Let's assume we have a plane wave $\psi(x,t)= A_{0}e^{i(kx-wt)}$ in position space. To find the momentum representation of this wave we'd apply the Fourier transform. However, I don't see how this is mathematically allowed, since the function we want to transform must be Lebesgue-integrable for the Fourier integral to exist, right? As far as I can see this isn't the case here, i.e. $ \int^{+\infty}_{-\infty} \left|\psi(x,t)\right|dx = \left|A_{0}\right| \int^{+\infty}_{-\infty} dx = \infty$. Where am I going wrong?
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2 Answers
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You are not going wrong anywhere. The plane wave is not Lebesgue-integrable.
However, it is Fouriertransformable as a distribution.
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$\begingroup$ So if one takes a look at eq. (202) and (203) for the momentum and position representation, what is meant by $\phi(p,t)$ and $\psi(x,t)$ in the resp. integrands? This looks awfully lot like a Fourier transform of wave functions to me. Or are they perhaps eigenstates? In that case they ought to be square-integrable as inhabitants of the Hilbertspace and the issue would be solved. $\endgroup$– playdisCommented Dec 31, 2018 at 12:32
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$\begingroup$ @playdis. The author jumps freely between Fourier transforms of $L^2$ functions and Fourier transforms of distributions. This is common behavior among physicists who don't want to deep-dive into the theory. $\endgroup$– md2perpeCommented Dec 31, 2018 at 13:02
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$\begingroup$ To me it looks like one is simply expanding the wavefunctions on the eigenfunctions of the respective operator in eq. (202) and (203) to obtain a change of representation. I can't make out where the tempered distribution is since to my knowledge none of these functions are (necessarily) Schwartz functions. To recap my question: given a wavefunction in position space how do you represent it in momentum space if it's just a plane wave? I can't really remember coming across distributions in this context even though I understand why it would make sense if your function isn't in $L^{2}$. $\endgroup$– playdisCommented Dec 31, 2018 at 16:30
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$\begingroup$ It can in fact be seen as a change of basis. That's a common view among physicists, but perhaps not as common among mathematicians. $\endgroup$– md2perpeCommented Dec 31, 2018 at 18:09
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$\begingroup$ Eqs. (201) and (203) are two variants of a Fourier transform of the $L^2$ function $\psi$. Letting $\psi(x) = \delta(x-x_0)$ we instead have a Fourier transform of the tempered distribution $\delta(x-x_0).$ $\endgroup$– md2perpeCommented Dec 31, 2018 at 18:13
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I think you were being confused by learning math definitions from physics books.
For the Fourier transform on L^2, you might look in Lieb and Loss's book "Analysis". (There the Fourier transform is defined on L^2 to be the unique unitary operator which restricts to the operator given by the usual integral formula on L^1 intersect L^2, which is a dense subspace of L^2. Note that for an arbitrary L^2 function, the Fourier transform is NOT given by the usual integral formula because that integral may simply fail to exist.) More generally, the Fourier transform on tempered distributions is defined by integration by parts, see for example Hormander's "analysis of linear partial differential operators".
