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

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 $$|\psi\rangle$$?

• clarification: are $A,B,C,D$ complex numbers? What are the dimensions of the two Hilbert spaces? Nov 18, 2019 at 13:39
• Please take note of the formatting edits I have made for future posts. It looks like you were making things way more complicated than they needed to be. If you have to split up your expressions with many \$'s, then there is probably a simpler way :) Nov 18, 2019 at 14:02

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
$$$$\boldsymbol{|}\psi_1\boldsymbol{\rangle}\boldsymbol{=} \begin{bmatrix} \xi_1 \vphantom{\dfrac{a}{b}} \\ \xi_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \in \mathcal{H}_1 \tag{01a}\label{01a}$$$$ and the Hamiltonian by the hermitian matrix $$$$\mathsf{H}_{1}\boldsymbol{=} \begin{bmatrix} a_1 & b_1\vphantom{\dfrac{a}{b}} \\ c_1 & d_1\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{01b}\label{01b}$$$$ Similarly in the two dimensional Hilbert space $$\,\mathcal{H}_2\,$$ $$$$\boldsymbol{|}\psi_2\boldsymbol{\rangle}\boldsymbol{=} \begin{bmatrix} \eta_1 \vphantom{\dfrac{a}{b}} \\ \eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \in \mathcal{H}_2 \tag{02a}\label{02a}$$$$ and $$$$\mathsf{H}_{2}\boldsymbol{=} \begin{bmatrix} a_2 & b_2\vphantom{\dfrac{a}{b}} \\ c_2 & d_2\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{02b}\label{02b}$$$$ One way to represent the product state $$\,\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\,$$ is $$$$\boldsymbol{|}\psi_1\psi_2\boldsymbol{\rangle}\boldsymbol{\equiv}\boldsymbol{|}\psi_1\boldsymbol{\rangle}\boldsymbol{\otimes}\boldsymbol{|}\psi_2\boldsymbol{\rangle}\boldsymbol{=} \begin{bmatrix} \xi_1 \vphantom{\dfrac{a}{b}} \\ \xi_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{\otimes} \begin{bmatrix} \eta_1 \vphantom{\dfrac{a}{b}} \\ \eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \xi_1 \begin{bmatrix} \eta_1 \vphantom{\dfrac{a}{b}} \\ \eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix}\\ \xi_2 \begin{bmatrix} \eta_1 \vphantom{\dfrac{a}{b}} \\ \eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \xi_1\eta_1 \vphantom{\dfrac{a}{b}} \\ \xi_1\eta_2 \vphantom{\dfrac{a}{b}} \\ \xi_2\eta_1 \vphantom{\dfrac{a}{b}} \\ \xi_2\eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \in \mathcal{H} \tag{03}\label{03}$$$$ Motivated by the following ZeroTheHero's comment
the Hamiltonian in the product space is $$$$\mathsf{H}\equiv \mathsf{H}_{1}\boldsymbol{\otimes}\,\mathsf{I}_{2}\boldsymbol{+}\mathsf{I}_{1}\,\boldsymbol{\otimes}\mathsf{H}_{2} \tag{04}\label{04}$$$$
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} $$$$\mathsf{H}\equiv \mathsf{H}_{1}\boldsymbol{\otimes}\,\mathsf{I}_{2}\boldsymbol{+}\mathsf{I}_{1}\,\boldsymbol{\otimes}\mathsf{H}_{2} \boldsymbol{=} \begin{bmatrix} a_1\boldsymbol{+}a_2 & b_2 & b_1 & 0 \vphantom{\dfrac{a}{b}}\\ c_2 & a_1\boldsymbol{+}d_2 & 0 & b_1 \vphantom{\dfrac{a}{b}}\\ c_1 & 0 & d_1\boldsymbol{+}a_2 & b_2 \vphantom{\dfrac{a}{b}}\\ 0 & c_1 & c_2 & d_1\boldsymbol{+}d_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{06}\label{06}$$$$ Note that if in both spaces we make use of the basis of eigenkets of the respective Hamiltonian for the matrix representations then, since $$\,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\,$$ while from \eqref{06} $$$$\mathsf{H}\equiv \mathsf{H}_{1}\boldsymbol{\otimes}\,\mathsf{I}_{2}\boldsymbol{+}\mathsf{I}_{1}\,\boldsymbol{\otimes}\mathsf{H}_{2} \boldsymbol{=} \begin{bmatrix} a_1\boldsymbol{+}a_2 & 0 & 0 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & a_1\boldsymbol{+}d_2 & 0 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & 0 & d_1\boldsymbol{+}a_2 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & 0 & 0 & d_1\boldsymbol{+}d_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{07}\label{07}$$$$ 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.