Is a function always defined with respect to a particular coordinate system I've come across a small misconception while trying to calculate the expectation values of some operators in quantum mechanics.
Suppose we have some function that we have defined as $f(x)$. Let us say, there is some operator $ \hat{A}$, and in the Schroedinger formalism, this is represented as the function $f(x)$.
We know, $$\langle \hat{A}\rangle = \int \psi^*(x,t)f(x)\psi(x,t)dx$$
Here, the wave function is assumed to be normalized.
However, let us now take a different wavefunction $\psi(u,t)$, where $u=g(x)$. We have to take the expectation value of $\hat{A}$ with respect to this new wave function.
My question is, is $\hat{A}$ still $f(x) ?$ Or does it trivially become $f(u) ?$ I'm inclined to believe that it is still $f(x)$, and  the integral would now become the following :
$$\langle \hat{A}\rangle = \frac{\int \psi^*(u,t)f(x)\psi(u,t)dx}{\int\psi^*(u,t)\psi(u,t)dx} = \frac{\int \psi^*(u,t)[f\space o\space g^{-1}(u)]\psi(u,t)\frac{du}{g'(x)}}{\int\psi^*(u,t)\psi(u,t)\frac{du}{g'(x)}}$$
Here I have transformed from $x$ to $u$ and made $u$ the dependent variable in the integral.
My question is why do we do this exactly? Why don't we directly swap $x$ and $u$ ? Is it because, the operator is defined to be $f(x)$, and so, in any coordinate system, it would still be $f(x)$ and not $f(u)$ directly. In order to make $u$ the dependent variable, we would have to use $f(x)=f\space o\space g^{-1} (u)$. We can't trivially exchange $u$ and $x$.
So, when we are defining a function, is it always done with respect to a coordinate system $x$, in this example. If we move to a new coordinate system, the function is still $f(x)$ and we need to figure out some transformation to represent it in this new system. Is this correct, and is there a more intuitive explanation for this?
In classical mechanics, in the case of any coordinate transform, the same thing happens. In some coordinate systems, say $F=F(r)$, where $r$ is the distance from the origin. In some other systems, the distance from the origin is $r'$. The force in this new system is not $F(r')$. Instead it is still $F(r) = F(r'+d)$. Is this happening because we are defining the force with respect to a particular coordinate system? Is the same thing happening in the case of the quantum operator above ?
We define the position operator in the Schrodinger picture as $\hat{x}=x$. In some other coordinate system where $u=f(x)$, is the position operator still $\hat{x}=x=f^{-1}(u)$, or does it trivially become $\hat{u}=u$ ?
Any help would be highly appreciated.
 A: Your expression is close, properly reinterpreted (New wavefunction, invertible g(x), etc, all symbols meaningful when encountered; you are not fussing about coordinate singularities, one assumes).
You appear confused about the meaning of the coordinate representation.
Above, one strictly means
$$
\hat x |x\rangle = x |x\rangle , ~~~  \hat A |x\rangle = f(x) |x\rangle ,
~~~ \psi(x)= \langle x| \psi\rangle\\
\langle \hat{A}\rangle = \langle \psi|\hat{A}|\psi\rangle/  \langle \psi| \psi\rangle. 
$$
Normally one takes $ \langle \psi| \psi\rangle=1$, as you do.
Now, your new, different, wavefunction is
$$
\psi(u)= \psi(g(x)) \equiv \tilde \psi(x),
$$
but $\langle \tilde \psi|\tilde \psi \rangle \neq 1$ anymore, so one has to normalize expectations by its norm.
Work in the x basis system, so
$$\langle \hat{A}\rangle ={\langle \tilde \psi |\hat{A}|\tilde \psi \rangle \over \langle \tilde \psi |\tilde \psi \rangle}= \frac{\int \tilde \psi^*(x,t)f(x)\tilde \psi(x,t)dx}{\int\tilde \psi^*(x,t)\tilde \psi(x,t)dx}\\
 = \frac{\int \psi^*(u,t) ~f(g^{-1}(u))  ~ \psi(u,t)\frac{du}{g'(g^{-1}(u)}}{\int\psi^*(u,t)\psi(u,t)\frac{du}{g'(g^{-1}(u))}},$$
so close to your expression, suitably corrected.
This answers your question, picking your first alternative.

My question is, is ̂ still ()? Or does it trivially become ()?

Unlike your new wavefunction, it stays the same operator represented in the new coordinates/basis.
To keep track, you might consider a trivial linear scaling transformation, g(x)=a x.
