0
$\begingroup$

Consider an electron revolving around the nucleus in a circle of radius $r$. We need to apply Heisenberg's uncertainty principle on it. What exactly do we take $\Delta x $ to be?

One way to think about is that the electron position along a straight line varies from $ a + r $ to $ a-r$. So its position can be represented as $ a \pm r$ . This implies that the uncertainty is $ r$.

Another way to think is that the difference between the two extreme positions of electron is $2r$. So uncertainty should be $2r$.

Which view is correct and why?

$\endgroup$
  • 2
    $\begingroup$ An electron doesn't 'revolve' (i.e. orbit) around the nucleus. Acc. modern QM an electron 'resides' in a electron orbital. See Bohr v. Schrodinger models. $\endgroup$ – Gert Sep 10 '19 at 17:40
  • 2
    $\begingroup$ Well, since electrons don't revolve around the nucleus in a literal fashion, you are left to determine how to construct your uncertainty in any manner you see fit. $\endgroup$ – Jon Custer Sep 10 '19 at 17:40
  • 1
    $\begingroup$ At the level of "using the HUP as a back-of-an-envelope estimator" small integer factors like that don't really matter. If you are trying to hit target value (say $a_0$ from the Bohr model ground state energy, perhaps), you can always fudge it after you see how things come out. There is a way to be precise but you need more QM than the Bohr model. $\endgroup$ – dmckee --- ex-moderator kitten Sep 10 '19 at 17:42
  • $\begingroup$ At any time when the electron has a well defined position, the uncertainty in that position is of course zero. $\endgroup$ – WillO Sep 10 '19 at 18:06
1
$\begingroup$

The description of an electron in a circular orbits pretty much classical, and doesn't work well with the uncertainty principle.

In the normal treatment the hydrogen atom, the stationary states (energy eigenstates) with quantum numbers $n, l, m$ are:

$$\psi_{nlm}(\vec r) = R_{nl}(x)Y_{lm}(\theta, \phi) $$

which is factored into the standard spherical harmonics and a radial part:

$$R_{nl}(x)=\frac{N_{nl}}{(n+l)!}x^le^{\frac{-x}{2}}L_{n-l-1}^{2l+1}(x) $$

with $x = \omega r $, $\omega=2\delta$, and $\delta = 1/n$. $L_a^b(x)$ are the associate Laguerre polynomials.

The standard deviation of $\vec r$ is then the uncertainty in position.

The Fourier transform gives the wave functions in the momentum rep:

$$\psi_{nlm}(\vec p) = (i)^lN_{nl}\frac{(l)!}{\sqrt{2\pi}} \frac{n(4\delta)^{l+1}}{(p^2+\delta^2)^{(l+2)}} C^{l+1}_{n-l-1}\big(\frac{p^2-\delta^2}{p^2+\delta^2} \big ) Y_{lm}(\vec p)$$

where $C^a_b$ are the Gegenbauer polynomials. The momentum uncertainty is the standard deviation of $\vec p$ is then the uncertainty in momentum.

The point is, it looks nothing like a classical estimate of $\pm r$.

$\endgroup$
0
$\begingroup$

The answer is more tricky than you think. JEB does a good describing the wave functions you need to calculate the size and shape of the atom. The part he overlooks is what that wave function describes - it describes the separation between the electron and the nucleus. If we assume that the position of the atom's center of mass is completely determined, then that is the tool we use to calculate the uncertainty of the position of the electron. In reality, the position of the proton and electron are given by the function $$\Psi(\mathbf{r}_p, \mathbf{r}_e) = \Psi_{COM}\left(\frac{m_e \mathbf{r}_e + m_p \mathbf{r}_p}{m_e + m_p}\right)\, \psi_{nlm}(\mathbf{r}_e - \mathbf{r}_p).$$

To calculate the uncertainty in the position of the electron you would need to calculate the multi-integrals \begin{array} \,\langle \mathbf{r}_e\rangle & = \int \Psi^*(\mathbf{r}_p, \mathbf{r}_e)\, \mathbf{r}_e\, \Psi(\mathbf{r}_p, \mathbf{r}_e) \ \mathrm{d}^3 r_e \, \mathrm{d}^3 r_p \\ \langle r_e^2\rangle & = \int \Psi^*(\mathbf{r}_p, \mathbf{r}_e)\, r_e^2\, \Psi(\mathbf{r}_p, \mathbf{r}_e) \ \mathrm{d}^3 r_e \, \mathrm{d}^3 r_p \end{array} which feed into the formula that $\sigma_{re}^2 = \langle r_e^2\rangle - \langle \mathbf{r}_e\rangle \cdot \langle \mathbf{r}_e\rangle$, which is the variance in the electron's position.

Working out how much $\Psi_{COM}$ and $\psi_{nlm}$ contribute to the uncertainty in the electron's position would be a bit involved on the algebra side, but it would definitely be non-trival.

$\endgroup$

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.