Quantum-Visualizing Molecular Vibrations

This post is about a work of mine which was published in June 2019. Do you have an idea what happens if a molecule vibrates? If you have been exposed to molecular vibrations you might think of little balls (representing atoms) oscillating around their equilibrium position, being held together by bonds that work like springs. For a water molecule, you may remember something like this:

Unfortunately those are not the vibrations of a water molecule. Those are normal modes and they represent certain displacements of the nuclei (not atoms) from their equilibrium position.¹

To understand a molecule microscopically, you can look at it as an isolated system and solve the quantum-mechanical equation that describes its state, the Schrödinger equation. Molecules are composed of nuclei and electrons, e.g. a water molecule consists of three nuclei (one oxygen and two hydrogen nuclei) and 10 electrons.² With some clever tricks you can reduce the molecular (or electron-nuclear) problem to a problem for the nuclei alone.³ However, even the nuclear problem alone is pretty hard: You need to find a state function \(\psi\) which depends on the coordinates of \(N\) nuclei. Each nucleus has an \(x\)-, \(y\)-, and \(z\)-coordinate, thus your \(\psi\) depends on \(3N\) coordinates. For a water molecule, these are already \(3 \times 3 = 9\) coordinates, while for ethanol (alcohol) we already have \(3 \times 9 = 27\) coordinates.⁴ Thus, the state function \(\psi\) “lives” on a high-dimensional space and is thus a tiny bit difficult to comprehend.

Normal modes are combinations of the \(x\)-, \(y\)-, and \(z\)-coordinates of your nuclei. In the animation above you see three such combinations. The normal modes are chosen such that the state function \(\psi\) is approximately a product of functions which each depend on only one coordinate. As equation:

\(\psi \approx \phi_1(q_1) \phi_2(q_2) \dots \phi_{3N}(q_{3N}) \)

For an ethanol molecule, for example, this means that we need not to find \(\psi\) by solving a 27-dimensional problem but we only need to find 27 \(\phi_j\)-functions by solving 27 one-dimensional problems. To give some numbers: Say, you want to find the state function \(\psi\) directly with a computer on a discrete grid with 10 points per coordinate. Then, to solve the full problem you need \(10^{27}\) points.⁵ By using normal modes, you only need \(27 \times 10 = 270\) points to get a pretty good approximation for \(\psi\). That is certainly something to consider.

By the way, does anything move? Not at all. This is because we treat our molecule as an isolated system, and to have a time parameter you need to include something else that you use as a clock. The picture above only shows the coordinates with the help of an animation, but those are really only the coordinates for the state function \(\psi\). It is like visualizing the \(x\)-coordinate of a cartesian coordinate system by moving a ball along this direction. However, the (absolute) square \(|\psi|^2\) represents the probability of finding the nuclei somewhere in the high-dimensional space when the position of the nuclei is measured.⁶⁷ We can excited vibrations, for example by absorption of photons (irradiation with light). Those excited states correspond to a different \(|\psi|^2\), but, again, nothing actually moves. It is only that if the molecule vibrates in some way (is in a certain vibrational state), a measurement would yield a different \(|\psi|^2\) than if it vibrates in another way. All the ways in which the molecule can vibrate are given by the solution of the Schrödinger equation.

We thus need to look at the state function \(\psi\) to understand molecular vibrations. How does it look like when we use normal modes? For a water molecule, for example, it is the product

\(\psi = \phi_1(q_1) \phi_2(q_2) \dots \phi_9(q_9)\).

Six of the functions in this product, say \(\phi_1\) to \(\phi_6\), correspond to translation and rotation of the molecule in space. For the remaining three, the normal modes \(q_7\), \(q_8\) and \(q_9\) are shown in the animation above. They carry the pictorial names “symmetric stretch”, “bending”, and “asymmetric stretch” and they correspond to vibrations.⁸ The corresponding functions \(\phi_7\), \(\phi_8\), and \(\phi_9\) are eigenstates of the harmonic oscillator. What this means is not so important except that we can “excite” each of these functions, for example by photon absorption (irradiation with light). Excited states typically extend a little further in space. If you imagine molecular vibrations as little balls connected by springs, this just means that the vibration is a bit stronger. However, excited states also have nodes, that is, regions where the nuclei have zero probability to be. This feature cannot be explained in the classical picture. To see what I am talking about, have a look at the following picture and its explanation.

OK, but how does a vibrational state of a molecule look like? The probability density \(|\psi|^2\) in its high-dimensional space is not so useful.¹⁰ Fortunately, there is the possibility of visualizing the vibrational state in a three-dimensional picture, although we will loose some information compared to the function \(|\psi|^2\). The quantity which we need to look at is the marginal one-nucleus probability density of finding some nucleus somewhere independent of where the others are.¹¹ Easy said, easy done. Just a little bit of algebra, a bit of coding, and a few assumptions later¹² we can visualize vibrations! Let us do that for the water molecule, for four vibrational states:

The good thing about the water molecule is that it only consists of three nuclei, so we can look at the one-nucleus density in the molecular plane¹³ and need not worry about how to visualize it in three dimensions. The top-left panel of the above figure shows the ground state. Unsurprisingly, it looks pretty boring, with three featureless blobs at the equilibrium position of the nuclei. As the oxygen nucleus is much heavier than the hydrogen nuclei, the probability density is much more localized and we need the inset in the panel to see the one-nucleus density without having to bring out our microscope.

More interesting are states that correspond to excitations of vibrational modes. A good example is the state corresponding to the first excitation of the bending normal mode, shown in the bottom-right panel. The probability density has a node along this mode (this rhymes!) which is clearly seen in the one-nucleus density: For the hydrogen nuclei, the probability to be found at their equilibrium position is zero. However, while we also see that the probability to find the oxygen nucleus at its equilibrium position is less than for the ground state, it is not zero. This shows us that something interesting happens when we perform the reduction from the full probability density \(|\psi|^2\) of all nuclei to the one-nucleus probability density.

In my article, I gave some rules for predicting features of the one-nucleus probability density from the normal modes. If you are still reading this text and your head has not yet exploded, you either already know something about molecular vibrations, you are extremely bored, or you can (at least partially) follow me. So, let me challenge you once more with the following figure:

Fig. 4 corresponds to the bottom-right panel of Fig. 3, the first excitation of the bending mode B. Although you get the one-nucleus density by doing a series of non-trivial steps,¹⁴ note that the B mode displaces the hydrogen nuclei orthogonal to the S and A mode. For example, if you look at the left hydrogen nucleus in the three panels of Fig. 4, you see that S and A move it to the top left, while B moves it to the top right. If a normal mode is the only normal mode displacing some nucleus into a certain direction, its excitation will clearly be visible in the one-nucleus density of that nucleus. In contrast, excitation of the B mode is not so clearly visible at the oxygen nucleus: From Fig. 4 we see that the B mode displaces the oxygen nucleus into the same direction as the S mode, hence the two modes “combine” to give the final one-nucleus density.

The supplemental information of my article has further examples. In fact, it turns out that we might not even see qualitatively in the one-nucleus probability density if the nuclei are in the vibrational ground state or in a vibrationally excited state. Especially for large molecules, the one-nucleus probability density of vibrationally excited states looks like that of the ground state, because for most nuclei there are many normal modes displacing it in the same direction.¹⁵ This is a bit unfortunate, as the one-nucleus probability density is, in principle, much better accessible for large molecules than the full many-dimensional nuclear probability density, but it is not totally unexpected.¹⁶

Truth be told, measuring the one-nucleus density is still hard. Neutron scattering could be a way, but it is still far from applications like electron tunneling microscopy where we can get nice pictures of (the one-electron density of) molecules. My study is more of educational interest, as it provides a new angle to the topic of molecular vibrations and a quantum-mechanical way to visualize them. In any case, I have published my code to compute one-nucleus densities in the normal mode approximation. The code is reasonably general and user-friendly, but don’t expect to be able to use it if you have no idea of Python coding or the theory of molecular vibrations.

So, in summary:

• Normal modes are not molecular vibrations.
• Molecular vibrations are a complicated topic and looking at them quantum-mechanically is quite different from classical mechanics.
• Molecular vibrational states can be visualized even for large molecules with the one-nucleus density.
• The one-nucleus density can be predicted qualitatively from the normal mode.
• The one-nucleus density seems to be of limited practical use.

Good to know.

1. The equilibrium configuration is the relative configuration of the nuclei which correspond to minimal energy. In this post I will not use the word “configuration” but talk of positions, just to make it all sound a bit less technical.
2. If you want to think of molecules as being composed of atoms (which is approximately true), 8 electrons come from the oxygen atom and one electrons comes from each of the hydrogen atoms.
3. The classic trick to separate the nuclei from the electrons is the Born-Oppenheimer approximation. However, the separation can be achieved exactly by separating the state function (wavefunction) of the molecule into a marginal state function for the nuclei and a conditional state function of the electrons (that depends conditionally on the coordinates of the nuclei). This is called the Exact Factorization and this is what I use for most of my research. In any case, the full electron-nuclear problem is reduced to a nuclear problem where the effect of the electrons is encoded in effective interactions (scalar and vector potentials).
4. In general, you can remove 3 coordinates for translation of the molecule in space and you can approximately separate 3 (2 for linear molecules) further coordinates for rotation of the molecule in space. As the system is assumed to be isolated, translation of the whole molecule to a different place or rotation of the whole molecule changes nothing.
5. A 1 TB hard drive can hold ca. \(1.7 \times 10^{10}\) double-precision floating point numbers. So you need 58 000 000 000 000 000 of such hard drives just to store the (actually pretty small) grid.
6. Have a look at the TiP 3 post to learn why we cannot work a three-dimensional space but have to work with high dimensional spaces in quantum mechanics.
7. Such a measurement is a coincidence measurement where you determine the relative position (configuration) of the nuclei with respect to each other.
8. Vibrations cannot be fully decoupled from rotations, hence the product form of latex]\psi[/latex] is an approximation. It only works if the nuclei are close to their equilibrium position.
9. There is also no time in the quantum-mechanical picture because the system is treated as an isolated system, which makes the interpretation more difficult than it already is. Quantum mechanics is not so intuitive.
10. A problem related to the high dimensionality is that we typically work with unintuitive relative coordinates to reduce the necessary number of dimensions.
11. For the “experts”: Coincidence measurements, where the location of multiple particles relative to each other is measured, allow access to the full \(|\psi|^2\). However, we often work with marginal densities: For example, electron microscopy of a molecule on a surface provides an image of the marginal one-electron density, i.e., of the density of finding any electron somewhere (relative to the surface). Hence, I talk about the marginal one-nucleus density, even ignoring the type of nucleus (but you know which nucleus it is from the spatial arrangement).
12. We need to localize and orient the molecule in space. Both is a bit questionable as we need to construct a wavepacket for this purpose. We can, however, also think of the marginal probabilities to be discusses to some extent as conditional probabilities: Given my molecule is somewhere and somehow oriented, how does the marginal vibrational density look like?
13. The plane is defined by the equilibrium positions of the nuclei.
14. Multiply the one-dimensional densities of all normal modes, transform back to Cartesian coordinates of the nuclei, and integrate over all but the coordinates of one nucleus. Do this for all nuclei and sum the densities.
15. There are counter-examples if we find a normal modes that is the only one displacing a nucleus in a certain direction of if a normal mode displaces a nucleus much more than other normal modes.
16. Again for the “experts”: The problem is much more severe for electrons as they are much more delocalized compared to the nuclei. Thus, the one-electron density of an electronically excited state differs little from that of the electronic ground state. This is a major obstacle for density functional theory (DFT) as it is based on the one-electron density: Clearly, this object can be handled much better computationally than the many-electron density, but excited-state DFT is a challenge.