Question about the general potential operator The Hamiltonian operator is given by $$\hat{H} = \frac{\hat{\textbf{p}}^2}{2m} + \hat{V}.$$ For this reason it's of interest to know how to find $\hat{V}$ in any representation. Is the potential operator---independent of a chosen basis---given by $$\hat{V} = V(\hat{\textbf{r}}, t)$$ where $V(\textbf{r}, t)$ is the potential energy at position $\textbf{r}$ and time $t$? I know this works when dealing with the position representation.
 A: The equation you write does not assume a basis. In other words, the equation:
$$\hat H=\frac{\hat p^2}{2m}+\hat V(\hat r,t)$$
can be expressed in any basis we choose.
For example, in 1D we can choose to work in the position basis:
$$\hat H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\hat V(x,t)$$
or we can work in the momentum basis:
$$\hat H=\frac{p^2}{2m}+\hat V\left(i\hbar\frac{\partial}{\partial p},t\right)$$
or whatever other basis you choose to work in.
A: Yes, usually the potential is written as a polynomial through a Taylor series which easily extends to the operator analog needed for quantum mechanics. 
A: In general, in the position representation:
$$
\hat p\psi(x)=-i\hbar \frac{d}{dx}\psi(x)\, ,\qquad 
\hat x\psi(x)=x\psi(x)\, ,
$$
and $V(\hat x)\psi(x)=V(x)\psi(x)$.  
Of course in the momentum representation
$$
\hat x\psi(p)=+i\hbar \frac{d}{dp}\psi(p)\, ,\qquad 
\hat p\psi(p)=p\psi(p)\, .
$$
Usually it's not terribly useful to use $V(\hat x)$ in the momentum representation because it leads to a function of a differential operator.  However, there are two obvious and more interesting cases:
The linear potential $V(x)=k x$ becomes a simple derivative
$V(x)\psi(p)\to ik\hbar \frac{d}{dp}\psi(p)$.  This kind of linear potential finds applications when metals are plunged into magnetic fields (see Wanner, M., R. E. Doezema, and U. Strom. "Far-infrared surface-Landau-level spectroscopy in Bi." Physical Review B 12.8 (1975): 2883), or because the potential between quarks contains a linear part (see Fig.10 from Drell, Sidney D. "The Richtmyer memorial lecture—when is a particle?." American Journal of Physics 46.6 (1978): 597-606.)
The linear potential is also useful in studying quantum effects in free-falling particles.
Of course in the case of a harmonic oscillator where $\hat H\sim \hat p^2/2m + k\hat x^2/2$, this becomes in the momentum representation basically the same as in the position representation up to some factor since $\hat p^2\psi(p)\to p^2\psi(p)$ but now $\hat x^2\psi(p)\sim -\frac{\hbar^2}{2m}\frac{d^2}{dp^2}\psi(p)$ 
A: The potential operator is $\hat V$, and it’s independent from the representation you choose (as any good representative of a vector space).
Nevertheless, pick up a basis $|\alpha\rangle$ of your Hilbert space: in this representation you can express the potential as
$$\hat V= \sum_{\alpha_1,\alpha_2} V_{\alpha_1\alpha_2}|\alpha_1\rangle\langle\alpha_2| $$
Where $\alpha_1,\alpha_2$ runs over all the vectors of the basis (possibly infinite, and if continuously infinite the sums may become integrals) and
$$V_{\alpha_1\alpha_2}=\langle\alpha_1|\hat V|\alpha_2\rangle$$
are the matrix elements of the potential in the basis you chose.
To reconnect this expression to what you already know, i.e. the expression in coordinate representation, we just have to do a little algebra, indeed if we apply the previous formula to the coordinate representation:
$$\hat V = \int d^3x_2\int d^3x_2 V_{\vec x_1\vec x_2} |\vec x_1\rangle\langle \vec x_2|$$
But as $\hat V$ is a function of the operator $\hat X$, it is diagonal in this basis so 
$$V_{\vec x_1\vec x_2}=\delta(\vec x_1-\vec x_2) V_{\vec x_1}$$ and the integral simplifies thanks to the Dirac delta:
$$\hat V = \int d^3x_1 V_{\vec x_1} |\vec x_1\rangle\langle \vec x_1|$$
Using the completeness of the coordinate basis, you can write:
$$V_{\alpha_1\alpha_2}=\langle\alpha_1|\hat V|\alpha_2\rangle= \langle\alpha_1|\int d^3x\hat V|\vec x\rangle\langle\vec x|\alpha_2\rangle=$$
$$=\int d^3x V_{\vec x} \langle\alpha_1|\vec x\rangle\langle\vec x|\alpha_2\rangle $$
So, knowing the matrix elements of an operator in a basis, and knowing the transition functions $\langle\vec x|\alpha_2\rangle$, you can express the potential in any other representation.
As others have already pointed out, for the momentum the transition functions are well known so this transition is almost harmless. 
