Why a minus sign in the position operator? I start by showing how I tried to obtain the position operator analogously to how
the momentum operator is obtained:
If we differentiate the wave function in one dimension $\Psi(x,t) = e^{i(px-Et)/\hbar}$, with respect to x:
$$
\frac{\partial}{\partial x}\Psi(x,t) = \frac{ip}{\hbar}~\Psi(x,t)
\tag1
$$
from which we get the momentum operator: $\hat{p}=-i\hbar\frac{\partial}{\partial x}$
But suppose I differentiate $\Psi$ with respect to momentum:
$$
\frac{\partial}{\partial p}\Psi(x,t) = \frac{i}{\hbar}(x-\frac{p}{m}t)~\Psi(x,t)
\tag2
$$
which gives $x-\frac{p}{m}t = -i\hbar\frac{\partial}{\partial p}$
Now, setting $t=0$, I suppose we would get the operator $$\hat x_0 = -i\hbar\frac{\partial}{\partial p}$$ Which would be something like an operator for initial position. But it doesn't look right, because I know that the position operator does not have a minus sign. And is there even such a thing as "initial position operator"?
So, what is wrong with this? The reason I'm asking is because I want to show that the expectation value for position satisfies the following relation (in accordance with the correspondance principle):
$$\left<x\right> = \left<x\right>_{t=t_0}+\frac{\left<p\right>}{m}(t-t_0)$$
and it was given as a hint to start as in $(2)$ and take it from there. I sort of know what to do, but the minus sign in the position operator confuses me.
The hint also suggests $(2)$ should lead to $x = i\hbar\frac{\partial}{\partial p} + \frac{p}{m}t$, but somehow the minus sign isn't there.
 A: It is a bit more subtle, and this subtlety is important here.
The definition of an operator is that by acting on a wave function the operator determines the expectation value of its corresponding physical quantity:
$$\langle \hat{O}\rangle =\int dx \Psi(x)^*\hat{O}\Psi(x).$$
Therefore, just like the wave function, the operator has different forms in different representations.
Let us start with the position expectation in the position representation:
$$\langle x\rangle = \int dx x|\Psi(x)|^2 = \int dx \Psi(x)^*x\Psi(x).$$ We easily read the position operator from this expression as
$$\hat{x}=x.$$ The momentum operator in this representation is given by $\hat{p}=-i\hbar\partial_x$, as can be verified by considering its action on a state with a definite momentum:
$$\psi_p(x)=\frac{1}{\sqrt{2\pi}}e^{i\frac{px}{\hbar}}.$$
Note that the minus sign in this operator is a matter of convention: if we defined plane waves as $\psi_p(x)=\frac{1}{\sqrt{2\pi}}e^{-i\frac{px}{\hbar}}$, we would have to choose $\hat{p}=i\hbar\partial_x$.
Let us now look at the momentum representation. The wave function in the momentum representation us given by
$$\Phi(p)= \int dx \psi_p(x)^*\Psi(x).$$ Now $|\Phi(p)|^2$ is the probability density for momentum states and the momentum operator is simply $\hat{p}=p$, as follows from
$$\langle p\rangle = \int dp p|\Phi(p)|^2 = \int dp \Phi(p)^*p\Phi(p).$$
For the position operator we have
$$\langle x\rangle = \int dp \Phi(p)^*\hat{x}\Phi(p)=
\int dp \int dx \Psi(x)^*\psi_p(x)\hat{x}\int dx'\psi_p(x')^*\Psi(x) = 
\int dx \Psi(x)^*x\Psi(x),$$
where the position operator has to be defined in such a way that
$$\int dp  \psi_p(x)\hat{x}\psi_p(x')^* = 
\frac{1}{2\pi}\int dp  e^{i\frac{px}{\hbar}}\hat{x}e^{-i\frac{px}{\hbar}}
x\delta(x-x').$$ Choosing $\hat{x} = i\hbar\partial_p$ we satisfy this condition. The sign of the position operator is different than the sign of the momentum operator. If, as I mentioned in the beginning, we defined the plane wave with momentum $p$ as $e^{-i\frac{px}{\hbar}}$, the signs of both operators would be different.
To summarize:

*

*In the position representation: $\hat{x}=x, \hat{p}=-i\hbar\partial_x$.

*In the momentum representation: $\hat{x}=i\hbar\partial_p, \hat{p}=p$

*The signs before the derivatives in the above expressions are always opposite, but depend on how we define the plane wave with definite momentum.

A: The position and momentum distributions are connected by the Fourier transform: $\frac{1}{(2\pi)^{1/2}}\exp(ixp /\hbar)$ is the base vector in the position space and $\frac{1}{(2\pi)^{1/2}}\exp(-ixp /\hbar)$ is the base vector in the momentum space.
Note the following relations from Fourier analysis and quantum mechanics:
$$-i\frac{d}{d x}f(x)=\frac{1}{(2\pi)^{1/2}}\int_{\mathbb{R^3}}kg(k)\exp(ixk)dk$$
and
$$p=\hbar k.$$
Now you can do the usual expectation value integral for momentum in momentum space and translate it into the position space.
A: Probably the clearest way to check the result is to write the operator explicitly in ket notation in terms of the momentum basis (with $\hbar=1)$
$$ X = \int d^3p |p\rangle i\frac \partial{\partial p} \langle p|$$
and apply this to a position state
$$ \begin{align}
X|x\rangle &= \int d^3p |p\rangle i\frac \partial{\partial p} \langle p|x\rangle \\
& = \frac 1{(2\pi)^{3/2}}\int d^3p |p\rangle i\frac \partial{\partial p} e^{-ip\cdot x}\\
& = \frac 1{(2\pi)^{3/2}}\int d^3p |p\rangle x e^{-ip\cdot x}\\
& = x \int d^3p |p\rangle \langle p|x\rangle\\
& = x |x\rangle\\
\end{align}$$
where the resolution of unity
$$ 1 = \int d^3 p|p\rangle \langle p| $$
has been used. It is seen that the minus sign comes from the conjugate, $\langle p|x\rangle$ rather than $\langle x|p\rangle$, as we would have for the momentum operator. More generally, the position operator can be written 
$$  X = \int d^3x |x\rangle x \langle x|$$
Then
$$ \begin{align}
X &= \int d^3 p \int d^3 q \int d^3x |p\rangle \langle p|x\rangle x \langle x |q\rangle \langle q|\\
&= \frac 1{(2\pi)^{3}}\int d^3 p \int d^3 q\int d^3x |p\rangle e^{-ip\cdot x}x e^{iq\cdot x} \langle q|\\
&= \frac 1{(2\pi)^{3}}\int d^3 p \int d^3 q\int d^3x |p\rangle i \frac\partial{\partial p} e^{i(q-p)\cdot x} \langle q|\\
&= \int d^3 p \int d^3 q |p\rangle i \frac\partial{\partial p} \delta (q-p) \langle q|\\
&= \int d^3 p |p\rangle i \frac\partial{\partial p} \langle p|\\
\end{align}$$
