The Orthogonality of Hermite Polynomials

[@MisterTutor2010]

When I figured this out before dinner yesterday, I thought I was wrong because what if m < m? This video pointed out to me that m and n can switched so I had figured the solution correctly all along.

[@FunnCubes]

I screamed at my computer at how easy that was. I tried to do it unsuccessfully on my own for a day...... omg..... Thank you so much.

[@physicsandmathlectures3289]

You're welcome!

[@sanderscheel]

Great Video and thank you for the clear explanation!

[@physicsandmathlectures3289]

Glad it was helpful!

[@ventrue1999]

At 5:29 , why is it just exp(- x^2)? I know the result will still be 0, but not quite following that jump. Thanks

[@physicsandmathlectures3289]

I'm skipping a few steps in that part. The idea is that each time we do integration by parts we generate a boundary term and a new integral where we take the derivative that is on one term (the e^(-x^2) in this case) and then move it over to the other term (the Hermite polynomial). If we do that a total of m times then we have the sum of boundary terms, as well as that term in red with e^(-x^2). The important fact is that all of the boundary terms go to zero because they all have a factor of e^(-x^2), so we really are just left with the integral involving only e^(-x^2)

[@Rdffuguihug]

Excellent video. Thanks for sharing.

[@MrMagraden]

One question, why the evaluation of the first term is 0 because of the gaussian? We don't have only to take into account the infinity but the central part, where the density is concentred

[@vasundarakrishnan4093]

You are sooo awesome. Legit.

[@captainfartolini4335]

you are assuming H_n is poly in the limit, but was that shown?

[@domenicagarzon6787]

why can you rewrite the hermitian pol as a derivative ?

[@calypo1491]

It follows from the Definition of the Hermitian Polynomial: H_m(x)=(-1)^m * exp(x^2)*d^n/dx^n(exp(-x^2)). The (-1)^m is pulled out of the integral bc of linearity. The exp(x^2) and the exp(-x^2) in the Integral cancel each other and the derivative of exp(-x^2) is the only term that is left.