So, why can't the uncertainty relations be violated in such a case, if I could, say, measure the position of the object with this wave function

That's the catch. You can't.

Or rather, you *can* measure the position, but the result you get will vary from one measurement to the next, because the wavefunction $\exp(x^2/2i - cx)$ is not an eigenstate of position. It's not an eigenstate of momentum, either, so if you tried to measure the momentum, you'd get similar variation between measurements. You could compute the standard deviations of the position and momentum measurements, $\sigma_x$ and $\sigma_p$ respectively, and you would find that they satisfy the uncertainty principle.

It's only if you measure the specific combination of position plus momentum that you would get the same result every time.

I) Well, the Legendre transformation can be e.g. seen as the leading classical tree-level formula of a formal semiclassical Fourier transformation.

This fact is e.g. used in QFT when relating the quantum action $S[\varphi]$, the partition function $Z[J]$, generating functional $W_c[J]$ for connected diagrams, and the effective action $\Gamma[\Phi]$.

II) To see the correspondence in detail, let $x$ and $p$ be the two dual/conjugate variables (in both senses!). Let

$$\tag{1} f(x;\hbar)~\equiv~\sum_{n=0}^{\infty}\hbar^n f_n(x)\quad\text{and}\quad g(p;\hbar)~\equiv~\sum_{n=0}^{\infty}\hbar^n g_n(p)$$

be two formal power series in $\hbar$ with function coefficients.
Consider their semiclassical exponentials

$$\tag{2} F(x;\hbar)~:=~e^{if(x;\hbar)/\hbar}\quad\text{and}\quad G(p;\hbar)~:=~e^{ig(p;\hbar)/\hbar}.$$

Now assume that

$$\tag{3} G(p;\hbar)~=~\int \! dx~ e^{-ipx/\hbar} F(x;\hbar)$$

is the Fourier transform of $F(x;\hbar)$. We can use the WKB/stationary phase approximation to deduce that the classical parts $f_0(x)$ and $-g_0(p)$ are then Legendre duals of each other, i.e.

$$\tag{4} g_0(p) ~=~ -px+f_0(x)\quad\text{where}\quad p~=~f_0^{\prime}(x),$$

for a sufficiently nice function $f_0(x)$.

## Best Answer

Practically speaking, the full machinery of Pontryagin duality is way more advanced than physicists need to understand the uncertainty principle. There are several ways to "derive" that the momentum-space wavefunction is the Fourier transform of the position-space wavefunction, which depend somewhat on your choice of starting postulates. Here's one common path:

One common starting fundamental postulate is the commutation relation $[\hat{x}, \hat{p}] = i \hbar.$ The most common position-space representation of this commutation relation is $\hat{x} \to x,\ \hat{p} \to -i \hbar \frac{\partial}{\partial x}$. In this representation, taking the inner product of $\langle x |$ and the eigenvalue equation $\hat{p} |p\rangle = p | p \rangle$ gives the differential equation $$-i \hbar \frac{d\, \psi_p(x)}{dx} = p\, \psi_p(x),$$ which has solution $\psi_p(x) = \langle x | p \rangle \propto e^{(i p x)/\hbar}$. Then to express an arbitrary state $| \psi \rangle$ in the momentum basis, we can use the resolution of the identity $$ \psi(p) = \langle p | \psi \rangle = \int dx\ \langle p | x \rangle \langle x | \psi \rangle \propto \int dx\ e^{-ipx/\hbar} \psi(x),$$ which is just the Fourier transform. This generalizes straightforwardly into higher dimensions.

BTW, the fact that position-space and momentum-space wavefunctions are Fourier transforms of each other (or more precisely, can be chosen to be Fourier transforms of each other) gives some nice intuition for the uncertainty relation but isn't actually necessary to derive it. All you need is the commutation relation, as I explain here.