Consider a Hilbert space (or just a vector space over $\mathbb{C}$), which is a tensor product of several smaller Hilbert spaces: $$ H = H_1 \otimes \cdots \otimes H_n, $$ and let $\mathcal{H}$ be a Hermitian operator on $H$.

Can $\mathcal{H}$ always be represented as a sum of tensor products of Hermitian operators on the smaller Hilbert spaces?
i.e. can we write $$ \mathcal{H} = \sum_i \mathcal{H}^{(1)}_i \otimes \cdots \otimes \mathcal{H}^{(n)}_i, $$ where $\mathcal{H}^{(k)}_i$ is a Hermitian operator on $H_k$?

I came up with this question in the context of Quantum Mechanics. I would expect that a Hamiltonian of a composite system can be represented as a sum of Hamiltonians, where each one is a tensor product of Hamiltonians of the smaller systems.


1 Answer 1


Yes, at least for finite dimensional vector spaces. First, I'll show you for two vector spaces $H_1$ and $H_2$ and you can proceed by induction. Use the fact that any linear operator can be represented as a sum of tensor product of linear operators, and if $H$ is Hermitian then $\frac{H + H^\dagger}{2} = H$. That is, $$H = \frac{\sum_{i = 1}^m A_i \otimes B_i + A_i^\dagger \otimes B_i^\dagger}{2}$$ for linear operators $A_i :H_1 \to H_1$ and $B_i:H_2 \to H_2$. Now $$ H = \sum_{i = 1}^m \left(\frac{A_i + A_i^\dagger}{2}\right) \otimes \left(\frac{B_i + B_i^\dagger}{2}\right) - \sum_{i = 1}^m \left(\frac{A_i - A_i^\dagger}{2i}\right) \otimes \left(\frac{B_i - B_i^\dagger}{2i}\right),$$ and each term is a tensor product of Hermitian operators.


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