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Raman Spectroscopy 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 hΨkj µ jΨni = Mkn + Mkn. Where the zeroth order term, 0 i!k;nt Mkn = µk;ne , is ignored because it is just a regu- The Raman effect is a light-scattering phenomenon lar transition dipole moment, 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 = 0e−i!kt, 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(!k;n−!)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~ !r;n − ! !r;n + ! man scattering, and is classified as either Stokes scat- r i(!k;n+!)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 ! + ! ! − ! scattered incident molecular ~ r r;n r;n in frequency is called the Raman shift: ∆ν = jνincident − (2) ν j. Section 2.1 describes the theory behind pre- scattered The first term in Eq.2 with angular frequency !kn − ! 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 !kn + ! 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 hΨnj to hΨkj from the Raman effect can be a transition from high to low frequency, !kn = (Ek − En)=~ > 0 (Stokes scattering), or from low to high fre- 2. RAMAN THEORY quency, !kn = (Ek − En)=~ < 0 (anti-Stokes scatter- ing). It is assumed that the incident photon, with fre- 2.1. Raman Transitions quency ! as sufficient energy to induce the dipole mo- mentum transition: ! − !kn > 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 !t, 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, ! 6= (E − E )= = ! for Looking at the kn-matrix element of this expression us- r k ~ rk 0 0 −i!kt any normal modes jΨ >, jΨ > of the sample, how- ing the first order wave functions Ψk = ke and r k ~ ~ ever it will induce an oscillating dipole moment that subbing in our oscillating electric field E = E0 cos !t 0 0 will re-radiate. Therefore transitions between normal then the expression becomes Mi;kn = hΨkj µi jΨni = 0 0 i(!kn−!)t i(!kn+!)t modes in the system are found by solving the transi- 1=2 h kj α^i;j j ni E0j(e + e ). As before tion dipole moment: Mr;k(t) = hΨrj µ jΨki. The nor- the term with frequency !kn +! is ignored and we're left mal modes are found by perturbatively solving the time- with: 0 0 i(!kn−!t) dependent multi-particle Schrodinger equation with the Mi;kn = 1=2 h kj α^i;j j ni E0je (3) with the electronic-dipole interaction perturbation, H^0 = Comparing Eqn. 3 with the kn-component of the Raman −µ·E0 cos !t, resulting from the applied electric field and part of Eqn. 2, i(!k;n−!)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 ! − ! ! + ! j ~ r rn rn 1 X µr;n · E0 µr;n · E0 1 0 −i(!n+!)t −i(!n−!)t (4) Ψn = r e + e : 2~ !r;n − ! !r;n + ! we find that the kn-polarizability matrix elements are: r (1) 1 X µ^i;krµ^j;rn µ^i;rnµ^j;kr αij;kn = + (5) Where the sum is over all of the time independent ~ !rn − ! !rk + ! 0 0 r zeroth-order solutions, µrn = h r j µ j ni, and !rn = (Er − En)=~. Using this first order correction we can approximate the dipole moment as: Mkn ≈ 0 1 0 1 0 0 0 1 hΨk + Ψkj µ jΨn + Ψni ≈ hΨkj µ jΨni + hΨkj µ jΨni + 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 µ^i jri hrj µ^j µ^j jri hrj µ^i α = h 0j + j 0 i representation is where nothing is done to the molecule, ij;kn k ! − ! ! + ! 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). h kj αij j ni = kαij ndτ 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 k, αij, and n pro- ∗ duces a representation which is or contains the symmetric k, αij and n 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 k and considered in order to understand this theorem. n 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.
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