Spectral Interferometry

Photons and phase

Light has properties of both particles (photons) and waves (electromagnetic fields). For time-resolved spectroscopy, the wave description is of particular interest and utility. Unfortunately, the detectors available for measuring visible light are fundamentally “photon detectors” (also known as “square-law” detectors¹) since they measure the intensity ($I \propto E^2$) of light rather than its field ($E$). This is because the frequency of visible light is too high (hundreds of terahertz) for detectors to respond to the extremely fast oscillations of the electric field like an antenna does for radio waves. Intensity is basically a measure of power, which means that information about the light wave’s phase is lost (as implied by squaring the field).

For many nonlinear spectroscopy techniques, we need to measure the complete electric field of the signal, including its phase. Luckily, there are methods for achieving this. One particularly powerful approach is known as spectral interferometry (SI).² SI involves spatially overlapping a known “reference” field with the unknown signal field and then measuring the spectrum of the combined fields (known as a spectral interferogram). The resulting spectral interference encodes the magnitude and phase of the signal field, which can be extracting using the Fourier-transform spectral interferometry (FTSI) algorithm.³

Mathematical basis of spectral interferometry

A spectral interferogram of spatially-overlapped signal $\hat{E}_s$ and reference $\hat{E}_r$ fields is measured by a spectrally-resolved square-law detector (i.e., $I \propto E^2$), yielding the intensity of the sum of the fields as a function of angular frequency ($\omega = 2\pi \nu$) $$ I_{sr}(\omega) = \left| \hat{E}_s(\omega) + \hat{E}_r(\omega) \right|^2 . \tag{1} $$ This can be expanded to yield $$ I_{sr}(\omega) = \left| \hat{E}_s(\omega) \right|^2 + \left| \hat{E}_r(\omega) \right|^2 + \hat{E}_s(\omega)\hat{E}_r^*(\omega) + \hat{E}_s^*(\omega)\hat{E}_r(\omega) \tag{2} $$ where the first two terms are the spectra of the signal and reference fields and the last two terms are the “interference terms”, which contain the phase information.

We can rewrite the first interference term as

$$ \begin{aligned} \hat{E}_s(\omega)\hat{E}_r^* (\omega) & = \left| \hat{E}_s(\omega) \right| \exp \left[ i \phi_s(\omega) \right] \times \left\{ \left| \hat{E}_r(\omega) \right| \exp \left[ i \phi_r(\omega) \right] \right\}^* \\
& = \left| \hat{E}_s(\omega) \hat{E}_r(\omega) \right| \exp \left[ i \left\{ \phi_s(\omega) - \phi_r(\omega) \right\} \right] \\
& = \left| \hat{E}_s(\omega) \hat{E}_r(\omega) \right| \exp \left[ i \Delta\phi(\omega)\right] \end{aligned} \tag{3} $$

where $\Delta\phi(\omega) = \phi_s(\omega) - \phi_r(\omega)$ is the relative phase of the signal and reference fields.⁴ We can express $\Delta\phi$ in terms of a linear phase (i.e., delay) and the sum of all other components:

$$ \Delta\phi(\omega) = \omega (t_s - t_r) + \Delta\phi^\prime(\omega) . \tag{4}$$

Using this new form for $\Delta\phi$, the interference term becomes

$$ \hat{E}_s(\omega)\hat{E}_r^*(\omega) = \left| \hat{E}_s(\omega) \hat{E}_r(\omega) \right| \exp \left[ i \omega (t_s - t_r) \right] \exp \left[ i \Delta\phi^\prime(\omega)\right] . \tag{5}$$

Therefore, by the Fourier modulation theorem⁵, when the reference field is delayed relative to the signal field (i.e., $t_s - t_r < 0$), the interference term $\hat{E}_s(\omega)\hat{E}_r^*(\omega)$ appears as a peak on the negative side of $t = 0$ once Fourier transformed to the time domain.

Fourier-transform spectral interferometry algorithm

Let’s simulate a spectral interferogram between an exponentially-damped sinusoidal “signal” field $E_s(t)$ and a “reference” field with a Gaussian temporal field envelope $E_r(t)$. The signal field is given by

$$ E_s(t) =\mathrm{Re}\left\lbrace \mathrm{exp}\left\lbrack \left(2\pi if_s^0 -\frac{1}{\tau_s }\right)t\right\rbrack H\left(t\right)\right\rbrace $$

where $\mathrm{Re}\lbrace\hat{z}\rbrace$ selects the real part of complex value $\hat{z}$, $f_s^0$ is the signal center frequency, $\tau_s$ is the exponential damping time constant, $t$ is time, and $H(t)$ is the Heaviside step function.⁶ The reference field is given by

$$ E_r(t) =\mathrm{Re}\left\lbrace G\left(t;T_r ,\mathrm{FWHM}\right)\;\mathrm{exp}\left\lbrack 2\pi if_r^0 \left(t-T_r \right)\right\rbrack \right\rbrace $$

where $G(t; T_r, \mathrm{FWHM})$ is a Gaussian function of $t$ with delay $T_r$ and full-width at half-maximum $\mathrm{FWHM}$, and $f_r^0$ is the reference center frequency. These fields are plotted below.

For the sake of example, we will use the following parameter values:

To simulate the measured spectral interferogram, we add the Fourier-transformed signal and reference fields together, square the result (since we measure intensity), and eliminate the negative frequency part using the Heaviside step function (since a spectrometer only resolves positive frequencies). The resulting spectral interferogram $|\hat{E}_s(f) + \hat{E}_r(f)|^2 H(f)$ is shown below. For simplicity, I’m only plotting the real part of complex data.

After Fourier transforming our interferogram, we get the complex-valued time-domain interferogram $\hat{I}_{sr}(t)$ plotted below.

To isolate $\hat{E}_s \hat{E}_r^*$, we need to select the peak at negative time (as we proved in the previous section) using the Heaviside step function (orange line, above). The isolated interference term $\hat{E}_s(t) \hat{E}_r^*(t) = \hat{I}_{sr}(t) H(-(t+200))$ is plotted below.

Fourier transforming back to the frequency domain yields the product of signal and reference fields $\hat{E}_s(f) \hat{E}_r^*(f)$ plotted below.

Notice that the spectrum is highly modulated. This is due to the time delay between the signal and reference fields and is an effect we’d like to remove. To fully isolate the signal field, we divide the signal–reference product $\hat{E}_s(f) \hat{E}_r^*(f)$ by the reference field $|\hat{E}_r(f)| \exp (-2\pi ifT_r)$. The factor $\exp (-2\pi ifT_r)$ eliminates the delay-induced modulation while $|\hat{E}_r(f)|$ corrects for the reference field’s finite spectrum (whose effect is primarily to attenuate the extremities of the signal).

Finally, we can Fourier transform back to the time domain to compare the recovered signal field $\hat{E}_s \hat{E}_r^* / |\hat{E}_r| \exp (-2\pi ifT_r)$ to the exact signal field $E_s$.

Notice that the rising edge of the signal field near $t = 0$ is somewhat smoothed in the recovered signal field compared to the exact signal field. This arises at least partially due to the reduction in resolution induced by Fourier filtering (when isolating $E_s E_r^*$). Contrary to the case of this simulation, experimental signals do not have an infinitely sharp rise and therefore will not exhibit this discrepancy.

Appendices

(A.I) Fourier transform⁷

By my convention, the (forward) Fourier transform is used to go from the frequency domain to the time domain while the inverse Fourier transform goes from time to frequency (i.e., $s \equiv t$ and $x \equiv \nu$).

(Forward) Fourier transform

$$ F(s) = \int_{-\infty}^\infty f(x) \exp \left( -2 \pi i s x \right) dx \tag{A.I.1}$$

Inverse Fourier transform

$$ f(x) = \int_{-\infty}^\infty F(s) \exp \left( 2 \pi i s x \right) ds \tag{A.I.2}$$

(A.II) Electric fields

Since electric fields are measurable, they must be real-valued in the time domain.

$$ \hat{E}(\omega) = e(\omega) \exp \left[ i \phi(\omega) \right] \tag{A.II.1}$$

$$ E(t) = e(t) \cos \left[ \phi(t) \right] \tag{A.II.2}$$

Based on the above Fourier transform conventions,

$$ E(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \hat{E}(\omega) \exp \left( - i \omega t \right) d\omega \tag{A.II.3}$$

$$ \hat{E}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty E(t) \exp \left( i \omega t \right) dt \tag{A.II.4}$$