Problem 1

Proton tunneling. Proton tunneling through energy barriers occurs in many hydrogen-bonded species (DNA, proteins). Propanedial (malonaldehyde) is one of the simplest molecules with intramolecular proton transfer. (1.1) Draw the condensed formula of propanedial and the structures of two of its isomers that can exist in equilibrium with it. (1.2) In aqueous solution propanedial is a weak acid comparable to acetic acid. Specify the acidic hydrogen atom and explain its acidity. The energy profile of the intramolecular proton transfer has a symmetric double-well form with two minima separated by

d = 0.06

nm. (1.3) Draw the structures corresponding to the two minima. The proton oscillates between the two minima L and R with angular frequency

\omega = 6.48\times10^{11}

^{-1}

. The probability density is

\Psi^2(x,t) = \frac{1}{2}[\Psi_L^2(x) + \Psi_R^2(x) + (\Psi_L^2(x) - \Psi_R^2(x))\cos\omega t]

where

\Psi_L, \Psi_R

describe a proton localized in the left and right wells. (1.4) Write expressions for the probability density at

t = 0

t = \pi/(2\omega)

and

t = \pi/\omega

. (1.5) Without calculation, determine the probability of finding the proton in the left well at

t = \pi/(2\omega)

. (1.6) How much time is required for a proton to transfer from one well to the other? What is the mean proton speed during transfer? (1.7) Estimate the position uncertainty

\Delta x

from the double well and calculate the minimal velocity uncertainty

\Delta v

from the Heisenberg relation

\Delta x \cdot m\,\Delta v \ge \hbar/2

; compare with the classical speed from (1.6) and draw a conclusion.

Step 2 of 4: Time-dependent probability densities

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\Psi^2(x,0) = \Psi_L^2(x);\quad \Psi^2\left(x,\frac{\pi}{2\omega}\right) = \frac{1}{2}(\Psi_L^2 + \Psi_R^2);\quad \Psi^2\left(x,\frac{\pi}{\omega}\right) = \Psi_R^2(x);\quad P_L\left(\frac{\pi}{2\omega}\right) = \frac{1}{2}

Analysis

Substituting $\cos(0) = 1$ , $\cos(\pi/2) = 0$ and $\cos(\pi) = -1$ gives: at $t = 0$ the proton is localized in the left well; at $t = \pi/(2\omega)$ it is delocalized equally between the two wells ( $P_L = 1/2$ by symmetry); at $t = \pi/\omega$ it is localized in the right well.