Learn Fourier Transforms: Books & Applications for QM

[kashokjayaram]

Can anybody helps in suggesting books on Fourier transforms and applications. I have seen many applications of Fourier transforms. But, I'm not able to visualize what's going on. Fourier transformations are there in Quantum mechanics also. It will be helpful in learning quantum mechanics.

Thanks

 

[sophiecentaur]

Many years ago I was really stumped for an understanding of the idea of transforms, convolution, correlation, modulation etc etc.
The thing that tipped the balance for me (to allow me to visualise the process) was when I looked again at the identity:

∫(-∞ to +∞) A Cos(ax) Cos(bx) dx is zero except for a = b, when it equals A

(If you avoid using the exponential notation, it is easier to explain.)
If you accept that any function f(t) can actually be represented by an infinite sum of simple harmonic functions then putting f(t) in the above gives you

∫(-∞ to +∞) f(t)Cos(bt) dt, which is the amplitude of the Cos(bx) component of f(x). That is what the Fourier transform does - it shows the correlation between your function (as a function of time) and the harmonic function with a particular angular frequency b over the whole range of b values - which shows it as a function of frequency.
I really must get to grips with writing equations better! but I think the above says what I mean. You need to tart it up a bit, with exponential notation but the basic idea is the correlation between the time function and what is, effectively, a swept frequency - to give the frequency function.

 

[DrClaude]

The book by Lanczos, Applied Analysis, has a nice chapter on harmonic analysis.

 

[ZombieFeynman]

If you've taken any Linear Algebra, you can think of the Fourier Transform of a function as an expansion in terms of plane waves.
Sophiecentaur's observation of can then be understood as a result of orthogonality.

 

[BruceW]

I think maybe it is easiest to start with Fourier series. Because then you can clearly see that you are really just writing out some periodic function as a sum of cosines and sines. Which is fairly intuitive in my opinion. Also, you can directly calculate the partial Fourier series (i.e. only use the first few terms), and you will see that the partial Fourier series will already approximately look like the periodic function you are trying to represent. And the more terms you add to your series, the more it will look like the function you are trying to represent.

And then, the Fourier transform can be seen as a generalization of the Fourier series. The Fourier transform can be used on non-periodic functions. So, roughly speaking, as the period of our function tends to infinity, we must use a continuum of frequencies, rather than discretely spaced frequencies. So we must use an integral, instead of a series.

 

[sophiecentaur]

I have no problem with the Maths but I like the almost physical process of 'scanning' the time domain test signal with a swept probe frequency and working out what!s there at each frequency step. It is a direct equivalent to the old spectrum analyser. The OP seemed to be after something tangible and my approach is along those lines.

 

[stevendaryl]

∫(-∞ to +∞) A Cos(ax) Cos(bx) dx is zero except for a = b, when it equals A

That's not quite right, as you can see in the case $a=b=0$. Then that integral gives infinity, rather than [itex].

The correct statement is this:

For $a > 0, b > 0$, the limit as $L \rightarrow \infty$ of

$\dfrac{1}{L} \int_{-L}^{+L} A cos(a x) cos(b x) dx$

is equal to $A$ if $a=b$
is equal to $0$ if $a \neq b$.
​

 

[sophiecentaur]

Thanks for adding the rigour. I had a feeling that it needed something to keep the integral finite. It's been a long time . . . . .
The hardware equivalent is a synchronous detector / homodyne receiver followed by an integrator with a period that relates to L in your expression - really not very different from a spectrum analyser (which looks at an intermediate frequency just above zero)

 

[dxy]

Linear algebra and Diffs Eq. is the key here!