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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    $\begingroup$ 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 Z Sep 6 '12 at 1:36
  • $\begingroup$ 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 Maimon Sep 6 '12 at 1:44
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    $\begingroup$ 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". $\endgroup$ – lai Sep 6 '12 at 2:07
  • $\begingroup$ Dear @RonMaimon, where can I get more information about it? $\endgroup$ – lai Sep 6 '12 at 2:24
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    $\begingroup$ 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. $\endgroup$ – user10001 Sep 7 '12 at 1:14

Yes, there is a very strong interconnection.

A particle in q.m. hasn't got 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, also the momentum of the particle 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 $w$ are in some sense the fourier transforms one of the 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.

Suppose to work in the $x$ basis. The $p$ operator is written $-i\hbar\partial/\partial x$. To find eigenstates of $p$, we can call $\langle x|p\rangle=f_p(x)$ $$ -i\hbar\frac{\partial}{\partial x}f_p(x)=pf_p(x)$$ which yields to $f_p(x) = e^{ipx/\hbar}$.

Now, to pass 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.

  • $\begingroup$ 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. $\endgroup$ – Stan Shunpike Feb 15 '15 at 6:52
  • $\begingroup$ Why does $\langle p | x \rangle= e^{-ipx/\hbar}$? $\endgroup$ – Stan Shunpike Feb 15 '15 at 6:54
  • $\begingroup$ Never mind. Answer is here physics.stackexchange.com/q/86824/66165 $\endgroup$ – Stan Shunpike Feb 15 '15 at 7:00
  • $\begingroup$ It's in my answer too: it's the solution to the differential equation (the second equation I wrote) "which yields..." $\endgroup$ – Bzazz Feb 16 '15 at 9:22
  • $\begingroup$ this is brilliant. Why aren't we taught with this? Can this be equivalent to classical mechanics? $\endgroup$ – Ooker Sep 24 '17 at 18:42

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