Why is uncertainty not a big problem in computational chemistry?

Question

The molecular Hamiltonian (or, for simplicity, the Fock operator) contains coulomb potentials as well as momentum operators. To evaluate the coulomb potential, we need to know where the electrons are. So basically we are messing with the position and momentum of the electron at the same time. Therefore, there should be problems with uncertainty, but somehow this does not seem to be a big issue. Why?

Answer 1

Following your arguments, we would see also a 'violation' of the Heisenberg uncertainty principle (HUP) in single-particle quantum mechanics, e.g. in a Hamiltonian of a particle in an external potential:

$$ H = \hat{p}^2/2m + V(\hat x) \quad .\tag{1}$$

But writing down such a Hamiltonian does not mean that we know the position or momentum (or both) of this particle. The HUP does not state that you can't write a Hamiltonian like this (or more generally a many-body Hamiltonian).

Similarly, writing the wave function as $\psi(x) \equiv \langle x|\psi\rangle$ does not mean that we know the particle is at position $x$, but instead the wave function gives the corresponding probability amplitude.

In fact, the Robertson uncertainty relations for the position and the momentum operators $ \hat x$ and $\hat p$ say that (under certain assumptions) their corresponding variances in a physical (normalized) state $\psi$ fulfill $$\sigma_\psi(\hat x) \, \sigma_\psi(\hat{p}) \geq \hbar/2 . \tag{2}$$

It tells you that if you prepare an ensemble of identical states $\psi$ and measure the position and momentum (separately) many times, then the respective variances of these measurements fulfill the above inequality.

Answer 2

Indeed, due to the uncertainty relation, the electrons in a molecule do not have a definite potential energy, nor do they have a definite kinetic energy (yet they have well-defined expectation values of potential energy and kinetic energy). But, assuming that the molecule is in an energy eigenstate, the two uncertainties exactly cancel out, and gives an unambiguous and definite total energy. So conceptually, the uncertainty relation does not make the total energy uncertain.

In practice, the uncertainty relation does make computational chemistry much more difficult than it would otherwise be. If a particle could have simultaneously a certain momentum and a certain position, then the ground state of the system will have all electrons absorbed into the nuclei, and there is no longer concepts like bonding, or electronic structure in general, so computational chemistry will become completely trivial. Thus, we generally do not view the uncertainly relation as a problematic issue, because without the uncertainly relation there is essentially no computational chemistry: not only the problems are gone, but the whole discipline will disappear too. (In fact, as the electrons collapse into the nuclei, we will all disappear too...)

Answer 3

A common textbook statement of the uncertainty principle is:

$$\tag{1} \Delta x \Delta p \ge \frac{\hbar}{2}, $$

where $\Delta x$ and $\Delta p$ are the uncertainties in measurements of the position $x$ and momentum $p$ respectively. What mathematically is an "uncertainty"? In this case it's the standard deviation:

\begin{align} \tag{2} \Delta x &\equiv \sqrt{\left\langle \hat{x}^2 \right\rangle - \Big\langle \hat{x} \Big\rangle^2} \\ \Delta p &\equiv \sqrt{\left\langle \hat{p}^2 \right\rangle - \Big\langle \hat{p} \Big\rangle^2},\tag{3} \end{align}

and from our quantum mechanics courses we know the following:

\begin{align} \Big\langle \hat{x} \Big\rangle &\equiv \int_{-\infty}^{\infty} \psi(x)^* x \psi(x)\textrm{d}x \tag{4}\\ \left\langle \hat{x}^2 \right\rangle &\equiv \int_{-\infty}^{\infty} \psi(x)^* x^2 \psi(x)\textrm{d}x \tag{5}\\ \Big\langle \hat{p} \Big\rangle &\equiv \int_{-\infty}^{\infty} \psi^*(x)\left( \frac{\hbar}{\textrm{i}} \right)\frac{\partial}{\partial x} \psi(x)\textrm{d}x \tag{6}\\ \left\langle \hat{p}^2 \right\rangle &\equiv \int_{-\infty}^{\infty} \psi^*(x)\left( \frac{\hbar}{\textrm{i}} \right)^2\frac{\partial^2}{\partial x^2} \psi(x)\textrm{d}x. \tag{7} \end{align}

Now in computational chemistry, when we're calculating energies for whatever reason (geometry optimization, singlet-triplet-gap, excitation energies, ionization energies, electron affinities, potential energy surfaces, reaction energies, etc.) we are calculating the following:

$$ \Big\langle H \Big\rangle \equiv \int_{-\infty}^{\infty} \psi(x)^* H\psi(x)\textrm{d}x. \tag{8} $$

Now you seem to be concerned about the fact that the $H$ in Eq. 8 involves both $x$ and $p$ and we can never have:

\begin{align} \Big\langle \hat{x}^2 \Big\rangle - \Big\langle \hat{x} \Big\rangle^2 &= 0, ~~\textrm{or} \tag{9} \\ \Big\langle \hat{p}^2 \Big\rangle - \Big\langle \hat{p} \Big\rangle^2 &= 0.\tag{10} \end{align}

In fact, you could also have asked, how can we ever get the quantity in Eq. 6? Because the right-hand-side involves both $x$ and $p$, just like Eq. 8 does.

The answer to your concern is that just because we cannot have Eqs. 9 or 10, does not mean that we can't calculate Eq. 8.

We do not need to "know" with some great accuracy, the position of the system to calculate Eq. 8: We are actually integrating over all possible values of position!