I am reading about Raman spectroscopy in general. And I learned that Raman spectra reflect the changes in the polarizability of a molecule, while the IR spectra reflect the changes in the dipole moment of a molecule. However, I do not have an intuitive feeling why that is so and what the key difference between the two is.

So, a dipole moment is just a displacement of charges, say, of electron cloud relative to the nuclei, after the molecule has been excited by a laser IR light. Dipole moment has an extent along one direction. Thus, it is denoted by a symbol $\mu_i,\ \ i=x,y,z.$

Polarizability is an intrinsic property of a molecule, showing the strength of the ability of the molecule to be polarized. The polarizability is represented by a tensor 3x3, $\ \ \alpha_{ij}, \ \ i,j=x,y,z$, reflecting the fact that the molecule can be polarized to a different extent in different directions. I guess, non-zero off-diagonal components (for example, for $\alpha_{xy}>0$) mean that exciting the molecule by $x$-polarized light, can induce dipole moments in both $x$ and $y$ directions. Or only in $y$ direction?

It seems that dipole moment is along one axis, and polarizability has two coordinate components.

However, by a rotation of coordinate system, we can move to the frame $(x',y',z)$, one can represent the dipole moment as superposition of two dipole moments: $\mu_x=\mu_{x'}+\mu_{y'}=\mu_{x'y'}$, right? Now it has two components, as polarizability does.

So polarizability is the potential ability of the molecule to get polarized, while the dipole moment is something real that has been already induced. Right?

And what does it mean that Raman signal probes the polarizability of a molecule then?


1 Answer 1


A polarizability tensor relates the vector components of the applied field to the vector components of the induced polarization by a scaling factor (nine relations in 3-space). For a field $\mathbf F$, polarization $\mathbf p$ and polarizability tensor $\alpha$, one can represent this is through the matrix equation:

$$ \quad \begin{bmatrix} p_x\\ p_y\\ p_z \end{bmatrix} \ = \ \begin{bmatrix} \alpha_{xx} & \alpha_{xy} & \alpha_{xz}\\ \alpha_{yx} & \alpha_{yy} & \alpha_{yz}\\ \alpha_{zx} & \alpha_{zy} & \alpha_{zz}\\ \end{bmatrix} \ \begin{bmatrix} F_x\\ F_y\\ F_z \end{bmatrix} $$

Tensors have some special (transformation) properties not shared by all matrices, but let's not worry about that now. To see which the tensor elements connect specific field and polarizability elements, just do the matrix multiplication:

$$p_x = \alpha_{xx}F_x + \alpha_{xy}F_y + \alpha_{xz}F_z\\ p_y = \alpha_{yx}F_x + \alpha_{yy}F_y + \alpha_{yz}F_z\\ p_z = \alpha_{zx}F_x + \alpha_{zy}F_y + \alpha_{zz}F_z$$

You may notice that the first index of $\alpha_{jk}$ always corresponds to the polarization direction, and the second index always corresponds to the field direction. For example, the component $\alpha_{xy}$ 'dots' into the $y$ direction of the applied field and results in a polarization in the $x$ direction, while $\alpha_{yx}$ 'dots' into the $x$ field direction and corresponds to a polarization in the $y$ direction. Imagining this 'dotting' with the closest component helps remember which index relates to the field/polarization.

Incidentally, the note above makes a good case for index notation:

$$p_j = \alpha_{jk} F_k.$$

A few more examples may be a helpful reference for relating index notation, matrix notation, and the physical system...

For an isotropic object/medium, the polarization always points in the same direction as the electric field scaled by a constant. This is given by a diagonal matrix:

$$\alpha = c\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix}$$

or in index notation: $p_k = c F_k$.

An object whose polarization is in the same direction as the applied field but is more polarizable in a given direction (say $z$) is given by:

$$p = \begin{bmatrix}c_1 & 0 & 0\\ 0 & c_1 & 0\\ 0 & 0 & c_2\end{bmatrix}$$

And any polarizability tensor with off diagonal elements result in polarization induced in a different direction than the applied field, as stated above.

  • 1
    $\begingroup$ And if there are two non-zero elements in the polarizability matrix, say, $\sigma_{xy}$ and $\sigma_{yy}$, then the field $E=(0,E_y,0)$ will create the polarization vector $p=(\sigma_{xy}E_y,\sigma_{yy}E_y,0)$. Correct? So, now the polarization of matter and the polarization of light are not in the same directions. The field is polarized along $y$ direction, and matter is polarized along the direction pointing in between $x$ and $y$ directions. $\endgroup$ Aug 29, 2016 at 12:20
  • $\begingroup$ @CapoPavelMestre it looks like you've got it. Have a look at the answer I've provided, it summarizes the key points you've picked up on. $\endgroup$
    – anon01
    Aug 29, 2016 at 12:36
  • $\begingroup$ Thanks a lot! I guess, I got this part. But what are the specific transformations for this matrix, which are not shared by all matrices? And it is still not clear why Raman signal tells information about polarizability of a medium, while IR spectra reflect the information about polarization of the medium, which is the main part of my question... $\endgroup$ Aug 29, 2016 at 12:47
  • $\begingroup$ Vectors have specific transformation properties (ie, change of basis representation, or rotation of coordinate labeling don't affect the modulus) and therefore an object that relates vectors must also obey these properties under suitable transformation. Google tensor transformations if you want to know more. $\endgroup$
    – anon01
    Aug 29, 2016 at 12:56
  • $\begingroup$ As for Raman signals, I suggest you submit a second question. $\endgroup$
    – anon01
    Aug 29, 2016 at 12:57

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