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 ?


1 Answer 1


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.

  • $\begingroup$ 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 ? $\endgroup$
    – Hans Wurst
    Commented Dec 10, 2021 at 13:29
  • $\begingroup$ @HansWurst yes. Eigenvalues of generators are additive but eigenfunctions are multiplicative. $\endgroup$ Commented Dec 10, 2021 at 18:10
  • $\begingroup$ @HansWurst Basically yes. To be slightly more formal, one would talk about representations of Lie groups and Lie algebras $\endgroup$ Commented Dec 10, 2021 at 20:02

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