How is tracing out a physical operation?

Question

Suppose $\rho_{AB}$ denotes the density matrix of a bipartite system. Reduced density matrix of A ($\rho_A$) is obtained by tracing out B $$\rho_A\equiv\sum_{i}\langle i_B |\rho_{AB}|i_B\rangle$$ where $\{|i_B\rangle \}$ is the basis of the subsystem B. It is said that $\rho_A$ is the physical state of the subsystem A. What justifies this claim?

Answer 1

The partial trace over $B$ of the quantum state of a bipartite system $AB$ corresponds to discarding $B$: that is, the reduced density matrix $\rho_A=\mathrm{Tr}_B(\rho_{AB})$ is the complete description of the state of the system for any and all measurements that are completely local to $A$.

This can be made precise by considering an arbitrary hermitian measurement operator $\mathcal O_A$ (which includes, among other things, eigenprojectors corresponding to the measurement of some other observable), whose expectation value is $$ \langle \mathcal O_A\rangle = \mathrm{Tr}\mathopen{}\left( \hat{\rho}_{AB} \ \hat{\mathcal O}_A\otimes \mathbb I\right)\mathclose{}. $$ Here the trace can be decomposed as $$ \langle \mathcal O_A\rangle = \mathrm{Tr}_A\mathopen{}\left( \mathrm{Tr}_B\mathopen{}\left( \hat{\rho}_{AB} \ \hat{\mathcal O}_A\otimes \mathbb I\right)\mathclose{}\right)\mathclose{}, $$ and since $\hat{\mathcal O}_A$ does not act on the $B$ sector, it can be factored out of the $B$ trace, giving $$ \langle \mathcal O_A\rangle = \mathrm{Tr}_A\mathopen{}\left( \mathrm{Tr}_B\mathopen{}\left( \hat{\rho}_{AB}\right)\mathclose{} \hat{\mathcal O}_A\right)\mathclose{}, $$ or in other words, $$ \langle \mathcal O_A\rangle = \mathrm{Tr}_A\mathopen{}\left( \hat{\rho}_{A} \hat{\mathcal O}_A\right)\mathclose{}. $$ Thus, if you want to predict the results of any possible experiment that only involves $A$, then you need no more, and no less, than $\hat{\rho}_{AB}$.

Answer 2

Axiomatisation

A map $\Phi \colon L^{1}(H_{1}) \to L^{1}(H_{2})$ (between the trace-class operators of two Hilbert spaces) is defined to be a Quantum operation if and only if it satisfies two properties:

1. $tr(\Phi(\rho)) = 1$ for all $\rho\in L^{1}(H_{1})$ with $\rho \geq 0$ and $tr(\rho)=1$.^[1]
2. $\Phi$ is completely positive. This means that $$\Phi \otimes id_{n} \colon L^{1}(H_{1}\otimes\mathbb{C}^{n}) \to L^{1}(H_{2}\otimes\mathbb{C}^{n})$$ maps positive operators to positive operators for each $n\in\mathbb{N}$.

So why these two axioms? Well 1. encapsulations energy preservation. Note ($\ast$) also, that if $\Phi$ satisfies 1., then the map $\Phi \otimes id_{n}$ clearly satisfies 1. too for each $n\in\mathbb{N}$. What we additionally want want is:

• ($\ast\ast$) the system with operation $\Phi$ separably coupled to any other quantum system (say with operation $\Psi$), then the resulting system (which has operation $\Phi\otimes\Psi$) should also be a 'quantum operation'. In particular, it should map states to states.

In light of ($\ast$), for the 'in particular' bit, it is sufficient to demand that positive operators (or states, if the trace is 1) to be mapped to positive operators. Property 2. achieves exactly this for the simplest examples of additional systems (i.e. where the underlying space is just finitely many points $\{1,2,\ldots,n\}$).

In fact, demanding just 1+2 results in the stronger property ($\ast\ast$).

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Examples

We can, of course, debate whether the above properties are too strict or too lax to characterise what constitutes a Quantum operation. Nonetheless, this is the definition used by (mathematical) physicists since over half a century.

One can now look at a bunch of examples and show that these satisfy 1+2 and therefore under the above definition qualify as Quantum operations:

• partial trace: Say in a system $H_{A} \otimes H_{B}$ we trace out system $B$. Then $\Phi$ is defined by $\Phi(\rho) = tr_{B}(\rho)$. Then for a state $\rho\in L^{1}(H_{A} \otimes H_{B})$ one has $$ \begin{eqnarray*} tr(\Phi(\rho)) &= &tr(tr_{B}(\rho))\\ &= &tr(tr_{B}(\rho)\cdot I_{A})\\ &= &tr(\rho\cdot (I_{A} \otimes I_{B}))\\ &= &tr(\rho) = 1\\ \end{eqnarray*} $$ so property 1 holds. Towards property 2, let $n\in\mathbb{N}$ and let $r = \sum_{ij}r_{ij} \otimes E_{ij} \in L^{1}(H_{A} \times H_{B} \times \mathbb{C}^{n})$ with $r \geq 0$. Let $x = \sum_{i=1}^{n}x_{i} \otimes e_{i}$ be an arbitrary vector in $H_{A} \otimes \mathbb{C}^{n}$. Then $$ \begin{eqnarray*} \langle x | (\Phi \otimes id_{M_{n}})(r)x\rangle &= &\sum_{ij} \langle x_{i} | \Phi(r_{ij}) x_{j} \rangle\\ &= &\sum_{ij} tr\Big( \Phi(r_{ij}) | x_{j} \rangle \langle x_{i} | \Big)\\ &= &\sum_{ij} tr\Big( tr_{B}(r_{ij}) | x_{j} \rangle \langle x_{i} | \Big)\\ &= &\sum_{ij} tr\Big( r_{ij} \cdot \Big( I_{A} \otimes | x_{j} \rangle \langle x_{i} | \Big) \Big)\\ &= &\sum_{ij} tr\Big( r_{ij} \otimes E_{ij} \cdot \Big( I_{A} \otimes | x_{j} \rangle \langle x_{i} | \otimes | e_{j} \rangle \langle e_{i} | \Big) \Big)\\ &= &\sum_{ij} tr\Big( r_{ij} \otimes E_{ij} \cdot \Big( I_{A} \otimes | x \rangle \langle x | \Big) \Big)\\ &= &tr\Big( \sum_{ij} r_{ij} \otimes E_{ij} \cdot \Big( I_{A} \otimes | x \rangle \langle x | \Big) \Big)\\ &= &tr\Big( r \cdot \Big( I_{A} \otimes | x \rangle \langle x | \Big) \Big)\\ &\geq &0\\ \end{eqnarray*} $$ whereby the final inequality follows from the positivity of $r$. A cleaner way to prove property to is to show that $\Phi \otimes id_{M_{n}}$ is just again $tr_{B}$ within the system $H_{A} \otimes H_{B} \otimes \mathbb{C}^{n}$ and to not that partial traces always turn positive operators into positive operators.
• The system with which couples a system $H_{A}$ to another system $H_{B}$ with fixed state $\omega$, i.e. $\Phi \colon L^{1}(H_{A}) \to L^{1}(H_{A} \otimes H_{B})$ defined by $\Phi(\rho) = \rho\otimes\omega$. It is a simple exercise to prove 1 and 2.
• Let $A_{i} \in B(H_{A},H_{B})$ with $\sum_{i}A_{i}^{\ast}A_{i}=I_{A}$. Then $L^{1}(H_{A}) \ni \rho \mapsto \sum_{i}A_{i}\rho A_{i}^{\ast} \in L^{1}(H_{A} \otimes H_{B})$ is a quantum operation. In fact, by Kraus's theorem, all quantum operations have a representation of this form.
• The operation $M_{N}(\mathbb{C}) \ni r \mapsto r^{T} \in M_{N}(\mathbb{C})$ is trace preserving and maps positive operators to positive operators, but is fails to be completely positive and thus is not a Quantum operation. (And you should ask if it makes sense from a physical point of view, why the transpose operation should not count as a physical operation.)
• The operation $M_{N}(\mathbb{C}) \ni r \mapsto \frac{1}{N-1}\Big(tr(r) \cdot I - r\Big) \in M_{N}(\mathbb{C})$ is trace preserving and maps positive operators to positive operators, but is not completely positive, thus not a Quantum operation.

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^{[1]: Sometimes 1 is weakened to $\leq 1$ instead of $=1$. to allow Energy to go out of a system.}