Adding two kets (and Hamiltonians) from two different hilbert spaces?

Question

The combined wavefunction for kets in two different Hilbert spaces

$$|\psi\rangle= c_{11}|11\rangle + c_{1r}|1r\rangle +c_{r1}|r1\rangle +c_{rr}|rr\rangle$$

Where $|ab\rangle = |a\rangle_1 \otimes |b\rangle_2$, and $|a\rangle$ is in a different Hilbert space than $|b\rangle$.

The Hamiltonian for each Hilbert space is say $$H_1 = H_2 = \left( \begin{matrix} A & B \\ C & D \end{matrix} \right)$$

How would you combine $H_1$ and $H_2$ to produce a Hamiltonian that can act on the above $|\psi\rangle$?

Answer 1

Consider that a state ket $\,\boldsymbol{|}\psi_1\boldsymbol{\rangle}\,$ in the 2-dimensional Hilbert space $\,\mathcal{H}_1\,$ is represented by its components with respect to a basis of this space by

\begin{equation} \boldsymbol{|}\psi_1\boldsymbol{\rangle}\boldsymbol{=} \begin{bmatrix} \xi_1 \vphantom{\dfrac{a}{b}} \\ \xi_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \in \mathcal{H}_1 \tag{01a}\label{01a} \end{equation} and the Hamiltonian by the hermitian matrix \begin{equation} \mathsf{H}_{1}\boldsymbol{=} \begin{bmatrix} a_1 & b_1\vphantom{\dfrac{a}{b}} \\ c_1 & d_1\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{01b}\label{01b} \end{equation} Similarly in the two dimensional Hilbert space $\,\mathcal{H}_2\,$ \begin{equation} \boldsymbol{|}\psi_2\boldsymbol{\rangle}\boldsymbol{=} \begin{bmatrix} \eta_1 \vphantom{\dfrac{a}{b}} \\ \eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \in \mathcal{H}_2 \tag{02a}\label{02a} \end{equation} and \begin{equation} \mathsf{H}_{2}\boldsymbol{=} \begin{bmatrix} a_2 & b_2\vphantom{\dfrac{a}{b}} \\ c_2 & d_2\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{02b}\label{02b} \end{equation} One way to represent the product state $\,\boldsymbol{|}\psi_1\psi_2\boldsymbol{\rangle}\boldsymbol{\equiv}\boldsymbol{|}\psi_1\boldsymbol{\rangle}\boldsymbol{\otimes}\boldsymbol{|}\psi_2\boldsymbol{\rangle}\,$ in the 4-dimensional product Hilbert space $\,\mathcal{H}\boldsymbol{\equiv}\mathcal{H}_1\boldsymbol{\otimes}\mathcal{H}_2\,$ is \begin{equation} \boldsymbol{|}\psi_1\psi_2\boldsymbol{\rangle}\boldsymbol{\equiv}\boldsymbol{|}\psi_1\boldsymbol{\rangle}\boldsymbol{\otimes}\boldsymbol{|}\psi_2\boldsymbol{\rangle}\boldsymbol{=} \begin{bmatrix} \xi_1 \vphantom{\dfrac{a}{b}} \\ \xi_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{\otimes} \begin{bmatrix} \eta_1 \vphantom{\dfrac{a}{b}} \\ \eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \xi_1 \begin{bmatrix} \eta_1 \vphantom{\dfrac{a}{b}} \\ \eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix}\\ \xi_2 \begin{bmatrix} \eta_1 \vphantom{\dfrac{a}{b}} \\ \eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \xi_1\eta_1 \vphantom{\dfrac{a}{b}} \\ \xi_1\eta_2 \vphantom{\dfrac{a}{b}} \\ \xi_2\eta_1 \vphantom{\dfrac{a}{b}} \\ \xi_2\eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \in \mathcal{H} \tag{03}\label{03} \end{equation} Motivated by the following ZeroTheHero's comment

the Hamiltonian in the product space is \begin{equation} \mathsf{H}\equiv \mathsf{H}_{1}\boldsymbol{\otimes}\,\mathsf{I}_{2}\boldsymbol{+}\mathsf{I}_{1}\,\boldsymbol{\otimes}\mathsf{H}_{2} \tag{04}\label{04} \end{equation}

Under the convention \eqref{03} \begin{align} \mathsf{H}_{1}\boldsymbol{\otimes}\,\mathsf{I}_{2} &\boldsymbol{=} \begin{bmatrix} a_1 & b_1\vphantom{\dfrac{a}{b}} \\ c_1 & d_1\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{\otimes} \begin{bmatrix} 1\hphantom{_1} & 0\hphantom{_1}\vphantom{\dfrac{a}{b}} \\ 0\hphantom{_1} & 1\hphantom{_1}\vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber\\ &\boldsymbol{=} \begin{bmatrix} a_1\begin{bmatrix} 1\hphantom{_1} & 0\hphantom{_1}\vphantom{\dfrac{a}{b}} \\ 0\hphantom{_1} & 1\hphantom{_1}\vphantom{\dfrac{a}{b}} \end{bmatrix} & b_1\begin{bmatrix} 1\hphantom{_1} & 0\hphantom{_1}\vphantom{\dfrac{a}{b}} \\ 0\hphantom{_1} & 1\hphantom{_1}\vphantom{\dfrac{a}{b}} \end{bmatrix} \\ c_1\begin{bmatrix} 1\hphantom{_1} & 0\hphantom{_1}\vphantom{\dfrac{a}{b}} \\ 0\hphantom{_1} & 1\hphantom{_1}\vphantom{\dfrac{a}{b}} \end{bmatrix} & d_1\begin{bmatrix} 1\hphantom{_1} & 0\hphantom{_1}\vphantom{\dfrac{a}{b}} \\ 0\hphantom{_1} & 1\hphantom{_1}\vphantom{\dfrac{a}{b}} \end{bmatrix} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} a_1 & 0 & b_1 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & a_1 & 0 & b_1 \vphantom{\dfrac{a}{b}}\\ c_1 & 0 & d_1 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & c_1 & 0 & d_1 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{05a}\label{05a} \end{align} and \begin{align} \mathsf{I}_{1}\,\boldsymbol{\otimes}\mathsf{H}_{2} &\boldsymbol{=} \begin{bmatrix} 1\hphantom{_1} & 0\hphantom{_1}\vphantom{\dfrac{a}{b}} \\ 0\hphantom{_1} & 1\hphantom{_1}\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{\otimes} \begin{bmatrix} a_2 & b_2\vphantom{\dfrac{a}{b}} \\ c_2 & d_2\vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber\\ &\boldsymbol{=} \begin{bmatrix} 1\begin{bmatrix} a_2 & b_2\vphantom{\dfrac{a}{b}} \\ c_2 & d_2\vphantom{\dfrac{a}{b}} \end{bmatrix} & 0 \begin{bmatrix} a_2 & b_2\vphantom{\dfrac{a}{b}} \\ c_2 & d_2\vphantom{\dfrac{a}{b}} \end{bmatrix} \\ 0\begin{bmatrix} a_2 & b_2\vphantom{\dfrac{a}{b}} \\ c_2 & d_2\vphantom{\dfrac{a}{b}} \end{bmatrix} & 1 \begin{bmatrix} a_2 & b_2\vphantom{\dfrac{a}{b}} \\ c_2 & d_2\vphantom{\dfrac{a}{b}} \end{bmatrix} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} a_2 & b_2 & 0 & 0 \vphantom{\dfrac{a}{b}}\\ c_2 & d_2 & 0 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & 0 & a_2 & b_2 \vphantom{\dfrac{a}{b}}\\ 0 & 0 & c_2 & d_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{05b}\label{05b} \end{align} Adding \eqref{05a},\eqref{05b} \begin{equation} \mathsf{H}\equiv \mathsf{H}_{1}\boldsymbol{\otimes}\,\mathsf{I}_{2}\boldsymbol{+}\mathsf{I}_{1}\,\boldsymbol{\otimes}\mathsf{H}_{2} \boldsymbol{=} \begin{bmatrix} a_1\boldsymbol{+}a_2 & b_2 & b_1 & 0 \vphantom{\dfrac{a}{b}}\\ c_2 & a_1\boldsymbol{+}d_2 & 0 & b_1 \vphantom{\dfrac{a}{b}}\\ c_1 & 0 & d_1\boldsymbol{+}a_2 & b_2 \vphantom{\dfrac{a}{b}}\\ 0 & c_1 & c_2 & d_1\boldsymbol{+}d_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{06}\label{06} \end{equation} Note that if in both spaces we make use of the basis of eigenkets of the respective Hamiltonian for the matrix representations then, since $\,c_1\boldsymbol{=}0\boldsymbol{=}b_1\,$, the eigenvalues of $\,\mathsf{H}_{1}\,$ are $\,a_1,d_1\,$ and since $\,c_2\boldsymbol{=}0\boldsymbol{=}b_2\,$ the eigenvalues of $\,\mathsf{H}_{2}\,$ are $\,a_2,d_2\,$ while from \eqref{06} \begin{equation} \mathsf{H}\equiv \mathsf{H}_{1}\boldsymbol{\otimes}\,\mathsf{I}_{2}\boldsymbol{+}\mathsf{I}_{1}\,\boldsymbol{\otimes}\mathsf{H}_{2} \boldsymbol{=} \begin{bmatrix} a_1\boldsymbol{+}a_2 & 0 & 0 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & a_1\boldsymbol{+}d_2 & 0 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & 0 & d_1\boldsymbol{+}a_2 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & 0 & 0 & d_1\boldsymbol{+}d_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{07}\label{07} \end{equation} that is the eigenvalues of the Hamiltonian in the product space are sums produced by combinations of the eigenvalues of the Hamiltonians in the factor spaces.