid, is that if we were to inspect the molecule, then the probability that it would be found with an ionic structure is proportional to $$\lambda^2$$. For instance, we might find that $$\lambda = 0.1,$$ so the best description of the bond in the molecule in terms of a wavefunction is a resonance structure described by the wavefunction $$\psi= \psi_\text{covalent} + 0.1\psi_\text{ionic}.$$ This wavefunction implies that the probabilities of finding the molecule in its covalent & ionic forms are in the ratio $$100:1.$$

This is what is known as wavefunction-collapse in Copenhagen interpretation which means $$\psi$$ collapses to one of its base state after the measurement; here the wavefunction is more prone to collapse to the covalent description. Hence, in this case, the state of the molecule has mostly covalent properties.

I've pretty much conceived the upper part; but speaking at the present context of the title of the question, let me start from Peter Atkins' description of benzene in his book Elements of Physical Chemistry:

One of the most famous examples of resonance is in the VB description of benzene, where the wavefunction is written as a superposition of the wavefunctions of the two covalent Kekulé structures: $$\psi = \psi_{\text{Kek}1} + \psi_{\text{Kek}2}$$ The two contributing structures have identical energies, so they contribute equally to the superposition. The effect of resonance in this case is to distribute the double-bond character around the ring and to make the lengths and strengths of all the carbon-carbon bonds identical. The wavefunction is improved by allowing resonance because it allows for a more better description of the location of the electrons, and in particular the description can adjust into a state of lower energy. ...

The statement that baffled me comes from Atkins in the last paragraph of resonance as:

Resonance is not a flickering between the contributing states: it is a blending of their characteristics, much as a mule is a blend of a horse and a donkey. It is only a mathematical device for achieving a closer approximation to the true wavefunction of the molecule than that represented by any single contributing structure alone.

The cause of my bafflement is due to the following excerpt of Feynman's lecture The benzene molecule:

There is a mystery about this benzene molecule. We can calculate what energy should be required to form this chemical compound, because the chemists have measured the energies of various compounds which involve pieces of the ring—for instance, they know the energy of a double bond by studying ethylene, and so on. We can, therefore, calculate the total energy we should expect for the benzene molecule. The actual energy of the benzene ring, however, is much lower than we get by such a calculation; it is more tightly bound than we would expect from what is called an “unsaturated double bond system.” Usually a double bond system which is not in such a ring is easily attacked chemically because it has a relatively high energy—the double bonds can be easily broken by the addition of other hydrogens. But in benzene the ring is quite permanent and hard to break up. In other words, benzene has a much lower energy than you would calculate from the bond picture.

Now we want to resolve these mysteries—and perhaps you have already guessed how: by noticing, of course, that the “ground state” of the benzene ring is really a two-state system. We could imagine that the bonds in benzene could be in either of the two arrangements.[...] You say, “But they are really the same; they should have the same energy.” Indeed, they should. And for that reason they must be analyzed as a two-state system shown in Fig.

Each state represents a different configuration of the whole set of electrons, and there is some amplitude $$A$$ that the whole bunch can switch from one arrangement to the other—there is a chance that the electrons can flip from one dance to the other.

In the blocked-letters Feynman told that if at one time the molecule be found in $$|1\rangle$$, at some time later, it may be found at $$|2\rangle$$ (& vice-versa) as the Hamiltonian elements $$H_{12},H_{21}$$ are not zero but $$-A$$, the probability of transition per unit time.

As we have seen, this chance of flipping makes a mixed state whose energy is lower than you would calculate by looking separately at either of the two pictures in Fig. Instead, there are two stationary states—one with an energy above and one with an energy below the expected value. So actually, the true normal state (lowest energy) of benzene is neither of the possibilities shown in Fig., but it has the amplitude $$1/\sqrt2$$ to be in each of the states shown. It is the only state that is involved in the chemistry of benzene at normal temperatures.

Why should one found the benzene at $$|1\rangle$$ at one time but at other time, at $$|2\rangle?$$ Because the transition-amplitudes are non-zero and 'the whole bunch can switch from one arrangement to the other—there is a chance that the electrons can flip from one dance to the other.'The electron can flip from one state to the other. But doesn't this statement is at dagger's drawn with the common statement 'Resonance is not a flickering between the contributing states.'? I am completely sure both Feynman and Atkins is telling right, I am misinterpreting both the statements but don't know how. Can anyone please, please explain me how does flipping of the electrons from one dance to the other doesn't contradict the statement Resonance is not a flickering between the contributing states?? Please help.

• – Wildcat Oct 14 '15 at 10:57
• – bon Jul 18 '16 at 12:52