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.

Yes, there is a very strong interconnection.

A particle in q.m. doesn't have 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, the momentum of the particle too 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 $w$, the space- and momentum-wavefunctions, are in some sense Fourier transforms of each 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.

Let's say we work in the $x$ basis. The $p$ operator is written $-i\hbar\partial/\partial x$. To find eigenstates of $p$, we can call $\langle x|p\rangle=f_p(x)$ and this has to obey the eigenvalue equation for $p$ in the $x$ basis representation: $$ -i\hbar\frac{\partial}{\partial x}f_p(x)=pf_p(x)$$ whose solution is the family of functions $f_p(x) = e^{ipx/\hbar}$.

Now, to change 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.