Overlap Integrals: Understand & Learn from a Source

[Ahmed123]

Hi
I study optics and many times i found a term called (Overlap integral ) as attached pictures .. I can't understand from where these expression comes (mathematically) and what these functions means in particular ( even from mathematical point of view)
I can't understand the nature of multiplying the two functions then integrating or making the integrals of individual functions then multiplying.. So I ask if anyone can help me to find a source for reading to understand .. or explain these expressions for me .. thank you in advance .

 

[Cutter Ketch]

If you have light focused down so it is imaged into a small spot on the tip of a fiber, this expression answers the question “how much of the light will go into the fiber?” (However this isn’t the only place in physics you will find an “overlap integral”). This function is a correlation of how well the field of the imaged spot (“light field”) overlaps the shape of the field the fiber can propagate (“mode field”).

The easiest way to describe it is to start with something simpler. Suppose you had a beam of light that made a perfectly circularly uniform spot on the wall. (Uniform intensity “top hat” profile). And suppose you had a circular window. In this scenario, all the light that hits the window goes through. If the circular beam partially overlaps the circular window, part of the beam goes through.
Mathematically how would you calculate how much gets through? One way to do it would be to write a function that describes the beam in some coordinate system: a constant value inside the circle, zero outside. Normalize this to one (divide by the integral of the function). Now in the same coordinates describe the window with a similar normalized function. If you multiply the two functions together and integrate you get the fraction of light that goes through the window. It’s how much gets through because the multiplied function is nonzero only where both functions are nonzero. (the “overlap”) It’s the fraction because the two functions are normalized to 1.
Of course you might have come up with a simpler geometric expression for that particular case, but this recipe is more generally useful. Suppose the beam isn’t uniform, but has a Gaussian intensity distribution. Same recipe: describe the intensity shape, normalize, multiply by the normalized window shape and integrate. Finally suppose the window isn’t completely transparent everywhere but has some pattern of transparency. You still use the same recipe.

One caveat, the simplified description above contemplates using the intensity throughout, but, really, when it comes to mode coupling, you overlap the fields. The intensity is proportional to the square of the field, and that is why your expression has all those conjugate squares.

So now let’s apply this to your expression. The light incident on the fiber tip has some field profile. Integrating the conjugate square of the field gives the total intensity. Dividing the field by the square root of the integrated conjugate square gives the normalized field. The fiber has a mode distribution that it can propagate. Dividing the description of the mode field by its integrated conjugate square gives the normalized mode field. Multiplying the two fields gives the field coupling: how much of the incident field couples into the fiber mode. Taking the absolute square of the whole expression puts this in terms of coupled intensity. The normalizers are real, so they just get squared and are once again intensities rather than fields.

 

[Ahmed123]

If you have light focused down so it is imaged into a small spot on the tip of a fiber, this expression answers the question “how much of the light will go into the fiber?” (However this isn’t the only place in physics you will find an “overlap integral”). This function is a correlation of how well the field of the imaged spot (“light field”) overlaps the shape of the field the fiber can propagate (“mode field”).

The easiest way to describe it is to start with something simpler. Suppose you had a beam of light that made a perfectly circularly uniform spot on the wall. (Uniform intensity “top hat” profile). And suppose you had a circular window. In this scenario, all the light that hits the window goes through. If the circular beam partially overlaps the circular window, part of the beam goes through.
Mathematically how would you calculate how much gets through? One way to do it would be to write a function that describes the beam in some coordinate system: a constant value inside the circle, zero outside. Normalize this to one (divide by the integral of the function). Now in the same coordinates describe the window with a similar normalized function. If you multiply the two functions together and integrate you get the fraction of light that goes through the window. It’s how much gets through because the multiplied function is nonzero only where both functions are nonzero. (the “overlap”) It’s the fraction because the two functions are normalized to 1.
Of course you might have come up with a simpler geometric expression for that particular case, but this recipe is more generally useful. Suppose the beam isn’t uniform, but has a Gaussian intensity distribution. Same recipe: describe the intensity shape, normalize, multiply by the normalized window shape and integrate. Finally suppose the window isn’t completely transparent everywhere but has some pattern of transparency. You still use the same recipe.

One caveat, the simplified description above contemplates using the intensity throughout, but, really, when it comes to mode coupling, you overlap the fields. The intensity is proportional to the square of the field, and that is why your expression has all those conjugate squares.

So now let’s apply this to your expression. The light incident on the fiber tip has some field profile. Integrating the conjugate square of the field gives the total intensity. Dividing the field by the square root of the integrated conjugate square gives the normalized field. The fiber has a mode distribution that it can propagate. Dividing the description of the mode field by its integrated conjugate square gives the normalized mode field. Multiplying the two fields gives the field coupling: how much of the incident field couples into the fiber mode. Taking the absolute square of the whole expression puts this in terms of coupled intensity. The normalizers are real, so they just get squared and are once again intensities rather than fields.

Woooow ,, then these denominator comes from normalization of these functions !
I can't describe how your answer is great and helpful ..

 

[Xinkuo Yu]

Hi, can you provide a reference for the first image you provided? Thank you so much for your help