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Raman Spectroscopy

Brie, Gaungnan, Krishan and Qian

A brief article on the Raman effect. We explore the theory behind predicting when Raman transitions occur, as well as how intense the signal from these transitions will be. We then compare this theory to various experimental results.

1 0 0 1 1. INTRODUCTION hΨk| µ |Ψni = Mkn + Mkn. Where the zeroth order term, 0 iωk,nt Mkn = µk,ne , is ignored because it is just a regu- The Raman effect is a light-scattering phenomenon lar , and therefore doesn’t con- used to gain understanding of the vibrational and ro- tribute to Raman scattering. Using the first order correc- tational modes of a material. When light strikes a tion to the wave function in Eqn.1, and orthonormality of the zeroth-order wave functions Ψ0 = ψ0e−iωkt, the material, at a frequency νincident, it can be scattered k k at its original frequency (Rayleigh scattering) or at a dipole moment transition that we are interested in is, i(ωk,n−ω)t   shifted frequency, due internal transitions within the 1 e X µk,r(µr,n · E0) µr,n(µk,r · E0) molecule, ν . This shifted scattering is called Ra- Mkn = + molecular 2~ ωr,n − ω ωr,n + ω man scattering, and is classified as either Stokes scat- r i(ωk,n+ω)t   e X µk,r(µr,n · E0) µr,n(µk,r · E0) tering, νscattered = νincident + νmolecular, or anti-Stokes + + . scattering, ν = ν − ν . The shift 2 ω + ω ω − ω scattered incident molecular ~ r r,n r,n in frequency is called the Raman shift: ∆ν = |νincident − (2) ν |. Section 2.1 describes the theory behind pre- scattered The first term in Eq.2 with angular frequency ωkn − ω is dicting when Raman transitions occur, section 2.2 de- associated with the Raman effect, and the second term scribes the theory behind predicting the intensity of a with angular frequency ωkn + ω is interpreted as a two- Raman signal, and finally section 3 compares this the- photon transition and therefore ignored when consid- ory to experimental Raman measurements under various ering the Raman effect. The induced transition from conditions. states hΨn| to hΨk| from the Raman effect can be a transition from high to low frequency, ωkn = (Ek − En)/~ > 0 (Stokes scattering), or from low to high fre- 2. RAMAN THEORY quency, ωkn = (Ek − En)/~ < 0 (anti-Stokes scatter- ing). It is assumed that the incident photon, with fre- 2.1. Raman Transitions quency ω as sufficient energy to induce the dipole mo- mentum transition: ω − ωkn > 0. For an applied elec- To model the effect of polarized light hitting a sample tric field that is not too strong the induced dipole mo- consider the application of an oscillating electric field of ment is proportional to the applied field: µi = αijEj, where the proportionality constant αij is characteris- the form E~ = E~0 cos ωt, with a wavelength much larger than the molecular dimensions. For Raman scattering tic of each molecule and is called the polarizability. the field is not in resonance, ω 6= (E − E )/ = ω for Looking at the kn-matrix element of this expression us- r k ~ rk 0 0 −iωkt any normal modes |Ψ >, |Ψ > of the sample, how- ing the first order wave functions Ψk = ψke and r k ~ ~ ever it will induce an oscillating dipole moment that subbing in our oscillating electric field E = E0 cos ωt 0 0 will re-radiate. Therefore transitions between normal then the expression becomes Mi,kn = hΨk| µi |Ψni = 0 0 i(ωkn−ω)t i(ωkn+ω)t modes in the system are found by solving the transi- 1/2 hψk| αˆi,j |ψni E0j(e + e ). As before tion dipole moment: Mr,k(t) = hΨr| µ |Ψki. The nor- the term with frequency ωkn +ω is ignored and we’re left mal modes are found by perturbatively solving the time- with: 0 0 i(ωkn−ωt) dependent multi-particle Schrodinger equation with the Mi,kn = 1/2 hψk| αˆi,j |ψni E0je (3) with the electronic-dipole interaction perturbation, Hˆ0 = Comparing Eqn. 3 with the kn-component of the Raman −µ·E0 cos ωt, resulting from the applied electric field and part of Eqn. 2, i(ωk,n−ω)t   induced dipole moment. The first order correction to the e X µi,krµj,rn µi,rnµj,kr M = + E , wave function is, i,kn 2 ω − ω ω + ω j   ~ r rn rn 1 X µr,n · E0 µr,n · E0 1 0 −i(ωn+ω)t −i(ωn−ω)t (4) Ψn = ψr e + e . 2~ ωr,n − ω ωr,n + ω we find that the kn-polarizability matrix elements are: r   (1) 1 X µˆi,krµˆj,rn µˆi,rnµˆj,kr αij,kn = + (5) Where the sum is over all of the time independent ~ ωrn − ω ωrk + ω 0 0 r zeroth-order solutions, µrn = hψr | µ |ψni, and ωrn = (Er − En)/~. Using this first order correction we can approximate the dipole moment as: Mkn ≈ 0 1 0 1 0 0 0 1 hΨk + Ψk| µ |Ψn + Ψni ≈ hΨk| µ |Ψni + hΨk| µ |Ψni + 2

ces. There are many different assignments of matrices to symmetry operators that will satisfy the group table mul- tiplication: each assignment is called a representation. Irreducible representations are the smallest possible ma- trices that satisfy the group table: they can’t be broken into smaller matrices that will also satisfy the group ta- ble multiplication. For example, a representation of the C2v group is: FIG. 1: The symmetry operators of H2O [3] 1 0 −1 0  1 0  −1 0 E = C = σ0 = σ = 0 1 v 0 −1 v 0 −1 v 0 1 (7) since the matrices satisfy the group table multiplication. However, it isn’t irreducible because it can be broken FIG. 2: The normal modes of H2O [4] into two irreducible representations: E = 1, C2 = −1, 0 σv = 1, and σv = −1 and E = 1, C2 = −1, σv = −1, and 0 σv = 1 which also satisfy the group table multiplication. All representations of a symmetry group are built from irreducible representations. FIG. 3: The group table of H O [5] 2 A normal mode of a molecule can be thought of as an irreducible representation. The three normal modes of H O are illustrated in Fig 2. The symmetric irreducible   2 1 X µˆi |ri hr| µˆj µˆj |ri hr| µˆi α = hψ0| + |ψ0 i representation is where nothing is done to the molecule, ij,kn k ω − ω ω + ω n every symmetry group contains this irreducible represen- ~ r rn rk (6) tation: it is the one where all symmetry operators are assigned to the constant 1 (E = 1, C = 1, σ = 1, and For nonzero dipole moment transitions, αij,kn = 2 v 0 0 R ∗ σ0 = 1). hψk| αij |ψni = ψkαijψndτ must be non-zero. In or- v der for this transition to be non-zero, ie Raman active, Now consider the theorem again, if the direct products ∗ the direct products of the irreducible representation of of the irreducible representation of ψk, αij, and ψn pro- ∗ duces a representation which is or contains the symmetric ψk, αij and ψn produces a representation which is or contains the symmetric irreducible representation, then irreducible representation, then Raman transitions will ∗ Raman transitions will occur [1]. Group theory must be occur. Where the irreducible representations of ψk and considered in order to understand this theorem. ψn will correspond to what normal mode of the molecule In group theory the symmetry of an molecule is de- they are, and the irreducible representations of αij are scribed by symmetry operations that leave the molecule assigned based on how the products of coordinates trans- 2 2 2 unchanged. Some possible symmetry operations are: form under the symmetry operators (x , y , z , xy, xz, identity E, rotation Cn (rotation of the molecule around and yz) since, from Eqn. 5, we can see it contains the an axis through the molecule by an angle of 2π/n), reflec- elements αxx, αxy... which transform like products of co- tion σ (reflection of the molecule across a plane through ordinates. For a more detailed description of the group the molecule). For an example, H2O has four symmetry theory of molecular structures please see Ref.[5]. operations, the identity E, and, as illustrated in Fig.1, a C2 rotation about the axis by π through the hydrogen 2.2. Intensity for Vibrational Raman Scattering atom, and two reflections about appropriate planes: σv 0 and σv. All of the symmetry operations on a molecule must form a closed group: a symmetry group. A group In general the polarizability of a molecule will be given table can be written up which describes the result of ap- by a 3 by 3 polarizability matrix with components αij. plying two symmetry operations on the molecule. For Since the atoms in the molecule will vibrate about their example, applying the symmetry operation C2 twice to equilibrium positions one can do a Taylor expansion, as- H2O is the same as applying the identity E, since two suming the vibrations are sufficiently small, of the com- rotations by π is a rotation by 2π, which is the same as ponents of the polarizability matrix: doing nothing to the molecule. The other entries of the ∂α (0) group table of H2O can be found in a similar way, by con- X ij 2 αij(Qk) = αij(0) + Qk + O(Q ) (8) sidering how the molecule changes after an application of ∂Qk k two symmetry operations and determining which single In the harmonic approximation we truncate the series symmetry operation achieves the same result. The group to linear order in the displacements. Here Qk are the nor- table for H2O is listed in Fig.3. This group table isn’t mal coordinates for the k − th normal mode (equilibrium unique to H2O, it describes any molecules with these four positions are at Qk = 0). If we assume that incoming symmetry operations, it is called the C2v group. radiation is monochromatic and polarized along the z- The symmetry operations can be thought of as matri- direction we get the following equation for the induced 3 dipole moment: 2  2 0 0 ˆ 2 E0z ∂αiz(0) | hn1, ..., nN | µi(Qk) |n1, ..., nN i | ∝ ! ωk ∂Qk X ∂αiz(0) µi = αiz(0) + Qk E0zcos(ωI t) (9) ∂Qk (12) k Where we time averaged over one cycle to get rid of 2 From electromagnetic theory we know that the follow- the cos (ωI t) function, the 1/ωk comes from evaluating ing proportionality relation holds for the power emitted the matrix element using ladder operators for the k − th by an electric dipole with dipole moment ~µ and ν is the normal mode: frequency of scattered photons: 2  2 E X ∂αiz(0) P ∝ (ω ± ω )4 0z (13) 4 2 4 2 2 2 k I k ω ∂Q P ∝ ν |~µ| = ν (|µx| + |µy| + |µz| ) (10) k x,y,z k

From this point we will treat µi as a quantum opera- This gives the scattered power for one molecule for under- tor. The normal coordinates will act as position opera- going a transition through the k − th normal mode. Here tors in the Hamiltonian. In particular since everything the plus corresponds to Anti-Stokes scattering and the is written in terms of normal coordinates we know that minus corresponds to Stokes scattering. However typical the unperturbed Hamiltonian will be a sum over inde- samples in a lab will have many molecules that can scat- pendent harmonic oscillators vibrating at their respec- ter the incoming photons. If we assume the sample to be tive normal frequencies. As an aside one should note in thermal equilibrium at a temperature T then we know that the number of normal modes for a given molecule from basic statistical mechanics that the relative proba- with M atoms, which we will denote as N, equals 3M −6 bility, compared to the ground state, of finding a molecule for non-linear molecules and 3M − 5 for linear molecules in some excited state will be suppressed by by the Boltz- [2]. For example a water molecule has three atoms so mann weight. This implies that at sufficiently low tem- it has 3 normal modes. The unperturbed states will peratures we expect most of the molecules to be in the be products of a harmonic oscillator associated with ground state. Hence if Raman scattering does occur it is each normal mode. We will represent these states us- more likely that a photon will excite the vibrational mode ing the ket notation |n1, n2, ..., nN i, where the ket rep- to higher energy states and come out with lower energy resents a state in which the k − th normal mode is ex- this is often seen in experiments. This means that the cited to to the nk level. The state has a total energy of ratio of the intensity of Anti-Stokes scattering and Stokes PN scattering through some Raman active mode vibrating at i=1[~ωi(ni + 1/2)] which is a sum of the collective nor- mal mode energies. Hence from a quantum point of view ωk will be roughly given as: transitions from one state to another occur due to the 4 IAnti−Stokes (ωI + ωk) time-dependent perturbation caused by the oscillating −β~ωk = 4 e (14) electric field inducing a dipole moment in the molecule. IStokes (ωI − ωk) −1 So to understand intensity of such transitions we should where β = (kbT ) . For a more precise treatment for 0 0 0 ˆ 2 calculating intensity please see Ref.[1]. calculate| hn1, n2, ..., nN | µi(Qk) |n1, n2, ..., nN i | . Since we are interested in Raman transitions this would corre- spond to having the initial state unequal to the state af- ter the interaction so the zeroth order term vanishes due 3. RAMAN EXPERIMENT to orthogonality (The zeroth order term actually would represent Rayleigh scattering). So we are left with: 3.1. Experimental Set-up

hn0 , n0 , ..., n0 | µ (Qˆ ) |n , n , ..., n i Fig.4 illustrates a typical Raman system. The system 1 2 N i k 1 2 N is mainly composed of three parts: incident light part, X ∂αiz(0) 0 0 ˆ sample part and scattered light part [6]. Fig.5 corre- = E0zcos(ωI t) hn1, ..., nN | Qk |n1, ..., nN i ∂Qk k sponds to the sample part of a Raman system, where (11) the red arrow stands for the direction of incident light, and the blue arrow stands for the scattered light, this We know from the the ladder for- picture was taken from Dr. Guangrui Xia's lab, located malism for SHO that the matrix element in the department of materials engineering at UBC. In 0 0 0 ˆ 0 hn1, n2, ..., nN | Qk |n1, n2, ..., nN i is non-zero iff nk = ' 0 Dr. Guangrui Xia s lab Raman spectra are collected on a nk ± 1 and ni6=k = ni6=k. We can only have transitions of backscattered Horiba Jobin Yvon HR800 Raman system the form |n1, n2, .., nk, .., nN i → |n1, n2, .., nk ± 1, .., nN i, with 442nm (2.81 eV) line from a He-Cd laser. The 442 assuming that ∂αiz (0) 6= 0, if it is zero then a transition nm incident laser beam is polarized. ∂Qk through that mode will not occur such modes are called To measure the Raman spectra, we first use the micro- Raman inactive. So we are left with the following type scope to focus on the sample, then turn on the laser. The of transition element which is going to correspond to a scattered light goes through various optical devices and transition through the k − th mode: finally the signal is collected by a CCD. Thus, Raman 4

FIG. 6: The stokes and anti-stokes Raman spectra of FIG. 4: Typical experimental set-up of a micro-Raman Rhodamine 6G (RH6G) under different temperatures spectrometer

FIG. 5: Sample part of a Horiba Jobin Yvon HR800 Raman system FIG. 7: The Raman spectra of BP in 4 different crystal orientations. The 3 peaks correspond to the 3 different normal modes. spectra will appear in the computer. A heating stage can be used to create a higher temperature environment for the sample. 3.3. Raman vs. Crystal Orientation

3.2. Raman vs. Temperature Fig.7 shows three typical Raman Modes in Black Phos- 1 2 2 phorus (BP):Ag,Ag,Bg [8]. In this example, for the Ag From the equation of intensity from Ref.[7] it is clear mode only has diagonal parts,  iσa  ~ν ae 0 0 that the Stokes Raman Intensity Is ∝ (ωI − ωk)e kT , ∂ ~αij = 0 beiσb 0 (15) differs from the Anti-Stokes Raman intensity, Ias ∝   ∂Qk iσc − ~ν −1 0 0 ce (ωI +ωk)(1−e kT ) . At low temperatures these inten- sity proportionalities match what was derived in section From Eqn.13 in section 2.2, we can see that the intensity 2.2. Fig.6 shows the Anti-Stokes (AS) (left) and Stokes of the scattered light, and therefore the electric field of (S) (right) spectra for the 610 cm−1 mode of Rhodamine the scattered light, satisfies, ∂ ~αij 6G (RH6G) under different temperatures. The x-axis E~scattered ∝ · E~incident, (16) is the differences between the frequency of the incident ∂Qk ~ light and the scattered light, which is in the units of (in section 1.2 the specific case of Eincident = Eincidentzˆ wave-number (cm−1). The y-axis stands for the relative was used).Usually we add a polarizer with the parallel po- intensity of the scattered light. Since the intensity de- larization of the incident light when measuring the scat- pends of the parameters of the set-up, the units differ for tered light. If we assume the parallel polarization of the ~ each experiment, and therefore are given the general la- incident light as Eincident = (sin θ 0 cos θ), the intensity bel “arbitrary units”. Note the change in scale between is, the AS and S spectra. Panels a and b show the AS and ∂ ~α Ik ∝ |(sin θ 0 cos θ) ij (sin θ 0 cos θ)T |2 Ag S modes respectively at 300 K (dark gray) and 170 K ∂Qk (light gray). It is clear that while there is little effect c c = |a|2[(sin θ2+| | cos φ cos θ2)2+| |2 sin φ 2 cos θ2]. on the intensity on the Stokes side there is a significant a ca a ca decrease in the intensity of the AS signal as the tem- (17) perature is decreased, as is expected from a decrease in Since the intensity is angularly resolved, we see the re- thermal population [7]. This result is expected from the sults in Fig.7, where intensity changes with changing an- theory described in section 2.2. gles [9]. 5

3.4. Raman Strain vs. Stress

The Raman scattering spectrum is a useful tool to re- solve internal strains and stresses in materials through the high resolution of determining normal mode frequen- cies, and it represents an effective and nondestructive technique. Tensile or compressive stress can induce the redshift or blueshift of each Raman peak. Once the strain/stress relationship is known for a specific mate- rial, it offers a convenient way, by Raman scattering, to detect the strain/stress distribution on a sample of that material. Silicon (Si) and Gallium Nitride (GaN) were used as the examples to illustrate the relationship between stress and Raman spectroscopy. Fig.8 shows typical Raman spectra measured from Si under tensile, free, and com- pressive stresses. According to the result, for silicon, a compressive stress will lead to a blueshift in spectra, FIG. 9: (a) The Raman peaks shift of Si under applying while a tensile strain will lead to a redshift. And the re- increasing compressive stress loadings (b) The Raman peaks sponse is almost linear in certain range of stress[10]. Fig.9 shift of GaN under the same condition illustrates the Raman spectra of Si and GaN under in- creasing compressive stress [11]. Curves “a”-“f” stand for one Raman spectra under increasing compressive stress. Each curve, except curve “a”, have been moved up in par- allel for comparison. Both the materials show a blueshift, and the shift almost increases linearly with the stress.

4. CONCLUSION

In this article we explored the experimental technique of the Raman effect. We began by describing the theory that predicts when Raman transitions will occur, as well as how intense the signal from those transitions should be. Then we looked into the experimental side of the Raman effect. We described how Raman spectra are measured, and analyzed various spectra from different experimental conditions, noticing how the theory tied in with what was experimentally observed.

FIG. 8: (a) Typical Raman spectra measured from Si under tensile, free, and compressive stresses (b) Relation between Si stress and measured Raman shift (peak position) 6

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