For $\rho \in \mathrm{D}(\mathcal{X} \otimes \mathcal{Y})$ denoting an arbitrary state of the pair $(\mathrm{X}, \mathrm{Y})$, how to prove the fact
$\text{Im} (\rho) \subseteq \text{Im} (\rho[X] \otimes \rho[Y])$ ?

This is an equation (5.109) on page 270 in John Watrous's book named "THE THEORY OF QUANTUM INFORMATION"

  • $\begingroup$ This feels like notational overload. How are $\rho[X]$ and $\rho[Y]$ defined if each $\rho$ is defined over a pair of systems? $\endgroup$ Commented Apr 28 at 16:32
  • $\begingroup$ $\rho[X]=Tr_{Y}(\rho), \rho[Y]=Tr_{X}(\rho) $, are the partial traces $\endgroup$
    – Aimin Xu
    Commented Apr 28 at 21:44
  • $\begingroup$ I suggest starting to think of the special case of $\rho$ being a pure, entangled state, and think about it in the Schmidt basis. What do $\rho[X]$ and $\rho[Y]$ look like? Then think about the spectral decomposition of $\rho$. $\endgroup$
    – DaftWullie
    Commented Apr 29 at 6:43

1 Answer 1


Let's consider some examples first to see what we're talking about:

  • Start thinking about a simple case: what does this mean if $\sqrt2|\Psi\rangle=|00\rangle+|11\rangle\in\mathbb{C}^2\otimes\mathbb{C}^2$? Given $\rho\equiv |\Psi\rangle\!\langle \Psi|$, the partial states are $\rho_X = \rho_Y = \frac I2.$ Therefore $\operatorname{im}(\rho_X)=\operatorname{im}(\rho_Y) = \mathbb{C}^2$, and $\operatorname{im}(\rho_X\otimes\rho_Y)$ is the just the whole 4-dimensional space. Contrast this with $\operatorname{im}(\rho)$, which is the one-dimensional subspace spanned by $|\Psi\rangle$ itself. It's clear that the image of $\rho$ (as any other subspace) is contained in the image of $\rho_X\otimes \rho_Y$ (which is the full space).

  • If $\sqrt3|\Psi\rangle=|00\rangle+\sqrt2|11\rangle$ is a two-qubit non-fully-entangled state, then $$\rho_X=\rho_Y= \frac13 |0\rangle\!\langle0| + \frac23 |1\rangle\!\langle1|.$$ Again, $\rho_X,\rho_Y$ have full-rank images, and thus so is their tensor product, and $\operatorname{im}(\rho_X\otimes\rho_Y)$ is again the entire space.

  • What if you have instead something like $\rho=|0\rangle\!\langle0|\otimes I/2$? In this case $\rho_X=|0\rangle\!\langle 0|$, $\rho_Y=\frac I2$, and thus $\rho=\rho_X\otimes\rho_Y$, and the statement is trivial.

  • Let $2\rho=\mathbb{P}_{00}+\mathbb{P}_{11}$, $\mathbb{P}_{ij}\equiv |ij\rangle\!\langle ij|$. Now $\rho_X=\rho_Y = I/2$, and thus $\rho_X\otimes\rho_Y=I/4$ is the identity matrix, whose image is again the full space $\mathbb{C}^4$. Contrast this with $$\operatorname{im}(\rho) = \operatorname{span}(\{|00\rangle,|11\rangle\}),$$ and we again see that the statement is trivially true.

As for the general statement:

  • A bit more generally, given an arbitrary pure bipartite state $|\Psi\rangle=\sum_i \sqrt{p_i} |u_i,v_i\rangle\in\mathbb{C}^n\otimes\mathbb{C}^m$, for some pair of orthonormal bases $\mathcal U\equiv \{|u_i\rangle\}_i$ and $\mathcal V\equiv \{|v_i\rangle\}_i$, then the images of the reduced states are spanned by $\mathcal U$ and $\mathcal V$, and thus the image of $\rho_X\otimes\rho_Y$ by vectors of the form $|u_i,v_j\rangle$ for all $i,j$.

    In particular, that means that if $|\Psi\rangle$ has full Schmidt rank, then $\operatorname{im}(\rho_X\otimes\rho_Y)=\mathbb{C}^n\otimes\mathbb{C}^m$ is the full space, making the statement automatically true. And even if the image of $\rho_X\otimes \rho_Y$ doesn't span the full space, $|\Psi\rangle$ is always in it, because $|u_i,v_i\rangle\in\operatorname{im}(\rho_X\otimes\rho_Y)$ for all $i$. More formally, we showed that $|u_i\rangle\in\operatorname{im}(\rho_X)$, $|v_i\rangle\in\operatorname{im}(\rho_Y)$, thus $|u_i,v_i\rangle\in\operatorname{im}(\rho_X\otimes\rho_Y)$, and thus $$|\Psi\rangle \in \operatorname{im}(\operatorname{tr}_Y\mathbb{P}_{\Psi} \otimes\operatorname{tr}_X \mathbb{P}_{\Psi}).$$

  • Finally, let $\rho$ be a generic state. Its image is always the span of its eigenvectors, so you can reduce the question to: are all eigenvectors of $\rho$ in the image of $\rho_X\otimes\rho_Y$? To answer in the positive, we observe that if $\rho=\sum_i p_i \mathbb{P}_{\Psi_i}$ for some set of (generally entangled) orthonormal states $|\Psi_i\rangle$, then $$\rho_X = p_1 \operatorname{tr}_Y[\mathbb{P}_{\Psi_1}] + (\text{positive stuff}).$$ In words, we're saying that doing the partial trace of (projections onto) eigenstates only ever gives positive semidefinite operators. But also, for any pair of positive semidefinite Hermitians $P,Q\ge0$, we know that $P+Q\ge Q$, and that if $Q\le P$ then $\operatorname{im}(Q)\subseteq \operatorname{im}(P)$. It follows that $$|\Psi_1\rangle \in \operatorname{im}(\operatorname{tr}_Y\mathbb{P}_{\Psi_1} \otimes\operatorname{tr}_X \mathbb{P}_{\Psi_1}) \subseteq \operatorname{im}(\rho_X\otimes\rho_Y),$$ where I used $$\operatorname{tr}_Y\mathbb{P}_{\Psi_1} \le \rho_X, \qquad \operatorname{tr}_X\mathbb{P}_{\Psi_1} \le \rho_Y.$$ The same identical reasoning can be applied to show that all eigenvectors $|\Psi_i\rangle$ are in the image of $\rho_X\otimes\rho_Y$, thus the conclusion.


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