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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.

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you'll still need the spin state, only x-representation is equivalent to p-representation the $\chi $ is the spin state. so you'll need to write $\langle p | \otimes \langle \chi | $

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  • $\begingroup$ Okay thanks for your response. See my additional question at the end, maybe you can assist with this. $\endgroup$ – Alex Jan 13 '17 at 11:57
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    $\begingroup$ The Hamiltonian can again depend on the spin and then your transformation won't be valid, but if it doesn't technically you can do the transformation, do you have a specific Hamiltonian to use ? $\endgroup$ – Ismasou Jan 13 '17 at 12:12
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    $\begingroup$ @Alex, you can use whatever basis you'd like. Whether it is an eigenbasis of $\hat X$, $\hat P$ or $\hat H$. And of course you can always express the elements of one basis in terms of another basis, e.g. $\langle p | \psi \rangle$ which is the fourier transform of the wave function $\langle x | \psi \rangle$. $\endgroup$ – Noiralef Jan 13 '17 at 12:15
  • $\begingroup$ @Ismasou No I don't but I understand your point. I have not come across any Hamiltonians which depend on spin at this stage hence I did not consider this. Am I right in stating that the Hamiltonian (which does not depend on spin) would be a operator acting on the space $L^2$ (as does the position and momentum operator) whereas spin is an operator on $\mathbb{C}^s$ (for some spin-$s$ particle)? $\endgroup$ – Alex Jan 13 '17 at 12:34
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    $\begingroup$ you'll be write because the Hamiltonian is written just as a function of p and x anyways. (H(x,p)) $\endgroup$ – Ismasou Jan 13 '17 at 12:35

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