12
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ This is not how quantum mechanics and uncertainty relation work. $\endgroup$
    – Greg
    Jul 30 at 2:07
9
$\begingroup$

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 their corresponding variances in a physical (normalized) state $\psi$ fulfill $$\sigma_\psi(\hat x) \, \sigma_\psi(\hat{p}) \geq \hbar/2 > 0\quad . \tag{2}$$ In particular there is no physical state $\psi$ which is an eigenstate of the position or momentum (or both), since this would violate the inequality.

It further 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.

$\endgroup$
1
  • 2
    $\begingroup$ +1 and excellent first answer! Welcome to our new community and thank you for contributing your answer here! We hope to see much more of you at MMSE in the future!! Your answer is very similar to mine (for example the part where I said the asker could have equally well asked how we can get the quantity in my Eq. 6 if the uncertainty principle is a problem). I had started writing my answer in the morning but then had to go to leave in order to arrive at my ultrasound appointment on time, and only got the chance to finish my answer now (after you've written something quite similar!). $\endgroup$ Jul 30 at 23:35
10
$\begingroup$

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...)

$\endgroup$
3
  • $\begingroup$ Are not the uncertainties in pairs related? I mean, the uncertainty in position is only correlated to momentum, where the uncertainty in energy is related correlated with time. There is no crossover between uncertainly in position/energy, for example. $\endgroup$
    – Camps
    Jul 30 at 13:17
  • $\begingroup$ That's true. In this reply I only mentioned the position-momentum uncertainty relation. The energy-time uncertainty relation means that the total energy is actually slightly uncertain as well, but for any state that is stable up to a macroscopic time scale, this uncertainty is for most purposes negligible. $\endgroup$
    – wzkchem5
    Jul 30 at 14:33
  • 1
    $\begingroup$ There is no time-energy uncertainty in a strict sense, because there is no time operator in quantum mechanics. $\endgroup$
    – Jakob
    Jul 30 at 20:10
6
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.