I am trying to get understand how different representations of a quantum state are equivalent: For example if we have our quantum state $$| \psi \rangle = \sqrt{\frac{1}{3}}|R_{21}Y_{1}^{0} \rangle \otimes |\chi_{+} \rangle + \sqrt{\frac{2}{3}}|R_{21}Y_{1}^{1} \rangle \otimes |\chi_{-} \rangle$$ then this is a representation of the quantum state in terms of a linear combination of energy eigenfunctions and spin eigenfunctions. But we could also consider showing the quantum state in terms of momentum eigenfunctions, say $$| \psi \rangle = \int dp \phi(p) | p \rangle.$$ Then is the idea that $$\sqrt{\frac{1}{2}}|R_{21}Y_{1}^{0} \rangle \otimes |\chi_{+} \rangle + \sqrt{\frac{1}{2}}|R_{21}Y_{1}^{1} \rangle \otimes |\chi_{-} \rangle = \int dp \phi(p) | p \rangle?$$ Even though the one in a tensor product and the other is not, in some general Hilbert state they are equivalent vectors and these are only different representations of them? Is this the idea?
Lastly in QM, both $\hat{X}$ and $\hat{P}$ are operators on the same space ($L^2(\mathbb{R})$), so their eigenbases span the same space and hence we have a transformation function $\langle x | p \rangle$. What about the Hamiltonian operator $\hat{H}$, is this also an operator on the space $L^2(\mathbb{R})$ and then do the transformation functions $\langle p | \psi \rangle$ make sense? Where $| \psi \rangle$ are the eigenfunction kets of the Hamiltonian.
