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?
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1$\begingroup$ 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). If you can be more specific about what you're asking, it will probably be fine. $\endgroup$– David ZCommented Sep 6, 2012 at 1:36
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$\begingroup$ 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. $\endgroup$– Ron MaimonCommented Sep 6, 2012 at 1:44
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1$\begingroup$ 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". $\endgroup$– laiCommented Sep 6, 2012 at 2:07
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$\begingroup$ Dear @RonMaimon, where can I get more information about it? $\endgroup$– laiCommented Sep 6, 2012 at 2:24
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2$\begingroup$ 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. $\endgroup$– user10001Commented Sep 7, 2012 at 1:14
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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.
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1$\begingroup$ 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. $\endgroup$ Commented Feb 15, 2015 at 6:52
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1$\begingroup$ Why does $\langle p | x \rangle= e^{-ipx/\hbar}$? $\endgroup$ Commented Feb 15, 2015 at 6:54
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$\begingroup$ Never mind. Answer is here physics.stackexchange.com/q/86824/66165 $\endgroup$ Commented Feb 15, 2015 at 7:00
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$\begingroup$ It's in my answer too: it's the solution to the differential equation (the second equation I wrote) "which yields..." $\endgroup$– MartinoCommented Feb 16, 2015 at 9:22
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1$\begingroup$ @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? $\endgroup$– StudentCommented Oct 7, 2020 at 11:36