# Wave functions of molecules?

1. Oct 6, 2007

### Mephisto

I am taking a second course in Quantum Mechanics right now, and one thing is bugging me... The wave function gives us the probability distribution of a particle being in some position. But which part of particle is it talking about? The center of the particle? But then how can you give a wave function for something more complex like a molecule? Shouldn't your wave function then kinda give you a probability distribution of the molecule occupying a set of points in space or something?

My guess was that you can't even define a wave function for a complex molecule like that, but they did Double Slit experiments with bucky-balls and observed interference patterns, so it must have something... And defining it to be just a sum of all the wave functions of things that it's made up of doesn't make too much sense either

...?

2. Oct 7, 2007

### michael879

I think you could get a reasonable approximation of the wave function of an atom (if not perfect) by assuming its a single object, and interpreting the results to mean the center of mass. Since nothing in the atom is moving (as long as nothing "observes" it), this should work. I could be wrong but that makes sense to me.

3. Oct 7, 2007

### meopemuk

Wave function is not the most general way to describe the state of the system (molecule) in quantum mechanics. The most general description is by a vector in the Hilbert space. In the case of molecule it is a N-particle Hilbert space, where N is the number of electrons + number of nuclei. In this Hilbert space one can define many different orthonormal bases (usually composed of eigenvectors of commuting operators of observables) and, correspondingly, different types of wave functions (coefficients of the state vector decomposition over these bases). For example, one can choose the basis of common eigenvectors of N particle position operators and define the molecule wave function in this position representation. Alternatively, one can chose a set of mutually commuting observables which includes the 3 components of the center-of-mass position. Then one would obtain a wavefunction for the molecule as a whole.

Eugene.

4. Oct 7, 2007

### jostpuur

Good question! I fought with the precisely same thing long ago, and I believe I did find the answer.

But there is always wave functions for those vectors.

Suppose you have two particles in one dimension that feel the delta potential $U(x_1,x_2)=-\Omega\delta(x_1-x_2)$. If you were given the task of "solving" this, this is how it most probably would be carried out:

We can assume that the other particle is fixed and use the reduced mass. Then the Shrodinger's equation is

$$\Big(-\frac{\hbar^2}{2m_{\textrm{red}}}\partial^2 - \Omega\delta(x)\Big)\psi(x) = E\psi(x).$$

By substituting an attempt $\psi(x)=e^{-A|x|}$ and using the identities $D|x|=2\theta(x) - 1$ and $D\theta(x)=\delta(x)$, you can verify that this attempt solves the SE with parameters

$$A=\frac{m_{\textrm{red}}\Omega}{\hbar^2}$$
$$E=-\frac{m_{\textrm{red}}\Omega^2}{2\hbar^2}$$

Having done this, we should still return to the problem and ask, well what if we don't want to fix the other particle? What is the full solution? It would be nice to know how to solve the system without reducing it to a one particle problem, because we cannot carry out the reduction anyway with 3,4... particles and so on.

With two particles moving in one dimension, the position eigenstates are represented by vectors $(x_1,x_2)\in\mathbb{R}^2$. Both of the particles can be anywhere in the space, so these vectors describe possible states of the system, and the wave function must assign complex amplitudes to these eigenstates. In other words, the wave function must have the form $\psi:\mathbb{R}^2\to\mathbb{C}$. In particular it is not just a pair of two one-particle wave functions.

Having two coordinates for the system, there is also two conjugate momentums $p_1$ and $p_2$. Just substitute the derivatives to these, and we get the Shrodinger's equation

$$\Big(-\frac{\hbar^2}{2m}\big(\partial_1^2 + \partial_2^2\big) - \Omega\delta(x_1-x_2)\Big) \psi(x_1,x_2) = E\psi(x_1,x_2)$$

Here it was assumed that the particles have the same mass m. The attempt

$$\psi(x_1,x_2) = e^{-A|x_1-x_2|} e^{B(x_1+x_2)}$$

works, and the solution can be written as

$$\psi(x_1,x_2) = \textrm{exp}\big(-\frac{m\Omega}{2\hbar^2}|x_1-x_2|\big)\textrm{exp}\big(\frac{ip}{\hbar}\frac{x_1+x_2}{2}\big)$$

with the energy

$$E=-\frac{m\Omega^2}{4\hbar^2} + \frac{p^2}{2(2m)}.$$

The are some features that we should draw attention to. It is a plane wave solution, if you consider the center of mass $(x_1+x_2)/2$ being a location of the molecule. The molecule is non-localized, and both of the particles have amplitudes to be anywhere. But the particles are very unlikely going to be far away from each others. The amplitude is peaked around the diagonal $x_1=x_2$. The energy spectrum is continuous, and the momentum $p$ can have any values, and there is the de'Broglie relation between the oscillation wave length and the momentum. The energy does not come only from the momentum, but it has been shifted downwards by the bounding energy. Notice, that when the bounding energy is written with the reduced mass $m_{\textrm{red}}=m/2$, it becomes the same energy as was solved first as a one particle problem.

To your original question. So this is how it goes. If you have 10 particles in three dimension, basically you have a wave function $\psi:\mathbb{R}^{30}\to\mathbb{C}$, and some sick PDE for it. If you have electrons in it, you should make sure that the solutions are antisymmetric in the correct sense, and well... it gets nasty.

Warning: The explanation I just gave, I something that I have figured out on my own. Feel free to be skeptical and try to explain why it would be wrong, if you want to.

Last edited: Oct 7, 2007
5. Oct 7, 2007

### christianjb

First off, the total wave-function for a molecule gives information about the prob distributions of both the electrons and the nuclei. However, to a very good approximation it's usually possible to use the 'Born Oppenheimer' approximation in which the electrons 'see' the nuclei as fixed, and the potential energy of the nuclei is assumed to be a function of the nuclear coordinates only. (Sorry if that's not a great description)

Anyway- forget about electrons and nuclei for the moment. Just think about what a wave-function means for more than one particle. For e.g. three particles (and ignoring spin for simplicity), the wave-function can be written as

psi=psi(r1,r2,r3)

where r1,r2,r3 are position vectors of particles 1,2 and 3, with each position vector containing 3 (x,y,z) components.

So in total psi is a function of 3N numbers, (x1,y1,z1,x2,y2,z2....xN,yN,zN), where the probability of observing a particular configuration in the region dx1dy1dz1....dxNdyNdzN is given by

Prob=|psi(x1,y1,z1,x2,y2,z2...xN,yN,zN)|^2 dx1dy1dz1....dxNdyNdzN

Naturally, the integral over all possible configurations is normalized

Int[psi(x1,y1,z1,x2,y2,z2...xN,yN,zN)|^2 dx1dy1dz1....dxNdyNdzN]=1

How you actually find what Psi is, is a difficult problem. The preceding post explains what S.E. looks like for a many particle system, but actually solving S.E. in several dimensions is quite challenging.

One 'trick' is to approximate the potential V as a harmonic system, e.g. V=1/2 q1^2+1/2q2^2+1/2q3^3..., where qi are linear combinations of coordinates. In the case that the potential is exactly harmonic, the solution is analytic- the wavefunction can be factored into 3N 1D harmonic oscillators.

If the system is close to harmonic it's quite easy to formulate the problem using a harmonic basis set, such that the numerical solution converges quite quickly.

The most important problem is when the potential goes as 1/r- i.e. the Coulomb potential. It's then not possible to use harmonic analysis- but you can still make some headway by using hydrogen atom like basis sets and then using perturbation theory to refine your answer. Physical chemists have been working on this problem for decades, and state of the art computer software can give good approximations to the wave-function for 100s-1000s of electrons.