Question about the true significance of the partial trace

Consider a composite system whose Hilbert space is $\mathcal{H}_{AB}=\mathcal{H}_A\otimes \mathcal{H}_B$, where $\{|0_A\rangle, |1_A\rangle\}$ and $\{|0_B\rangle, |1_B\rangle\}$ are orthonormal bases of $\mathcal{H}_A$ and $\mathcal{H}_B$, respectively.

Suppose the composite system is in the pure state $|\psi\rangle = \frac{1}{\sqrt{2}}(|0_A1_B\rangle - |1_A0_B\rangle)$ whose density matrix is given by \begin{align} \rho = \frac{1}{2}(|0_A1_B\rangle \langle 0_A 1_B| + |1_A0_B\rangle \langle 1_A0_B| - |0_A1_B\rangle \langle 1_A0_B| - |1_A0_B\rangle \langle 0_A1_B|) \end{align}

The state of the $A$-subsystem is then given by the partial trace of $\rho$ with regard to $B$, i.e. \begin{align} \rho_A = tr_B(\rho) = \frac{1}{2}(|0_A\rangle \langle 0_A| + |1_A\rangle \langle 1_A|). \end{align}

$\rho_A$ is a complete description of the $A$-subsystem, and using it we can predict the outcomes of all measurements one can perform on $A$.

Now, taking the partial trace of $\rho$ with regard to $B$ seems to be doing two things: (1) it tells us that the 'off-diagonal terms' in $\rho$ (i.e. the last two terms in the above equation for $\rho$) are irrelevant for the description of the $A$-subsystem; (2) it gives us a way to talk about the $A$-subsystem without mentioning the $B$-basis vectors via the above expression for $\rho_A$.

However, it seems (to me, at any rate) that (1) is what's really important and significant. Once we know which terms in $\rho$ are required to describe the $A$-subsystem completely, it doesn't matter whether we are mentioning some basis vectors of the $B$-subspace. In other words: it seems as if \begin{align} \rho_A' = \frac{1}{2}(|0_A1_B\rangle \langle 0_A 1_B| + |1_A0_B\rangle \langle 1_A0_B|) \end{align} is just as good a description of the $A$-subsystem as $\rho_A$; the fact that we're using the language of $\mathcal{H}_{AB}$ is unimportant.

Question: Is this correct? If not, where am I going wrong?

PS: On this line of thought, the operator $\rho_A'$ would also be a complete description of the $B$-subsystem. But this is just a result of the symmetries of the simple example I chose: both partial traces of $\rho$ over $A$ and $B$ enforce the deletion of exactly the same off-diagonal terms. But this is not generally the case.

If you take two quantum systems with joint state space $\mathcal H_{AB}=\mathcal H_A \otimes \mathcal H_B$, then any pure or mixed state of the joint system can be described by a joint density matrix $\rho_{AB}$, and that joint density matrix can be used to obtain the expectation values of any observable $\hat Q$ via the full trace $$\langle Q\rangle = \mathrm{Tr}\mathopen{}\left(\hat Q \rho_{AB}\right)\mathclose{} .$$ If you just have access to measurements on subsystem A, though, then you are restricted to measuring observables of the form $\hat Q = \hat R \otimes \mathbb I$, and for those the expectation value simplifies substantially: taking bases $\{|n_A\rangle\}$ and $\{|n_B\rangle\}$ for $\mathcal H_A$ and $\mathcal H_B$, respectively, it reads \begin{align} \langle Q\rangle & = \mathrm{Tr}\mathopen{}\left(\hat Q \rho_{AB}\right)\mathclose{} \\ & = \mathrm{Tr}\mathopen{}\left(\hat R\otimes \mathbb I\ \rho_{AB}\right)\mathclose{} \\ & = \sum_{n_A,n_B} \langle n_A| \langle n_B| \left(\hat R\otimes \mathbb I\ \rho_{AB}\right) |n_A\rangle|n_B\rangle \\ & = \sum_{n_A} \langle n_A| \left(\hat R\ \sum_{n_B}\langle n_B|\mathbb I\rho_{AB}|n_B\rangle\right) |n_A\rangle \\ & = \mathrm{Tr}_A\mathopen{}\left(\hat R \ \mathrm{Tr}_B\mathopen{}\left(\rho_{AB}\right)\mathclose{}\right)\mathclose{} . \end{align} In other words, the partial trace $\rho_A=\mathrm{Tr}_B\mathopen{}\left(\rho_{AB}\right)\mathclose{}$ is the density matrix that accounts for all of the experimental observations done on subsystem A that do not involve subsystem B.
• Just to clarify: I was taking for granted everything you explain. I guess what I was trying to say is: whenever you are considering an operator of the form $Q= R_A\otimes \mathbb{1}_B$, then $Tr(Q\rho_{AB})= Tr(Q\rho'_A)$ since $Tr_B(\mathbb{1}_B\rho_{AB}) = Tr_B(\mathbb{1}_B\rho'_{A})$, where $\rho'_A$ is obtained from $\rho_{AB}$ by subtracting all summands of the form $|n_An_B\rangle \langle n'_An'_B|$ with $n_B\neq n'_B$. If this is incorrect, what would be a counterexample? – Al_Gebra Aug 8 '18 at 10:45
• Your proposed $\rho_A'$ is not an operator on the state space of A like $\rho_A:\mathcal H_A \to \mathcal H_A$. There really isn't anything else to add to that - if it's not restricted to the system's state space, then it's not a state of the system. If that core point isn't clear then there isn't much to say. – Emilio Pisanty Aug 8 '18 at 11:49
• If your reasons for it are good enough, and they provide you with strong enough grounds to break the huge degeneracy (i.e. choose one method out of an infinity), it can be justified to express the content of $\rho_A$ in terms of an operator on $\mathcal H_{AB}$; however, you cannot call that new operator a reduced density matrix. (On the other hand - are you sure you're not actually looking for a complete-dephasing quantum channel on the $|n_B\rangle$ basis? just saying...). – Emilio Pisanty Aug 8 '18 at 12:05
• The canonical resource is Nielsen and Chuang's Quantum Computation and Quantum Information; I don't have a copy with me so I don't know how accessibly it covers that particular topic. If it's too complex, a resource-recommendations question on this site would be appropriate; you'll get best results by asking for an introductory text to quantum channels, with the dephasing channel as an example. But the basics fit in a comment: it is the operation that takes a single-system density matrix $\rho$ and sets to zero all of its off-diagonal coherences in a given basis. – Emilio Pisanty Aug 8 '18 at 12:26