mexicankrill

Scientist. Shitposter. Sleepy.

Raman Scattering Thermometry

In order to probe the physics of plasmons, we use a technique called Raman spectroscopy, which makes clever use of Raman scattering to measure the change in energy of scattered photons. During this process, an electron is excited to a virtual energy level, corresponding to the energy of the photon. From this state, the electron would normally decay to the ground state via photoemission, known as Rayleigh scattering. However, Raman scattering leaves the electron in a different vibrational state from where it began. This means the re-emitted photon will have a different energy and momentum from the initially incident photon. If the vibrational state increases, the scattering photon loses energy. We call this Stokes scattering, and the opposite is called anti-Stokes scattering. The change in the photon energy is \(\Delta{E}=\pm{h}\nu_{M}\), where \(h\) is Planck's constant, and \(\nu_{M}\) is the vibrational transition frequency. We choose the positive or negative solution depending on the type of Raman scattering.

Through the detection of scattered photons, we can plot a spectrum of intensity against frequency change. We call this a Raman spectrum, and it is becoming an increasingly more important technique in a wide range of fields. The frequency change, \(\Delta\nu\) is the Raman shift, and the spectrum will have some reflective symmetry in \(\Delta\nu=0\). This symmetry arises naturally since both processes involve the same two vibrational states, but with electron transfer occurring in opposite directions. The Raman spectrum is typically dominated by a background photoluminescence signal, with sharp peaks appearing on top of it due to Raman scattering between select vibrational states.

The photoluminescence process involves internal relaxation effects and internal energy redistribution before the electron relaxes back to the ground state. This comes in the short-lived form of fluorescence, typically acting over the order of nanoseconds. This energy is redistributed non-radiatively via vibrational and rotational modes of atoms, or radiatively, emitting a photon, where the electron relaxes to an excited state. It is understood that the anti-Stokes photoluminescence spectrum is well-approximated by a Boltzmann-like distribution, \begin{equation} I_{a}\left(\hbar\omega\right) \propto f\left(\hbar\omega\right)g\left(\hbar\delta\omega\right)\exp\left(-\hbar\delta\omega/k_{B}T_{e}\right) \end{equation} where $T_{e}$ is the effective electron temperature, \(\hbar\delta\omega = \hbar\omega - \hbar\omega_{pr}\) is the difference in energy between emitted photons and the probe. \(f(\hbar\omega)\) and \(g(\hbar\delta\omega)\) are the probability of emission and density of states respectively. For narrow plasmon resonances far from the interbad transitions of the metal, it is safe to assume that \(f(\hbar\omega)g(\hbar\delta\omega)\) varies much slower than the Boltzmann factor, hence \begin{equation} I_{a}\left(\hbar\omega\right) = A\exp\left(-\hbar\delta\omega/k_{B}T_{e}\right) = A\exp\left(-hc\tilde{\nu}/k_{B}T_{e}\right) \end{equation} for some constant \(A\). Previously other research has approximated that \(f(\hbar\omega)g(\hbar\delta\omega) \propto \omega^{3}\), however this is similar to our approximation, in that \(\omega^{3}\) varies much slower than the Boltzmann factor.

Scattering
Energy state diagram of Rayleigh scattering, both forms of Raman scattering, and fluorescence. In Raman spectroscopy the laser wavelength is chosen to be lower than the bandgap of the material to avoid noisy spectra due to excessive fluorescence. Image adapted from [1].

Be warned that the second half of this post is very math heavy. Feel free to skip past the following derivation, I'm simply putting it here for those who might find it interesting. Let us continue!


For a system in thermodynamic equilibrium, the lower vibrational state will be more populated than the higher state. This means that there will be more Raman scattering from electrons in the lower state, corresponding to Stokes transitions, relative to the anti-Stokes transitions. This means that on the Raman spectrum, the Stokes peaks will appear stronger than anti-Stokes. Increasing the temperature of the system will promote more electrons into the higher vibrational state, increasing the proportion of anti-Stokes scattering. Thus, the ratio of the two processes is temperature dependent. This relationship can be derived by consideration of the transition susceptibility (Raman) tensor, which describes probability of transition between two states, \begin{equation} \label{eq:susceptibility} \left[\chi_{mn}\right]_{fi} = \left(\chi_{mn}\right)_{0}\braket{..v_{fk}..|..v_{ik}..} + \sum_{k}\left(\frac{\partial\chi_{mn}}{\partial Q_{k}}\right)_{0}\bra{..v_{fk}..}\hat{Q}_{k}\ket{..v_{ik}..} \end{equation} where we have taken a weighted sum over all initial and final total vibrational wavefunctions in a normal coordinate space $Q_{k}$ and we express the total vibrational wavefunctions as a product of harmonic-oscillator wavefunctions. \begin{equation} \begin{split} \bra{v_{f1}, ..v_{fk}, ..v_{fn}} = \prod_{k}\bra{v_{fk}}\\ \bra{v_{i1}, ..v_{ik}, ..v_{in}} = \prod_{k}\bra{v_{ik}} \end{split} \end{equation}

For simplicity, we will ignore the Franck-Condon principle, and can safely assume that the nucleus does not undergo a vibration when realigning itself with the new electronic configuration. This means that the oscillators remain unshifted, hence \begin{equation} \braket{v_{fk}|v_{ik}} = \begin{cases} 0 & \text{for $v_{fk} \neq v_{ik}$}\\ 1 & \text{for $v_{fk} = v_{ik}$} \end{cases} \end{equation} and \begin{equation} \bra{v_{fk}}\hat{Q}_{k}\ket{v_{ik}} = \begin{cases} 0 & \text{for $v_{fk} = v_{ik}$}\\ \left(v_{ik}+1\right)^{1/2}\sqrt{\hbar/{2\Omega_{k}}} & \text{for $v_{fk} = v_{ik} + 1$}\\ \left(v_{ik}\right)^{1/2}\sqrt{\hbar/{2\Omega_{k}}} & \text{for $v_{fk} = v_{ik} - 1$} \end{cases} \end{equation} This means that the first term in eqn \ref{eq:susceptibility} is non-zero if and only if \(v_{fk} = v_{ik}\) for all \(k\), representing the case where the quantum state of the system has remained unchanged. Thus, \(\left(\chi_{mn}\right)_0\) encodes the process of Rayleigh scattering, whilst the second term represents Raman scattering, which is non-zero if it is such that for all \(k' \neq k\), \(v_{fk'} = v_{ik'}\), or for the vibrational mode \(k\), \(v_{fk} = v_{ik} \pm 1\). Hence in the case of Raman scattering, eqn \ref{eq:susceptibility} yields \begin{equation} \label{eq:raman_tensor} \begin{split} \left[\chi_{mn}\right]_{v_{ik}+1,v_{ik}} = \left(v_{ik} + 1\right)^{1/2}\sqrt{\hbar/{2\Omega_{k}}}\left(\frac{\partial\chi_{mn}}{\partial Q_{k}}\right)_{0}\\ \left[\chi_{mn}\right]_{v_{ik}-1,v_{ik}} = \left(v_{ik}\right)^{1/2}\sqrt{\hbar/{2\Omega_{k}}}\left(\frac{\partial\chi_{mn}}{\partial Q_{k}}\right)_{0} \end{split} \end{equation} for Stokes and anti-Stokes scattering processes respectively.

Now we consider how the intensity depends on the occupation number of the vibrational state, \(v_{k}\). For this we use the Boltzmann factor \begin{equation} W\left(\epsilon_{k}\right) = \frac{\exp\left(-\epsilon_{k}/k_{B}T\right)}{Z} = \frac{\exp\left[-\hbar\Omega_{k}\left(v_{k} + 1/2\right)/k_{B}T\right]}{\sum_{v_{k}}\exp\left[-\hbar\Omega_{k}\left(v_{k} + 1/2\right)/k_{B}T\right]} \end{equation} and take the thermal average as \(\sum_{v_{k}}\left(v_{k} + 1\right)W\left(\epsilon_{k}\right)\), in order to calculate the effective square of the susceptibility tensor in eqn \ref{eq:raman_tensor}. This gives an average of \(n_{k} + 1\) for Stokes, and \(n_{k}\) for anti-Stokes scattering. \(n_{k}\) is given by consideration of the Bose-Einstein distribution for mode \(k\), which we expect our vibrational phonon states to obey \begin{equation} n_{k} = f_{E}\left(\Omega_{k}\right) = \left[\exp\left(\hbar\Omega_{k}/k_{B}T\right) - 1\right]^{-1} \end{equation} Then it is possible to derive the scattering intensity per steradian [REF 16] to be \begin{equation} \frac{d^{\|}\phi_{\perp}}{d\bar{\Omega}} = \frac{\hbar\left(\omega - \Omega_{k}\right)^{4}\mathcal{V}_{u}\chi^{2}_{yx,k}\left(n_{k} + 1\right)I_{i}\mathcal{V}}{32\pi^{2}c^{4}_{0}\Omega_{k}} \end{equation} and a similar relation for anti-Stokes scattering for incident intensity \(I_{i}\), scattering volume \(\mathcal{V}\) and scattering per unit cell \(\mathcal{V}_{u}\). From this it is then possible to directly obtain the temperature dependence as \begin{equation} \frac{I_{a}}{I_{s}} = \frac{\phi'_{a}}{\phi'_{s}} = \left(\frac{\omega + \Omega_{k}}{\omega - \Omega_{k}}\right)^{4}\exp\left(\frac{-\hbar\Omega_{k}}{k_{B}T}\right) \end{equation} Since it is convenient to use the Raman shift, which is often measured in wavenumber \(\tilde{\nu}\), we use the conversion \(\omega = 2\pi c\tilde{\nu}\) to obtain the necessary result, \begin{equation} \label{eq:peak_height_ratio} \frac{I_{a}}{I_{s}} = \left(2\frac{\tilde{\nu}_L}{\Delta\tilde{\nu}} - 1\right)^{4}\exp\left(\frac{-hc\Delta\tilde{\nu}}{k_{B}T}\right) \end{equation} where \(I_{s}\) and \(I_{a}\) are the Stokes and anti-Stokes peak intensities, \(\tilde{\nu}_{L}\) is the wavenumber of the incident light and \(\Delta\tilde{\nu}\) is the Raman shift.


The derivation ends here, we will quickly summarise after this point and conclude the post.


Eq \ref{eq:peak_height_ratio} allows us to perform thermometry of the chemical species responsible for peaks in the Raman spectrum by comparison of their peak heights, an important technique in my research.

I mentioned earlier in the post that Raman spectroscopy has some real-world applications. Whilst I think this is very interesting, I likely won't go into any detail about it in a blog post, however you will find some sources in the references below discussing its use in chemical analysis in medicine, nanotechnology, and explosives detection [2,3,4]. In the following point I will discuss how we can optimise Raman spectroscopy with nanofabrication.


References

[1] G. W. Auner, S. K. Koya, C. Huang, et al., “Applications of raman spectroscopy in cancer diagnosis,” Cancer Metastasis Rev., vol. 37, no. 4, pp. 691–717, Dec. 2018.

[2] M. Schütz, C. I. Müller, M. Salehi, C. Lambert, and S. Schlücker, “Design and synthesis of Raman reporter molecules for tissue imaging by immuno‐SERS microscopy,” Journal of Biophotonics, vol. 4, no. 6. Wiley, pp. 453–463, Feb. 07, 2011.

[3] P. Pistor, A. Ruiz, A. Cabot, and V. Izquierdo-Roca, “Advanced Raman Spectroscopy of Methylammonium Lead Iodide: Development of a Non-destructive Characterisation Methodology,” Scientific Reports, vol. 6, no. 1. Springer Science and Business Media LLC, Oct. 27, 2016.

[4] A. K. Misra, S. K. Sharma, T. E. Acosta, J. N. Porter, and D. E. Bates, “Single-Pulse Standoff Raman Detection of Chemicals from 120 m Distance during Daytime,” Applied Spectroscopy, vol. 66, no. 11. SAGE Publications, pp. 1279–1285, Nov. 2012.


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