Before stating my question I present my motivation: to learn more about the tensor product.

Now, quantum mechanics assigns a Hilbert space to each physical system as a postulate of the theory. Suppose that we have two physical systems $A$ and $B$, whose state is described by density operator $\rho^{AB}$. Before characterizing density operators, let's ascribe the Hilbert space $\mathcal{H}_{A}$ to system $A$, and the Hilbert space $\mathcal{H}_{B}$ to system $B$. The composite system is the tensored space $\mathcal{H}_{A} \otimes \mathcal{H}_{B}$. The density operator $\rho^{AB}$ is characterized as a nonnegative trace class operator of trace 1 on $\mathcal{H}_{A} \otimes \mathcal{H}_{B}$.

Definition: The reduced density operator for system $A$ is described by $\rho^{A} \equiv Tr_{B} (\rho^{AB})$, where $Tr_{B}$ is a map of operators known as the partial trace over system $B$. The partial trace is defined by $$Tr_{B} (a_{1}a_{2}^{T} \otimes b_{1}b_{2}^{T}) \equiv a_{1}a_{2}^{T} \: Tr(b_{1}b_{2}^{T}),$$ where $a_{1}$ and $a_{2}$ are any two vectors in $\mathcal{H}_{A}$, $b_{1}$ and $b_{2}$ are any two vectors in $\mathcal{H}_{B}$, and $a_{1}^{T}$ is the transposed vector (row vector). So, the trace on the right hand side is the usual trace, i.e. $$Tr(b_{1}b_{2}^{T})=b_{2}^{T}b_{1}.$$

Given the above does the following claim hold: $$a_{1}a_{2}^{T} \otimes b_{1}b_{2}^{T}=(a_{1} \otimes b_{1})(b_{2}^{T} \otimes a_{2}^{T})$$

  • $\begingroup$ In general the answer is no, but for some particular cases it can be true. One question: How is your question related to the partial trace? $\endgroup$
    – Daniel
    Mar 24, 2015 at 14:59
  • $\begingroup$ @Daniel I want to take the partial trace with respect to system B of the right hand side of the claim, and show that this equals the partial trace with respect to system B of the left hand side of the claim. $\endgroup$
    – sunspots
    Mar 24, 2015 at 15:12
  • $\begingroup$ @Daniel what are the particular cases in which the claim can be true? $\endgroup$
    – sunspots
    Mar 24, 2015 at 15:13
  • 1
    $\begingroup$ In the matrix representation, $\otimes$ is the Kronecker product which obeys the followng mixed-product rule $$\large{a_1a_2^T\otimes b_1b_2^T = (a_1\otimes b_1)\,(a_2^T\otimes b_2^T)}$$ so the second factor on the RHS is reversed. $\endgroup$
    – greg
    Mar 22, 2023 at 12:36


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