Suppose one has two single-particle Hilbert spaces $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ and consider the tensor product of these such that $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ is a two-particle Hilbert space. As I understand it, the inner product of two state vectors $\lvert\phi_{1},\psi_{2}\rangle,\lvert\chi_{1},\xi_{2}\rangle\in \mathcal{H}_{A}\otimes\mathcal{H}_{B}$ (where $\lvert\phi_{1}\rangle,\lvert\chi_{1}\rangle\in\mathcal{H}_{A}$ and $\lvert\psi_{2}\rangle,\lvert\xi_{2}\rangle\in\mathcal{H}_{B}$) is defined by $$\langle\xi_{2},\chi_{1}\lvert\phi_{1},\psi_{2}\rangle =\langle\chi_{1}\lvert\phi_{1}\rangle_{A}\langle\xi_{2}\lvert\psi_{2}\rangle_{B}$$ However, I'm unsure how their outer product would be defined? Is it something like $$\lvert\phi_{1},\psi_{2}\rangle\langle\xi_{2},\chi_{1}\rvert =\lvert\phi_{1}\rangle\langle\chi_{1}\rvert\otimes\lvert\psi_{2}\rangle\langle\xi_{2}\rvert$$ Also, say I have two operators $\hat{A}\in L(\mathcal{H}_{A})$ and $\hat{B}\in L(\mathcal{H}_{B})$ (where $L(\mathcal{H}_{i})$ is the set of linear operators acting on the single-particle Hilbert space $\mathcal{H}_{i}$). Then I can consider an operator $\hat{\tilde{C}}\in L(\mathcal{H}_{A}\otimes\mathcal{H}_{B})$ defined as $$\hat{\tilde{C}}=\hat{A}\otimes\mathbf{1}_{B}+\mathbf{1}_{A}\otimes\hat{B}+\hat{A}\otimes\hat{B}$$ where $\mathbf{1}_{i}$ is the identity operator for the single-particle Hilbert space $\mathcal{H}_{i}$ ($i=A,B$). If I had another operator $\hat{\tilde{D}}\in L(\mathcal{H}_{A}\otimes\mathcal{H}_{B})$ defined similarly as $$\hat{\tilde{D}}=\hat{F}\otimes\mathbf{1}_{B}$$ for simplicity (again with $\hat{F}\in L(\mathcal{H}_{A})$ and $\hat{G}\in L(\mathcal{H}_{B})$) then what would be the commutator of these two operators, i.e. what is $$\left[\hat{\tilde{C}},\hat{\tilde{D}}\right]=\left[\hat{A}\otimes\mathbf{1}_{B}+\mathbf{1}_{A}\otimes\hat{B}+\hat{A}\otimes\hat{B}, \hat{F}\otimes\mathbf{1}_{B}\right]\qquad\qquad\qquad\qquad\qquad\qquad\quad\\ =\left[\hat{A}\otimes\mathbf{1}_{B},\hat{F}\otimes\mathbf{1}_{B}\right]+\left[\mathbf{1}_{A}\otimes\hat{B},\hat{F}\otimes\mathbf{1}_{B}\right]+\left[\hat{A}\otimes\hat{B},\hat{F}\otimes\mathbf{1}_{B}\right]$$ equal to? In essesnce, what is a product of operators in the two-particle Hilbert space, e.g. $(\hat{A}\otimes\mathbf{1}_{B})(\hat{F}\otimes\mathbf{1}_{B})=\,\,?$


1 Answer 1


Yes, their outer product is defined as you said. Further, the product of operators is given by $$ (A \otimes 1_B)(F \otimes 1_B) = (AF) \otimes (1_B1_B) = (AF) \otimes 1_B $$ Therefore, $$ [A \otimes B, F \otimes 1_B] = (A \otimes B)( F \otimes 1_B) - ( F \otimes 1_B) (A \otimes B) = [A,F] \otimes B $$

  • $\begingroup$ Great. Does that apply in general then? i.e. can I have something like $\left(A\otimes B\right)\left(C\otimes D\right)=\left(AC\right)\otimes\left(BD\right)$? Is there any particularly good literature on this that I can read up on? $\endgroup$
    – Will
    Apr 26, 2016 at 16:12
  • 2
    $\begingroup$ I don't think the commutator part of this is right. Think about an explicit example with spins and expand the commutator: $[\sigma_z \otimes 1, \sigma_x \otimes 1] = (\sigma_z \otimes 1)(\sigma_x \otimes 1) - (\sigma_x \otimes 1)(\sigma_z \otimes 1) = [\sigma_z,\sigma_x] \otimes 1$. Not zero because the identities commute. $\endgroup$
    – RGWinston
    Jan 16, 2017 at 21:37

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