Fourier's Trick and calculation of Cn

[I_laff]

I understand that the solutions to the time-independent Schrodinger equation are complete, so a linear combination of the wavefunctions can describe any function (i.e. $f(x) = \sum_{n = 1}^{\infty}c_n\psi_n(x) = \sqrt{\frac{2}{a}} \sum_{n=1}^{\infty} c_n\sin\left(\frac{n\pi}{a}x\right)$ for the infinite well case).

I don't understand how using Fourier's trick allows you to calculate $c_n$. Also, I don't understand how $\sum_{n=1}^{\infty}c_n\delta_{mn} = c_m$. I get that the Kronecker delta comes out of the fact that $\psi_m^\ast(x)$ and $\psi_n(x)$ are orthogonal. I understand that when $m = n$, then the Kronecker delta equals 1, but how does this allow us to find $c_m$ and $c_n$ .
$$ \int_{0}^{a} \psi_m^\ast(x) f(x)\mathrm dx = \sum_{n=1}^{\infty}c_n \int_{0}^{a} \psi_m^\ast (x)\psi_n(x)\mathrm dx = \sum_{n=1}^{\infty}c_n\delta_{mn} = c_m $$

 

[BvU]

I don't understand how using Fourier's trick allows you to calculate $c_n$.

Read up on Fourier transforms in any textbook of your liking.

Also, I don't understand how $\sum_{n=1}^{\infty}c_n\delta_{mn} = c_m$

That's because $\delta=0$ unless m = n; then it's 1.

Fourier analysis is a bit more than just a 'trick'

 

[I_laff]

Fourier analysis is a bit more than just a 'trick'

I referred to it as a trick because that's what the material I was reading referred to it as. I'll take a look into Fourier transforms then.

 

[Nugatory]

I referred to it as a trick because that's what the material I was reading referred to it as. I'll take a look into Fourier transforms then.

What was that material? You'll generally get better answers if you identify your sources - chances are that someone here is already familiar with it.

(I remember that exact phrase "Fourier's trick" appearing in a thread here a while back)

 

[I_laff]

It was from Griffiths, Introduction to quantum mechanics second edition chapter 2

 

[BvU]

Found it on p 28. (and you are not responsible for his wording )
Now I'm puzzled about your question: you reproduce the line that works out $c_m$, so that should answer it !?

($c_n$ does not come out, though! n is just a summation variable)

 

[I_laff]

I understand the properties of the Kronecker delta, what I didn't understand was why $c_m$ came out. So, let me get this right. When applying the Kronecker delta to $\sum_{n = 1}^{\infty} c_n$, you will get a coefficient value that both $\psi_n$ and $\psi_m$ share (due to the properties of $\delta_{mn}$)?

What I misunderstood previously, was that I thought $\psi_m$ consisted of multiple coefficients, and so $\delta_{mn}$ when applied would work out the mutual coefficients between $\psi_n$ and $\psi_m$, however only one coefficient ($c_m$) was stated.

 

[phyzguy]

Think of it like a dot product. If I have a 3D vector $\overrightarrow{V} = V_x \overrightarrow{e_x} + V_y \overrightarrow{e_y} + V_z \overrightarrow{e_z}$, and I take the dot product with a unit vector, I get $\overrightarrow{V} \centerdot \overrightarrow{e_x} = V_x$. this is because the unit vectors are orthogonal, so the dot product "picks off" the component along the x-axis. It's exactly the same in Fourier analysis, even though there are an infinite number of "unit vectors".

 

[I_laff]

Ah, that's a nice way to think about it