# Prove that $\rho_{AB} \leq |B|(\rho_A\otimes I_B)$ for any bipartite state $\rho_{AB}$

I'm trying to prove the following statement but am lost on how to show it. For a quantum state $$\rho_{AB}$$ with marginal $$\rho_A$$, how can one show that

$$\rho_{AB} \leq|B|(\rho_A\otimes I_B)$$

where $$A\leq B$$ means that $$B-A$$ is positive semidefinite, $$|B|$$ is the dimension of the register $$B$$ and $$I_B$$ is the identity matrix.

• what does $|B|$ mean? Jan 3 at 10:31
• @narip, it's the dimension of the Hilbert space $B$. Have edited to clarify
– polp
Jan 3 at 10:45
• One thing to note is that it is sufficient to prove this for pure states. You can this apply the identity to the purification and trace out the purifiying system (noting this preserves the PSD inequality) to get the result for general mixed states, i.e., $\rho_{A'AB} \leq \rho_{A'A} \otimes I_B |B| \implies \rho_{AB} \leq \rho_{A} \otimes I_B |B|$ . Probably the Schmidt decomposition can then help. Jan 3 at 11:59

There's a very quick proof if you can use the properties of the Choi-Jamiołkowski isomorphism.

Define a map that acts on subsystem $$B$$ as $$\Lambda(\rho_B) = \mathrm{Tr}(\rho_B) |B| I_B - \rho_B.$$ The Choi operator of this map is $$J(\Lambda)_{BB'} = |B| I_{BB'} - |B| \Phi^+_{BB}$$, where $$\Phi^+$$ is the maximally entangled state. It follows that $$J(\Lambda)\geq 0$$, and so the map is completely positive.

Then $$[I \otimes \Lambda](\rho_{AB}) = |B| (\rho_A \otimes I_B) - \rho_{AB} \geq 0$$ for any subsystem $$A$$ due to the complete positivity of $$\Lambda$$.

(Case of pure states) Let $$\rho=|\psi\rangle\!\langle\psi|$$ be a pure bipartite state, suppose the underlying space is $$\mathbb C^n\otimes\mathbb C^m$$, and write as $$|\psi\rangle=\sum_{k=1}^r \sqrt{p_k}(|u_k\rangle\otimes|v_k\rangle)$$ the Schmidt decomposition of $$|\psi\rangle$$, with $$r\le \min(m,n)$$. Then $$\rho$$ has a single nonzero eigenvalue equal to $$1$$, while $$\rho_A=\sum_k p_k |u_k\rangle\!\langle u_k|$$ has eigenvalues $$(p_k)_k$$. Note that also $$\rho_A\otimes I_B$$ then has eigenvalues $$p_k$$, with a possible choice of eigenvectors being $$|u_k\rangle\otimes |v_j\rangle$$ for all $$j,k$$. We then have $$(\langle u_j|\otimes\langle v_k|)\Big[ |B| (\rho_A\otimes I_B) - \rho \Big] (| u_j\rangle\otimes | v_k\rangle) = (|B| -\delta_{jk})p_j,$$ and for all $$j,k,m,n$$, $$(\langle u_j|\otimes\langle v_k|)\Big[ |B| (\rho_A\otimes I_B) - \rho \Big] (| u_m\rangle\otimes | v_n\rangle) = |B| p_j \delta_{jm}\delta_{kn} - \sqrt{p_j p_m} \delta_{jk}\delta_{mn}.$$ Decomposing an arbitrary vector in this basis, $$x=c_{jk}(|u_j\rangle\otimes|v_k\rangle)$$, we then get $$x^\dagger P x\equiv x^\dagger\Big[ |B| (\rho_A\otimes I_B) - \rho \Big] x = |B| \sum_j p_j \sum_k |c_{jk}|^2 - \left| \sum_j \bar c_{jj} \sqrt{p_j} \right|^2.$$ To show that this quantity is always positive, we use the following identity: $$r \sum_{j=1}^r p_j |c_{jj}|^2 =\sum_{k=1}^{r}\left| \sum_{j=1}^r \omega_r^{(j-1)(k-1)} \bar c_{jj}\sqrt{p_j} \right|^2,$$ where $$\omega^{jk}\equiv e^{2\pi i jk/r}$$ are the elements of the QFT matrix (up to normalisation). Notice how the $$k=1$$ term in the RHS equals $$\left|\sum_j \bar c_{jj}\sqrt{p_j}\right|^2$$, and that $$r\le |B|$$, and thus $$x^\dagger P x \ge r\sum_{j=1}^r p_j |c_{jj}|^2 - \left|\sum_{j=1}^r \bar c_{jj} \sqrt{p_j}\right|^2 \\ = \sum_{k=2}^{r}\left| \sum_{j=1}^r \omega_r^{(j-1)(k-1)} c_{jj}\sqrt{p_j} \right|^2 \ge0.$$

(General case) For a generic $$\rho$$, consider its eigendecomposition $$\rho=\sum_j p_j \rho_j$$ with $$\rho_j$$ pure states and $$p_j\in[0,1], \sum_j p_j=1$$. Notice that $$\rho_A=\sum_j p_j (\rho_j)_A$$, and thus $$|B|(\rho_A\otimes I_B)-\rho =\sum_j p_j \left[|B|( (\rho_j)_A\otimes I_B) - \rho_j\right]\ge0,$$ which is positive semidefinite because a convex combination (more generally, a linear combination with nonnegative coefficients) of positive semidefinite operators remains positive semidefinite.

• I don't see why this is a proof (I'm probably missing something though). $M$ is PSD if $x^* M x \geq 0$ for all vectors $x$ but here you have shown this only for a family of vectors. In particular you proof seems to indicat that $x^* M x \geq 0$ whenever $|B| \geq 1$ but the result would fail if we replace $|B|$ with $1$ for example. Jan 3 at 16:52
• @Rammus indeed, I just kinda assumed it to work for any $x$, but that turned out to be much more involved than I expected it to be. Hopefully the current argument works: I'm showing $x^\dagger Mx\ge0$ for all $x$ in the support of $M$
– glS
Jan 3 at 22:38