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The Fourier transform $$F: L^2\rightarrow L^2 \qquad \hat{f}(\omega) \equiv (Ff)(\omega) \equiv \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$$ can be viewed as a unitary operator with inverse (and hence, adjoint) $$(F^{-1}Ff)(t) = (F^\dagger F f)(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\hat{f}(\omega)e^{i\omega t}d\omega \, .$$ All unitary operators $U$ can be written as the exponential of a Hermitian operator $H$, i.e. $U = \exp(iH)$. What is the Hermitian operator $H$ that satisfies $F = \exp(iH)$, and does it have any physical significance or uses?

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Aug 17 '16 at 20:29
  • $\begingroup$ Not all unitary operators are a single exponential. In some cases you might need to take a product of many exponentials, hence many self-adjoint operators. $\endgroup$ – Phoenix87 Aug 17 '16 at 21:00
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    $\begingroup$ You are asking about the logarithm of the Fourier transform operator. You must first generalize to continuous rotations in phase-space, thus Fractional Fourier transforms first explored by Condon in the 1930s in the context of canonical transformations of quantum mechanics. $\endgroup$ – Cosmas Zachos Aug 17 '16 at 21:59
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    $\begingroup$ This hermitian generator is simply the Hamiltonian of harmonic oscillator $H=\frac{1}{2}(x^2+p^2)$. $\endgroup$ – Blazej Aug 18 '16 at 3:38
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    $\begingroup$ Dang. This is one of the neatest questions I've ever seen on Phys.SE! I sort of wish it stayed over there. $\endgroup$ – knzhou Aug 18 '16 at 6:04
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I found that question quite interesting and found a very nice paper on the arXiv: Classical Structures in Quantum Mechanics and Applications. It deals with the Fourier transform in the framework of quantum mechanics (QM). In QM the Fourier transform is used frequently to transform from position to momentum space: $$\hat F|q(x)\rangle=|p(x)\rangle,$$ where $|q(x)\rangle$ and $|p(x)\rangle$ are quantum eigenkets of the position and momentum observables and $\hat F$ is the operator of the Fourier transform. $\hat F$ can be defined elegantly as $$\hat F=\int^\infty_{-\infty}dx|p(x)\rangle\langle q(x)|.$$

$\hat F$ is unitary and therefore can be expressed via the exponential of a hermitan operator. For the Fourier transform in this setting this hermitian generator is the number operator defined with the ladder operators: $ \hat N=\hat a^\dagger \hat a$:$$\hat F=\exp(i(\pi/2)\hat N).$$ This implies a nice bridge between second quantisation: ladder operators, number operator,.. and the canonical represantaion in position or momentum states.

For a more detailed explanation of this, the consequences and possible generalisations I highly recommend reading that paper. Some points coming up in that paper related are "Weyl-Wigner Formalism" and "Quantum Symplectic Group".

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    $\begingroup$ Could you show why the number operator is the right thing to exponentiate? It's not even clear what $a$ and $a^\dagger$ mean in this context. I guess they take each energy eigentstate to the next one? $\endgroup$ – DanielSank Aug 17 '16 at 21:58
  • $\begingroup$ @DanielSank Yeah, it's not clear from the context, but that's the number operator from the harmonic oscillator, $N=a^\dagger a$ for $a=x+ip$ (essentially). The connection to the harmonic oscillator is not obvious but it becomes apparent once you recall that the HO time evolution is simply a rigid rotation of phase space - hence running it for a quarter period will interchange the $x$ and $p$ axes. It would be good if M.J. can include this information in his/her answer, though - it's not at all obvious from the get-go. $\endgroup$ – E.P. Aug 17 '16 at 22:17
  • $\begingroup$ This also shows that the inverse is completely nonunique; any $(2k+\frac12)\pi\hat N$ for integer $k$ will also do essentially the same. $\endgroup$ – E.P. Aug 17 '16 at 22:20
  • $\begingroup$ I gave a brief summary of the expressions given in the paper: Yes of course it is not clear if you do not put in some time to read it and maybe even look at their references if some points remain unclear. Maybe the answer is lacking a bit content but once you start fleshing those expressions out: where do I stop? If you look in the paper they should become clear or at least clear enough to search for specific unclear points. I am no expert on the subject and just found the paper and read some passages briefly. $\endgroup$ – M. J. Steil Aug 17 '16 at 22:30
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This hermitian generator is simply the Hamiltonian of harmonic oscillator $H=\frac{1}{2}(x^2+p^2)$. This is simple to understand. In Heisenberg picture evolution with this Hamiltonian is just rotations in $x,p$ plane. Fourier transform is transformation which maps $x \to p$ and $p \to -x$ so it is a particular rotation with angle $\frac{\pi}{2}$. Hence $\mathcal F = \exp (-i \frac{\pi H}{2})$.

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  • $\begingroup$ Shouldn't you have an $i$ in the exponential (as the Hamiltonian is self-adjoint)? $\endgroup$ – gented Aug 18 '16 at 9:10
  • $\begingroup$ Thank you, of course you are right. $\endgroup$ – Blazej Aug 18 '16 at 9:12

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