I Tensor products and simultaneous eigenstates

[LightPhoton]

In A Modern Approach to Quantum Mechanics, Townsend writes:



One of the most evident features of the position-space representations (9.117), (9.127), and (9.128) of the angular momentum operators is that they depend only on the angles $\theta$ and $\phi$, not at all on the magnitude $r$ of the position vector. Rotating a position eigenstate changes its direction but not its length. Thus we can isolate the angular dependence and determine $\langle \theta, \phi | \ell, m \rangle$, the amplitude for a state of definite angular momentum to be at the angles $\theta$ and $\phi$. These amplitudes, which are functions of the angles, are called the spherical harmonics and denoted by

$$\langle \theta, \phi | \ell, m \rangle = Y_{\ell, m}(\theta, \phi) \tag{9.132}$$

where

$$\hat L_z\rightarrow\frac\hbar i\frac\partial{\partial
\phi}\tag{9.117}$$ $$\hat L_x\rightarrow\frac\hbar
i\bigg(-\sin\phi\frac\partial{\partial\theta}-\cot\theta\cos\phi\frac\partial{\partial\phi}\bigg)\tag{9.127}$$
$$\hat L_y\rightarrow\frac\hbar
i\bigg(\cos\phi\frac\partial{\partial\theta}-\cot\theta\sin\phi\frac\partial{\partial\phi}\bigg)\tag{9.128}$$


Then he states that since an eigenstate of a hydrogen-like system can be written as $\vert E,l,m\rangle$, we have


$$\langle r,\theta,\phi\vert E,l,m\rangle=R(r)Y_{\ell, m}(\theta, \phi)$$



Does this mean that he means one could write $\vert E,l,m\rangle= \vert E\rangle\otimes\vert l,m\rangle$ and $\langle r,\theta,\phi\vert=\langle r\vert\otimes\langle\theta,\phi\vert$?


If yes, then does this mean that when we say we have a simultaneous eigenstate of three variables $\vert x,y, z\rangle$, what we really mean is that we have a state $\vert x\rangle\otimes\vert y\rangle\otimes\vert z\rangle$?

 

[div_grad]

Then he states that since an eigenstate of a hydrogen-like system can be written as $\vert E,l,m\rangle$, we have
$$\langle r,\theta,\phi\vert E,l,m\rangle=R(r)Y_{\ell, m}(\theta, \phi)$$

This expression is wrong, because the radial part $R_l(r)$ of the wave function depends on the quantum number $l$ (but it does not depend on $m$). This is because after the separation of variables the "leftover" one-dimensional radial Schrödinger equation still contains a term $\propto l(l+1)$.

Does this mean that he means one could write $\vert E,l,m\rangle= \vert E\rangle\otimes\vert l,m\rangle$ and $\langle r,\theta,\phi\vert=\langle r\vert\otimes\langle\theta,\phi\vert$?

This depends on what the symbols $|r \rangle$ and $|\theta, \phi \rangle$ mean. Are they the eigenvectors of Hermitian operators? If so, what are those operators and what are the associated eigenvalues? Further, if one can apparently write $|r, \theta, \phi \rangle = |r \rangle \otimes |\theta, \phi \rangle$, then can one also write $|\theta, \phi \rangle = |\theta \rangle \otimes |\phi \rangle$?

The point is, if you solve a given problem and thus know the correct steps leading to the solution already, then you can introduce any arbitrary form of (Dirac) notation and then present your reasoning to other people using this notation - but then you risk others not understanding you. Does this textbook's author defines what they mean by the expression $|r, \theta, \phi \rangle$ somewhere in the text?

(...) when we say we have a simultaneous eigenstate of three variables $\vert x,y, z\rangle$, what we really mean is that we have a state $\vert x\rangle\otimes\vert y\rangle\otimes\vert z\rangle$?

Yes - but see my above comment on notation. For example, is it true that $\vert x\rangle\otimes\vert y\rangle\otimes\vert z\rangle = \vert y\rangle\otimes\vert x\rangle\otimes\vert z\rangle$?