How a molecule forgets its earlier frequency

Make frequency memory quantitative

Define the zero-mean angular-frequency fluctuation $\delta\omega(t)=\omega(t)-\langle\omega\rangle$ . Multiply its value after a wait by its starting value and average over molecules or time origins. The correlation $C(T)$ has units of angular frequency squared.

Frequency offsets with the same sign contribute positively to the correlation. Independent late frequencies give a correlation approaching zero. A simple model combines decaying memory with a contribution that stays effectively static during the observation. Here $b$ is the static fraction of variance and $\tau$ is the decay time.

C(T)=\langle\delta\omega(T)\delta\omega(0)\rangle,\qquad \frac{C(T)}{C(0)}=b+(1-b)e^{-T/\tau}

One stationary classical Gaussian model. A very slow finite process can resemble a static contribution within a short observation window.

Reading: Fenn and Fayer · MIT Spectroscopy

A Gaussian frequency pair gives a straight centre line

Take an ideal isolated positive band, separated from excited-state absorption. Let $x$ be initial frequency offset and $y$ later offset, with equal variances and correlation $\rho=C(T)/C(0)$ . At fixed excitation frequency, a detection-axis slice peaks at $y=\rho x$ . Connecting those maxima gives slope equal to the correlation.

Differentiate the exponent below with respect to detection frequency and set the derivative to zero. Real spectra also contain homogeneous broadening, pulse selection and overlapping pathways; converting their slopes into a correlation requires the corresponding calibration.

p(x,y)\propto\exp\!\left[-\frac{x^2-2\rho xy+y^2}{2\sigma^2(1-\rho^2)}\right]

A nonsingular bivariate Gaussian requires correlation magnitude below one; perfect memory is its limiting case. Fix excitation frequency and cut parallel to detection.

\left.\frac{\partial\log p}{\partial y}\right|_x=0\quad\Longrightarrow\quad y_{\max}=\rho x

This defines our centre-line convention. Swapping the slicing direction can exchange a slope with its inverse in the literature.

Original ideal Gaussian contours at strong and weak frequency correlation. These are model illustrations, not measured spectra.

Reading: Fenn and Fayer

Read covariance from complex time signals

When frequencies remain nearly frozen during each coherence interval, both envelopes are characteristic functions of the same frequency-pair distribution. Let $v$ be the angular-frequency variance after known pulse selection and $q(T)$ its waiting-time covariance. The two pathways differ only in the sign of their cross term.

Both share coherence rate $\gamma$ and population amplitude $A(T)$ . Their ideal magnitude ratio cancels those factors. In practice, calibrated early samples with adequate signal-to-noise constrain the quadratic envelope, with a separate common log-amplitude offset. Taking logarithms of weak noisy values amplifies noise and introduces bias.

R_{\mathrm{R/NR}}=A(T)\exp\!\left[-\frac{v}{2}(t_1^2+t_3^2)\pm q(T)t_1t_3-\gamma(t_1+t_3)\right]

The upper sign is rephasing. The selected positive band is separated from excited-state absorption; this is not a complete multimode vibrational response.

\log\frac{|R_{\mathrm R}|}{|R_{\mathrm{NR}}|}=2q(T)t_1t_3

A noiseless identity explaining the covariance information. Known finite bandwidth changes both variance and covariance and must be inverted before interpreting the underlying correlation.

Worked example of a decaying memory

Choose $b=0.2$ and $\tau=1\,\mathrm{ps}$ . The normalized correlation is approximately $0.4943$ after $1\,\mathrm{ps}$ and $0.2398$ after $3\,\mathrm{ps}$ . In the ideal Gaussian limit, these are also centre-line slopes.

A wavenumber standard deviation of $18\,\mathrm{cm}^{-1}$ corresponds to approximately $3.39\,\mathrm{rad\,ps}^{-1}$ . Substituting 18 directly as an inverse-picosecond angular frequency would give the wrong time decay. Bandwidth and windowing also change displayed slopes, which must be distinguished from these underlying values.

Spectral diffusion and chemical exchange

Continuous environmental fluctuations can round an initially elongated peak. Chemical exchange can move molecules between states with distinct frequencies and build cross features during the wait. Coupling and energy transfer can also create cross peaks, so a single map rarely assigns their origin uniquely.

For a simple two-state example, start entirely in state A. Use forward rate $k_{AB}$ and reverse rate $k_{BA}$ . The conditional B population follows the expression below. Measured peak heights also contain dipole, lifetime and pulse-coverage factors.

P(B,T\mid A,0)=\frac{k_{AB}}{k_{AB}+k_{BA}}\left[1-e^{-(k_{AB}+k_{BA})T}\right]

Rates are per picosecond. This separate example explains exchange kinetics; the isolated-band diffusion experiment does not fit exchange rates.

Reserve some waiting times for prediction

Fit earlier waits and predict later records that were withheld. A single exponential plus a static term and a two-exponential model may both follow the training points but separate at late times. Held-out errors test whether an extra time scale improves prediction.

Good prediction supports a model over the measured range. It does not identify a unique molecular structure. A slow exponential can resemble a constant within a short window; polarization, temperature or independent structural measurements can supply additional information.

How a molecule forgets its earlier frequency

Make frequency memory quantitative

A Gaussian frequency pair gives a straight centre line

Read covariance from complex time signals

Worked example of a decaying memory

Spectral diffusion and chemical exchange

Reserve some waiting times for prediction

Try the prediction

Test memory models on withheld records

Exercises and explanations

Sources and calculation scope