Suppose we have two systems with density matrices $\rho_1$ and $\rho_2$. Initially they are non-interacting, and so their composite density matrix looks like: $$\rho_t = \rho_1 \otimes \rho_2$$ I have seen in some places that,if $\rho_1 = |\psi\rangle\langle\psi|$ then $\rho_t$ can be written as: $$\rho_t = |\psi\rangle\rho_2\langle\psi|$$ How can this manipulation be done and what is the general identity that allows such a manipulation?

I found this in Braginsky, Khalili - Quantum Measurement, page 44

  • $\begingroup$ Please provide a source where you see this. $\endgroup$ – user1936752 Apr 3 '19 at 15:20

Due to the definition $\rho_t = \rho_1 \otimes \rho_2$ the two partial operators $\rho_1$ and $\rho_2$ live in separate Hilbert spaces $H_1$ respectively $H_2$ and therefore their matrix product commutes within the combined Hilbert space $H_1 \otimes H_2$. For this to be more obvious you could equivalently write the tensor product as a matrix product after the constitunt operators are mapped to the combined space: $$ \rho_1 \otimes \rho_2 = (\rho_1 \otimes \mathbb{1}_2)(\mathbb{1}_1 \otimes\rho_2) = (\mathbb{1}_1 \otimes \rho_2)(\rho_1 \otimes \mathbb{1}_2) $$ where the matrix product can be evaluated as $$ (A \otimes B)(C \otimes D) = (AC) \otimes (BD) $$ The identity operators are usually omitted if it is clear what Hilbert space the individual operators belong to.The part, which you don't understand, is indeed notationaly confusing when someone is not familiar with the tensor product. It would be more correct to write it as $$ \rho_t = (|\psi\rangle \otimes \mathbb{1}_2)(\mathbb{1}_1 \otimes \rho_2)(\langle \psi | \otimes \mathbb{1}_2) $$ and applying the lazy notation yields $$ \rho_t = |\psi\rangle(\mathbb{1}_1 \otimes \rho_2)\langle \psi | = |\psi\rangle \rho_2 \langle \psi |. $$ Due the commutation this is also equivalent to $\rho_t = |\psi\rangle \langle \psi | \rho_2 = \rho_2 |\psi\rangle \langle \psi |$, illustrating the pecularities of tensor product notation further.

Another way to understand this expression is by assuming that the writer implicitly assumed the identities $$ \rho_1 = \rho_1 \otimes \mathbb{1}_2 \\ \rho_2 = \mathbb{1}_1 \otimes \rho_2 $$ without explicitly stating so. When using this notation yourself, you should make sure that it is clear from the context, how to interpret everything.

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  • $\begingroup$ How can you say that the tensor product is commutative on Hilbert spaces? As far as I know, the tensor product is in general non-commutative. $\endgroup$ – Dharanish Rajendra Apr 5 '19 at 8:44
  • $\begingroup$ I didn't say that the tensor product itself is commutative and you are right that it isn't. Only the separable constituents of $\rho_t$, which are $\rho_1$ and $\rho_2$, do commute within the combined Hilbert space. I'll update my answer later to elaborate more on that. $\endgroup$ – Halbeard Apr 6 '19 at 12:30
  • $\begingroup$ Now it's clear to me. I get what you are trying to say by commutativity. $(\rho_1 \otimes \mathbb{1}_2)(\mathbb{1}_1 \otimes\rho_2) = (\mathbb{1}_1 \otimes \rho_2)(\rho_1 \otimes \mathbb{1}_2)$ is true only because identity commutes with any matrix in normal matrix multiplication. Thanks. $\endgroup$ – Dharanish Rajendra Apr 8 '19 at 5:23

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