The emission and absorption transitions are only a mirror image if they relate to the electronic transitions between the same two states and there are no major molecular configuration changes in the excited state. Often additional transitions are present.

The "shoulder" peak you mention is probably an unresolved vibrational-electronic (vibronic) transition. Look at Lakowicz, Principles of Fluorescence Spectroscopy, 3rd Ed. pages 7,8 for information.

Search for Jablonski diagram.

Jablonski diagrams are used in a variety of forms, to illustrate various molecular processes that can occur in excited states.

The emission and absorption transitions are only a mirror image if they relate to the electronic transitions between the same two states and there are no major molecular configuration changes in the excited state. Often additional transitions are present.

The "shoulder" peak you mention is probably an unresolved vibrational-electronic (vibronic) transition. Look at Lakowicz, Principles of Fluorescence Spectroscopy, 3rd Ed. pages 7,8 for information.

I agree with Peter, but there are also much more trivial reasons, such as reabsorption if the concentration is too high or energy transfer to a trace impurity in the polymer chain, excimer or exciplex formation etc. THe mirrow image prevails in riigid, aromatic molecules in dilute solution.

There are many, excellent text books on this and related topics.

Walter Klöpffer

@Walter

Reabsorption lowers the intensity of the blue-side of the emission spectrum. The questioner probably refers to the red-sided shoulder in the emission spectrum of the polydiacetylene, for which dr Tanner gave the answer

@ Harry

If blue side emission is completely vanished due to reabsorption but, still there exist red side emission. Then what is the reason for red side emission?  

In spectroscopy terms only the O"-O' in absorption and emission overlap.

This notation means the origin of the vibronic spectrum: it is the transition without change in vibration

I like to refer to a standard fluorophore anthracene with its progression of vibronic bands in the stretching mode.

Therefore only the "first" band around 375 nm can be reabsorbed. The rest of the fluorescence is not absorbed. The spectrum of a highly concentrated solution therefore is more red

https://www.google.nl/search?q=anthracene+fluorescence+spectrum&tbm=isch&imgil=xZMYLG7a9PRGkM%253A%253B8K_HQm4wOyNj_M%253Bhttp%25253A%25252F%25252Fcnx.org%25252Fcontents%25252F81bb0311-98ee-4cfc-b3c8-0eab6aeace37%2525402%25252Fphotoluminescence-spectroscopy-and-its-applications&source=iu&pf=m&fir=xZMYLG7a9PRGkM%253A%252C8K_HQm4wOyNj_M%252C_&usg=__-9GST1QhH7HGqivW5w6EvwYaJyc%3D&biw=1350&bih=636&ved=0ahUKEwi6ufql6_fUAhXFK1AKHTZUA8oQyjcINQ&ei=8NdfWbqBM8XXwAK2qI3QDA#imgrc=txXo_X8fdRNU5M:&spf=1499453471380

In some cases, the red shift may be due to excimer emission in addition to reabsorption effects. Be very careful in explaining such effects and always compare with spectra in high dilution.

UV/VIS absorption/emissiion absorption /emission spectroscopy (including proton transfer and excimer formation) is nevertheless great fun.

Kind regadrs

Walter Kloepffer.