Symmetric stretching vibrations cause the greatest changes in the polarizability derivative with respect to the vibrational coordinate and therefore are showing up in Raman as the strongest (and usually sharply polarized) lines. This means that it is absolutely safe to ascribe the most intense line in the Raman spectrum of a substance (either liquid or solution or crystal) just to THE symmetric stretching vibration. 

This conclusion follows from Wilson et al and from Turrell adn from any other textbook on vibrational spectroscopy.

Boson peaks manifest themselves at much lower frequencies, ca. 20-150 cm-1. They have a cheracteristic asymmetric profile and due to this circumstanse they cannot be mistakenly ascribed to molecular (or lattice) vibrations themselves.

Not sure what you mean in this instance by 'symmetric' or 'antisymmetric.'  I'm also not sure you can characterize the vibrational mode as a "In-O-In" stretching mode, considering that the unit cell (described later) has two In atoms and six O atoms.  You could still refer to it as a vibrational mode with the majority contributions from In-O stretching (assuming your assignment of the 386 cm-1 to In-O stretch is correct).  However, maybe you already did the vibrational calculations and know that this is a In-O-In stretching mode.  The rest of this posting assumes that the 386 cm-1 mode is a vibrational mode with majority contribution from In-O stretching motions.

The LaInO3 has a Pnma space group (http://ssirl.snu.ac.kr/web/data/Lanthanium%20indium%20oxide%20from%20x-ray%20powder%20diffraction.pdf)  where you have two La, two In, and six O atoms in a unit cell.  A drawing of the unit cell looking down on the b-axis is in the PDF report.

The unit cell has an inversion center (since the space group is Pnma), which means that the Raman and infrared modes are mutually exclusive.  In your specific case, a 'symmetric' mode (which in the usual sense means a vibrational mode that preserves the symmetry of the space, point, or line group in question) would be Raman active and not infrared active.   Now, the In-O point group is octahedral (Oh), see http://www.webqc.org/symmetrypointgroup-oh.html, and there is only one infrared active In-O stretching mode possible, but three different Raman modes possible for In-O stretch.  Only one of them will be a "polarized" band whereas two will be "depolarized."  You may be able to distinguish polarized vs. depolarized by using a polarizer analyzer on the Raman signal, as long as there are no Mie scattering events (i.e. as long as your sample is a "large" crystal compared to your laser beam size).

Dear Dr. Kaoru

Thanks for the answer. I have found from the literature that for LaInO3, the peak at 386 cm-1 is due to vibrational mode and you rightly said that the contribution is due to stretching. http://www.sciencedirect.com/science/article/pii/S0022369700002201. 

The question still remains that is this vibration mode is symmetric. Anyways thanks for the detailed explanation. 

Maybe you can record  the polarized Raman spectra of your sample, if the depolarization ratio of this peak is ~0.75, it is antisymmetrical mode, and if the depolarization ratio is less than 0.75, it is symmetrical mode.

can you apply model compounds known assigement to find your solution?

Mr. Kapil, it helped that you cited the paper that you were reading.  The doping levels cited there are upwards of 10 mol%, and the paper abstract mentions that new modes are appearing as Raman active.  When new modes appear in a crystal which previously has an inversion center, then you are seeing mode that were previously only Infrared active (let's loosely call that "antisymmetric").

But you also say that the 386 cm-1 mode is strong.  So before insisting on my suggestion of doing polarized Raman spectroscopy experiments, I would ask:  (1) did you see the 386 cm-1 peak for the pure LaInO3, or (2) was it only starting to appear when you start doping the Ca? 

If you say "yes" to #1, then it is a "symmetric" mode.  If you say "yes" to #2 instead, then you're looking at an "antisymmetric" mode.

Also, to address Ke Lin's comment: that depolarization ratio of 0.75 applies specifically for polarized incident light into an isotropic liquid where there is no macroscopic scattering (which would destroy polarization information).  So it's not appropriate for crystals.  Now, you can still do polarized Raman and get qualitatively accurate information, you just need to make sure you have a large enough crystal (powders or tiny crystals will cause macroscopic scattering and destroy polarization information) and have some idea which way the a, b, and c axes are pointing on your sample.

Dear all Thanks for the contribution.

Dear Dr. Kaoru

The Raman mode at 386cm-1 is the strongest peak and also appeared in undoped sample (LaInO3). As Ca is doped, the peak intensity of all Raman modes start decreasing uniformly with Ca-doping. It may be due to increase in disordering in sample as Ca-can substitute at both La- and/or In-site creating oxygen vacancies and lattice distortions.  Also, for highest intensity peak (386cm-1) the full width at half maximum  (FWHM) uniformly increases, which could be due to weakening of In-o bond with doping. 

As per your advice, (yes to #1) the mode at 386 cm-1 is symmetric, but please refer me some tutorials to get better understanding to this study.  Thanks for the healthy discussion.

As you know in this region Bozon peack is located. This band  can be assigned to this tipe of  band? You can verify that.

I am agree with Kaoru Aou

Not sure if there is a good tutorial (I would be interested in it as well for pedagogical purposes).  Here are two books I learned from a long time ago:

Molecular Vibrations (classic text by E B Wilson - maybe a hard read at first, with many equations, but teaches almost everything you need) ---http://books.google.com/books/about/Molecular_Vibrations.html?id=CPkvsDrPiv0C

Infrared and raman spectra of crystals by George Turrell -- has lots of pictorial examples with worked out solutions for vibrational analysis.  This might be a better book to start with for quick solutions. http://books.google.com/books/about/Infrared_and_raman_spectra_of_crystals.html?id=u_KF20knLf0C

Dear Dr Kaoru

Thanks for the books and your suggestions.

We can get the data of symmetric or antisymmetric from the software. We use ACD Lab Software.

Symmetric stretching vibrations cause the greatest changes in the polarizability derivative with respect to the vibrational coordinate and therefore are showing up in Raman as the strongest (and usually sharply polarized) lines. This means that it is absolutely safe to ascribe the most intense line in the Raman spectrum of a substance (either liquid or solution or crystal) just to THE symmetric stretching vibration. 

This conclusion follows from Wilson et al and from Turrell adn from any other textbook on vibrational spectroscopy.

Boson peaks manifest themselves at much lower frequencies, ca. 20-150 cm-1. They have a cheracteristic asymmetric profile and due to this circumstanse they cannot be mistakenly ascribed to molecular (or lattice) vibrations themselves.