Consider the vector space L2x of square integrable functions.  We have earlier identified two orthonormal, continuously labeled bases for L2x,
{vp(x) = (2πħ)-1/2 exp(ipx/ħ)},
with Ψ(x) = ∫ Ψ(p) vp(x) dp  =  (2πħ)-1/2-∞ Ψ(p) exp(ipx/ħ) dp,
and
xo = δ(x - x0)} with Ψ(x) = ∫-∞ Ψ(x0) δ(x - x0) dx0.
We can generalize to three dimensions.  Then the bases are
{vp(r) = (2πħ)-3/2 exp(ipr/ħ)}
with Ψ(r) = ∫ Ψ(p) vp(r) d3p  =  (2πħ)-3/2-∞ Ψ(p) exp(ipr/ħ) d3p,
and
ro = δ(r - r0)} with Ψ(r) = ∫-∞ Ψ(r0) δ(r - r0) dx0.

Let us associate the kets |p0> with vp0(r) and |r0> with δro, and the ket |Ψ> with the wave function Ψ(r).
Then Ψ(p) = <p0|Ψ> denotes the components of |Ψ> in the {|p0>} basis and Ψ(r) = <r0|Ψ> denotes the components of |Ψ> in the {|r0>} basis.
Recall:
For the {|p>} basis we have  <p|p'> = δ(p - p'),  ∫ d3p |p><p| = I,
and for the {|r>} basis we have  <r|r'> = δ(r - r'),  ∫ d3r |r><r| = I.

The scalar product of two vectors |Ψ> and |Φ> can be calculated in the {|r>} basis and the {|p>} basis.
<Φ|Ψ> = ∫ d3r <Φ|r><r|Ψ> = ∫ d3r Φ*(r)Ψ(r)
and
<Φ|Ψ> = ∫ d3p <Φ|p><p|Ψ> = ∫ d3p Φ*(p)Ψ(p)
We use the transformation matrix U to change from the {|r>} basis to the {|p>} basis.
Ur,p =  <r|p> = vp(r) = (2πħ)-3/2 exp(ipr/ħ).
For any change of basis with U we have shown di = ∑jUijaj,
where di denotes the coordinates in the new basis and aj denotes the coordinates in the old basis.
Therefore
Ψ(p) = ∫d3r (2πħ)-3/2 exp(-ipr/ħ) Ψ(r) = ∫ d3r Ur,p* Ψ(r).
Ψ(p) denotes the coordinates of |Ψ> in the {|p>} basis and Ψ(r) denotes the coordinates of |Ψ> in the {|r>} basis.  If we denote the matrix elements of the operator A, <r|A|r’>, by A(r,r’) and the matrix elements <p|A|p’> by A(p,p’) then
A(p,p’) = (2πħ)-3∫d3r∫d3r' exp(-i(p - p')∙r/ħ)A(r,r’),
which corresponds to Ωi'j' (new basis) = (UΩU)ij (old basis).

The R and P operators
Let X be the operator whose eigenvectors are {|r>}.
Let X|r> = x|r>, with x a real number.
Let Px be the operator whose eigenvectors are {|p>}.
Let Px|p> = px|p>, with px a real number.
Make similar definitions for Y an Z and for Py and Pz.  Let |Ψ> and |Φ> be arbitrary kets.
Calculate
<Φ|X|Ψ> = ∫ d3r x<Φ|r><r|Ψ> = ∫ d3r Φ*(r) x Ψ(r) = <Ψ|X|Φ>*.
X is Hermitian.  In the {|r>} representation operating with the operator X just amounts to multiplying by the number x.
Similarly:
<Φ|Px|Ψ> = ∫ d3p Φ*(p) px Ψ(p) = <Ψ|Px|Φ>*.
Px is Hermitian.  In the {|p>} representation operating with the operator Px just amounts to multiplying by the number px.
But we may also write
<Φ|Px|Ψ> = ∫ d3r <Φ|r><r|Px|Ψ> = ∫ d3r Φ*(r) ∫ d3p <r|p><p|Px|Ψ>.
But ∫ d3p <r|p><p|Px|Ψ> = (2πħ)-3/2∫px  Ψ(p) exp(ipr/ħ) d3p,
since Px is Hermitian.
Ψ(p) is the Fourier transform of Ψ(r).
Ψ(p) = (2πħ)-3/2∫Ψ(r) exp(-ipr/ħ) d3r,

But what is pxΨ(p)?
The Fourier transform of ∂Ψ(r)/∂x is
(2πħ)-3/2-∞+∞∂Ψ(r)/∂x exp(-ipr/ħ) d3r
= (2πħ)-3/2∫∫dy dz exp(-i(pyy + pzz)/ħ )∫dx exp(-ipxx/ħ)∂Ψ(x,y,z)/∂x.

-∞+∞dx exp(-ipxx/ħ)∂Ψ(x,y,z)/∂x
= exp(-ipxx/ħ) Ψ(x,y,z)|-∞+∞ - ∫dx (-ipx/ħ) exp(-ipxx/ħ) Ψ(x,y,z),
using ∫dx u (dv/dx) = uv - ∫dx (du/dx)v.
exp(-ipxx/ħ) Ψ(x,y,z)|-∞+∞ = 0.
We therefore have
(2πħ)-3/2∫(ħ/i)∂Ψ(r)/∂x exp(-ipr/ħ) d3r
= px(2πħ)-3/2∫Ψ(r) exp(-ipr/ħ) d3r = pxΨ(p).
Therefore <r|Px|Ψ> = ∫ d3p<r|p><p|Px|Ψ>
= (2πħ)-3/2∫px Ψ(p) exp(ipr/ħ) d3p
= (2πħ)-3∫d3r'∫d3p exp(ip∙(r - r')/ħ) (ħ/i)∂Ψ(r')/∂x'
= (ħ/i)∂Ψ(r)/∂x,
since (2πħ)-3∫d3p exp(ip∙(r - r')/ħ) = δ(r - r').
We have <r|Px|Ψ> = (ħ/i)∂Ψ(r)/∂x.

In the {|r>} representation the operator Px coincides with the differential operator (ħ/i)∂/∂x.
Similarly, Py --> (ħ/i)∂/∂y,  Pz --> (ħ/i)∂/∂z,  P --> (ħ/i).
In the {|p>} representation the operator X coincides with the differential operator (-ħ/i)∂/∂px, etc.

The commutator [X,Px]
<r|X|ζ> = x<r|ζ> for any |ζ>,  <r|Px|ζ> = ((ħ/i)∂/∂x)<r|ζ> for any |ζ>.
<r|[X,Px]|ζ> = <r|XPx|ζ> - <r|PxX|ζ> = x <r|Px|ζ> - ((ħ/i)∂/∂x)<r|X|ζ>
= x ((ħ/i)∂/∂x)<r|ζ> - ((ħ/i)∂/∂x)(x<r|ζ>) = -(ħ/i)<r|ζ>,
for any |ζ> and any |r>.  Therefore [X,Px] = iħ I.  We proceed similarly for the y and z components.  We write
[Ri,Pj] = iħδij,    [Ri,Rj] = [Pi,Pj] = 0.  (i, j = 1, 2, 3)
The operators Ri and Pi do not commute.  A common eigenbasis of Ri and Pi does not exist.