In Fourier theory, the pair composed of a variable and its Fourier transform is called conjugate variables, and one crucial property between the two is the uncertainty relation. This relation tells us that e.g. if one variable/function has a bounded (or compact?) support, its Fourier transform cannot have a bounded support as well. Now in the context of Quantum Mechanics, as an example, the physical variables are represented by linear operators, and there we have the famed Heisenberg relations for special pairs of quantities, such as position and momentum, or two different spin components. The question is, do we have the same bounded and unboundedness consequence of Fourier transforms when we talk about linear operators such as the ones in QM? (so Hermitian ones).

$\begingroup$ It's a little hard to figure out what you mean by "physical variable". Do you mean an observable? In the Schrodinger picture, observables are indeed represented by Hermitian operators  but what would it mean to take the Fourier transform of an operator? The application to quantum mechanics comes when you apply the uncertainty principle from Fourier analysis to functions representing quantum states. $\endgroup$ – Steven Landsburg Jan 11 '18 at 16:16
Yes, the Heisenberg uncertainty principle literally is the Fourier uncertainty relation. For simplicity consider the onedimensional case.
The Hilbert space of a spinless onedimensional particle is $L^2(\mathbb{R})$ and position is represented by $Q = M_x$, the operator of multiplication by $x$. Whereas momentum is represented by $P = i\hbar \frac{d}{dx}$, which is just to say that $P = \hbar F^{1}M_xF$ where $F: L^2(\mathbb{R}) \to L^2(\mathbb{R})$ is the $L^2$normalized Fourier transform. Heisenberg's uncertainty principle is then a fact about the tradeoff between the variance of a wave function $\psi \in L^2(\mathbb{R})$ and the variance of its Fourier transform, because the position variance of $\psi$ is $$\M_x\psi\_2^2  \langle M_x\psi,\psi\rangle^2 = \int x^2\psi(x)^2  \left(\int x\psi(x)^2\right)^2$$ and its momentum variance is $$\left\i\hbar\frac{d}{dx} \psi\right\_2^2  \left\langle i\hbar\frac{d}{dx}\psi,\psi\right\rangle = \hbar^2\int x^2\hat{\psi}(x)^2  \hbar^2\left(\int x\hat{\psi}(x)^2\right)^2.$$
You can also see from this that, yes, a particle cannot simultaneously have a bounded range of possible position values and a bounded range of possible momentum values  that is just saying that $\psi$ and $\hat{\psi}$ cannot both be supported on bounded intervals.

$\begingroup$ (But probably this is not a research level question.) $\endgroup$ – Nik Weaver Jan 11 '18 at 17:42

$\begingroup$ Yes, I think it's fine to talk about the Fourier transform of an operator as its conjugation by $F$. But the uncertainty principle is a fact about states the system can be in, so I think there you have to work with Fourier transforms of vectors. $\endgroup$ – Nik Weaver Jan 11 '18 at 18:52

$\begingroup$ Makes sense thanks a lot. Explicitly, how do we define $F$ operator in the equation for $P$? $\endgroup$ – user929304 Jan 11 '18 at 20:02

$\begingroup$ $F: \psi \mapsto \hat{\psi}$ where $\hat{\psi}(t) = \frac{1}{\sqrt{2\pi}}\int\psi(x) e^{itx}\, dx$. Chapter 4 of my book Mathematical Quantization might (or might not) be helpful for what you need. $\endgroup$ – Nik Weaver Jan 11 '18 at 20:18

$\begingroup$ Sure, $F$ is a unitary operator and if you conjugate any bounded operator by a unitary the result will be bounded. I think what you are looking for here is the canonical commutation relation, $QP  PQ = i\hbar$. $\endgroup$ – Nik Weaver Jan 11 '18 at 21:05