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I found that some theories about quantum theory is similar to Fourier transform theory. For instance, it says "A finite-time light's frequency can't be a certain value", which is similar to "A finite signal has infinite frequency spectrum" in Fourier analysis theory. I think that a continuous frequency spectrum cannot be measured accurately, which is similar to Uncertainty principle by Hermann Weyl. How do you think about this?

DanielSank
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lai
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    Hi user1297181, and welcome to Physics Stack Exchange! Your question is essentially "How do you think about this?" and that suggests that you might be looking for a discussion, not an answer, which suggests that this may not be appropriate for this site in its current form (see the [FAQ#dontask]). If you can be more specific about what you're asking, it will probably be fine. – David Z Sep 06 '12 at 01:36
  • This is generally true--- Fourier analysis is a fundamental part of quantum mechanics and quantum field theory, but it is taken for granted, you are supposed to have internalized it. – Ron Maimon Sep 06 '12 at 01:44
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    Thanks @DavidZaslavsky for your good suggestion. I think I should have studied the FAQ first. But I still wonder the key to this question, not just meaning "How do you think this". – lai Sep 06 '12 at 02:07
  • Dear @RonMaimon, where can I get more information about it? – lai Sep 06 '12 at 02:24
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    Functions $exp(ikx)$ are eigenfunctions of momentum operator $-i\partial/\partial x$; that is the main (and perhaps only) link between QM and Fourier analysis. – user10001 Sep 07 '12 at 01:14

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Yes, there is a very strong interconnection.

A particle in q.m. doesn't have a defined position. Instead, there is a function describing the probability amplitude distribution for the position: the wavefunction $u(x)$. This is always told even in books for the general public. However, the momentum of the particle too isn't, in general, well defined: for it also we have a probability amplitude distribution, let's call it $w(p)$. It happens that $u$ and $w$, the space- and momentum-wavefunctions, are in some sense Fourier transforms of each other. The reason is the following. In Dirac's notation, $$u(x) = \langle x|\psi\rangle,\quad w(p)=\langle p|\psi\rangle $$ where $|\psi\rangle$ is the state of the particle, $|x\rangle,|p\rangle$ are respectively the eigenstates of the position and momentum operators.

Let's say we work in the $x$ basis. The $p$ operator is written $-i\hbar\partial/\partial x$. To find eigenstates of $p$, we can call $\langle x|p\rangle=f_p(x)$ and this has to obey the eigenvalue equation for $p$ in the $x$ basis representation: $$ -i\hbar\frac{\partial}{\partial x}f_p(x)=pf_p(x)$$ whose solution is the family of functions $f_p(x) = e^{ipx/\hbar}$.

Now, to change from a basis to the other we can write $$\langle p|\psi\rangle= \int \langle p|x\rangle\langle x|\psi\rangle dx$$ or $$ w(p) = \int e^{-ipx/\hbar}u(x)dx$$ which is a Fourier transform! The $\hbar$ factor is to give the correct dimensionality.

Nice, isn't it? As you pointed out, the fact that if $u$ is "spread" then $w$ is "peaked" and vice versa is typical of Fourier transformed functions. So Heisenberg's principle can be thought to come from here.

This holds for a lot of conjugated quantum variables.

Martino
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    So this is where $e^{ipx}$ comes from! Assuming of course convention of $\hbar = 1$. I have been wanting to know thus for ages but I thought too trivial to ask on SE. – Stan Shunpike Feb 15 '15 at 06:52
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    Why does $\langle p | x \rangle= e^{-ipx/\hbar}$? – Stan Shunpike Feb 15 '15 at 06:54
  • Never mind. Answer is here http://physics.stackexchange.com/q/86824/66165 – Stan Shunpike Feb 15 '15 at 07:00
  • It's in my answer too: it's the solution to the differential equation (the second equation I wrote) "which yields..." – Martino Feb 16 '15 at 09:22
  • this is brilliant. Why aren't we taught with this? Can this be equivalent to classical mechanics? – Ooker Sep 24 '17 at 18:42
  • @StanShunpike I have been confused for about 6 years too: Why does $<p|x>=\exp(-ipx/h)$? The link you posted is helpful, but it reduces that question back to: Why do we define $\hat{p} \sim \frac{\partial}{\partial x}$? I just found a nice answer (the accepted one), which starts from defining that the momentum is the (up to a scalar) quantity that's conserved under translational symmetry. I find it so satisfying. – Student Oct 07 '20 at 11:26
  • @Student you have derived that result from Noether's theorem, which is considered one of the most important and beautiful results in theoretical physics - for a reason! – Martino Oct 07 '20 at 11:33
  • @Bzazz Exactly. That's why I think it's better to start with the definition of position + Noether theorem, instead of the definitions of position+momentum. Though in the end two descriptions are equivalent, I think it's more elegant, as I view symmetry as more fundamental. And this is the answer I was searching for.. all physics friends start with the latter postulates :( – Student Oct 07 '20 at 11:35
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    @Bzazz I also commented there, and think it's related to our conversation, so here's a re-comment: ... This could provide a nicer explanation of Fourier transform: (roughly,) whenever I have an operator O and a system S symmetric under O, would there always be a conserved quantity (corresponding to operator O^)? Will O^ always be the Fourier transform of O? – Student Oct 07 '20 at 11:36