One spin-1/2 particle in B field in a 3D harmonic potential (Part III) Consider a spin-1/2 particle in a magnetic field (say in z direction) and in a harmonic potential. For the 3D harmonic oscillator component, The Hamiltonian $H_1= \frac{p^2}{2m}+\frac{1}{2}m\omega ^2r^2.$ For the spin component, the Hamiltonian $H_2=-\gamma B_z S_z$, where $\gamma$ is the gyroscopic ratio.
Questions:


*

*Is it possible to represent the an eigenstate of the system as tensor product of eigenstate of each of the two Hamiltonian? I.e. Is the eigenstate $\left|n_x,n_y,n_z\right>\otimes \left|1/2,m_s\right>$, where $\left|n_x,n_y,n_z\right>$ is an eigenstate of $H_1$, and $\left|1/2,m_s\right>$ is an eigenstate of $H_2$? Is the explicit form of a state, for example, 
$\left|0,0,0\right>\otimes \left|1/2,+1/2\right>=(\frac{m\omega}{\pi\hbar})^{1/4}exp(-\frac{m\omega}{2\hbar} r^2)\otimes \begin{pmatrix}1\\0\end{pmatrix}$?

*If answer to 2 is yes, are the states $\left|0,0,0\right>\otimes \left|1/2,+1/2\right>$ and $\left|0,0,0\right>\otimes \left|1/2,-1/2\right>$ orthogonal? My guess would be that the inner product of the two tensor states is the inner product of each of the component, and since for the second component $\left<1/2,+1/2\middle| 1/2,-1/2\right>=0$, the two tensor states should be orthogonal?
I have been trying to answer the last question, but different reasoning (some might be too naive) seems to give contradicting answers, so I want to see if the above reasoning is valid.
 A: Your overall Hamiltonian for this system is actually
$$
H = H_{1} \otimes \mathbb{I} + \mathcal{I} \otimes H_2
$$
where your overall Hilbert space is $L_{2}(\mathbb{R}_{3}) \otimes \mathbb{C}^2$, and $\mathcal{I}$ is the identity on $L_{2}(\mathbb{R}_{3})$, while $\mathbb{I}$ is the identity on $\mathbb{C}^2$.
Let $|\mathbf{n} \rangle \in L_{2}(\mathbb{R}_{3})$ be the eigenstates of $H_1$ such that
$$
H_1 |\mathbf{n} \rangle = E^{(1)}_{\mathbf{n}} |\mathbf{n} \rangle
$$
and $| s \rangle \in \mathbb{C}^2$ be the eigenstates of $H_2$ such that
$$
H_2 | s \rangle = E^{(2)}_{s} | s \rangle \ .
$$
It is a fact that $|\mathbf{n} ; s \rangle : = |\mathbf{n} \rangle \otimes | s \rangle$ are eigenstates of the overall Hamiltonian $H$. You can calculate the eigenvalues corresponding to such a state by using the rules of how tensor products work (see Tensor product of linear maps):
\begin{align}
H |\mathbf{n} ; s \rangle & = \big( H_{1} \otimes \mathbb{I} + \mathcal{I} \otimes H_2 \big) \big( |\mathbf{n} \rangle \otimes | s \rangle \big) \\
& =  \big( H_1 |\mathbf{n} \rangle \big) \otimes \big( \mathbb{I} | s \rangle \big) + \big( \mathcal{I} |\mathbf{n} \rangle \big) \otimes \big( H_2 | s \rangle \big) \\
& =  \big( E_{\mathbf{n}}^{(1)} |\mathbf{n} \rangle \big) \otimes \big(  | s \rangle \big) + \big( |\mathbf{n} \rangle \big) \otimes \big( E_{s}^{(2)} | s \rangle \big) \\
& =  E_{\mathbf{n}}^{(1)} \big( |\mathbf{n} \rangle \otimes | s \rangle \big) + E_{s}^{(2)} \big( |\mathbf{n} \rangle \otimes | s \rangle \big) \\
& =  \big( E_{\mathbf{n}}^{(1)} + E_{s}^{(2)} \big) |\mathbf{n} ; s \rangle 
\end{align}
so the eigenvalues are $E_{\mathbf{n}}^{(1)} + E_{s}^{(2)}$.
These states are certainly orthogonal. For two states $|\mathbf{n} ; s \rangle$ and $|\mathbf{m} ; r \rangle $, you can compute their inner-product as
$$
\langle \mathbf{n} ; s | \mathbf{m} ; r \rangle = \delta_{n_x, m_{x}} \delta_{n_y, m_{y}} \delta_{n_z, m_{z}} \delta_{s,r}
$$
which is zero if any of the labels on the states are different.
A: You have $[H_1, H_2] = 0$, so yes, it is possible to diagonalize $H = H_1 + H_2$ using simultaneous eigenkets of $H_1$ and $H_2$. And to the second question yes again, the two states are indeed orthogonal. You can use the reasoning that if one individual component is orthogonal then the whole state must be orthogonal or, since the two tensor products kets are labeled by different indexes, then they represent different eigenkets and due to the fact that $H$ is hermitian, its eigenkets are orthogonal to each other.
