# Combining two 1D Hamiltonians: How to construct the new Hamiltonian properly?

Lets say I have two 1D Hilbert spaces $$\mathcal H_A, \mathcal H_B$$, for example two 1D harmonic oscillators. Each space comes with an orthonormal basis $$B_A=\{\phi^A_n \}, \ B_B=\{\phi^B_m \}, \ n,m\geq 0$$ where each function is a eigenfunction of the respective 1D-Hamilton operator. Now I'd like to construct the combined system, without any coupling. My understanding is that we take the tensor product of both spaces to obtain the new space $$\mathcal H_c$$, $$\mathcal H_C = \mathcal H_A\otimes \mathcal H_B$$

A basis for our new space is $$B_C = \{\phi_n^A\otimes\phi_m^B \}$$ such that \begin{aligned} \hat H_C \phi^C_{nm} &= E_{nm}\phi^C_{nm} \\ (\hat H_A\otimes \hat 1 + \hat 1\otimes \hat H_B)(\phi_n^A\otimes\phi_m^B) &= (E^A_n+E_m^B)\phi_n^A\otimes\phi_m^B \\ \end{aligned}

For everything to work out like this we need $$\hat H_C = \hat H_A\otimes \hat 1 + \hat 1\otimes \hat H_B$$

But it is not a-priori clear to me that it should be so. Why is the operator not given by $$\hat H_C = \hat H_A\otimes \hat H_B \quad ?$$ When are operators in the new space of the form $$\hat O_A\otimes \hat 1+ \hat 1\otimes \hat O_B$$ and when do operators take on the form $$\hat O_A\otimes \hat O_B.$$ The parity operator for example is of this form. Is there a simple way to tell how operators are "transferred" to a product space ?

A first answer in the case of the Hamiltonian is dimensional analysis : $$H_A\otimes H_B$$ has dimension of energy squared, so it is not a good candidate hamiltonian.

A more profound answer is that unitary operators extend using the tensor product, while hermitian operator extend using the sum rule (like the Hamiltonian).

For example, the time evolution operator $$U(t) = e^{-i\hat Ht /\hbar}$$ solves the Schrödinger equation. If you are given two solutions $$|\psi_A(t)\rangle = U_A(t) |\psi_A(0)\rangle$$ and $$|\psi_B(t)\rangle= U_B(t) |\psi_B(0)\rangle$$, you expect (since you are not introducing any coupling between the two subsystems, that $$|\psi_A(t)\rangle \otimes |\psi_B(t)\rangle$$ is a solution of the Schrödinger equation for the combined system.

That is : $$U_{AB}(t) = U_A(t)\otimes U_B(t)$$ Since since $$i\hbar\frac{d}{dt} U(t)|_{t=0} = H$$, by taking a time derivative at $$t=0$$, you get : $$H_{AB} = \hat H_A \otimes \mathbb I_B + \mathbb I_A \otimes H_B$$

More generally, symmetry operators (eg translations, rotations, parity, etc.) will extend using the tensor product. For continuous symmetries, taking a derivative will mean that the generators (momentum, angular momentum, spin, etc.) will extend using the sum rule.

• Does this mean that operators corresponding to Lie group elements go to tensor products and the corresponding Hermitian operators based on the Lie algebra elements go to a sum of operators ? Commented Dec 10, 2021 at 13:29
• @HansWurst yes. Eigenvalues of generators are additive but eigenfunctions are multiplicative. Commented Dec 10, 2021 at 18:10
• @HansWurst Basically yes. To be slightly more formal, one would talk about representations of Lie groups and Lie algebras Commented Dec 10, 2021 at 20:02