The Exercise

Alice and Bob share a pure state $|\psi\rangle$ of two systems, $A$ on her side and $B$ on his. They want to turn it into another pure state $|\varphi\rangle$ of the same two systems, each working only in their own lab and with no phone call in between: no measurement result and no classical bit goes from one side to the other. Exercise 12.22, on page 577, asks when this is possible. Let $\lambda_\psi$ be the list of squared Schmidt coefficients of $|\psi\rangle$ and $\lambda_\varphi$ that of $|\varphi\rangle$. The conversion is possible if and only if $\lambda_\psi \cong \lambda_\varphi \otimes x$ for some probability vector $x$, where $\cong$ means equal non-zero entries, zeros dropped and order ignored.

The library states this as a theorem about density operators rather than about lists of coefficients. These notes go through that theorem and the definitions it is built on.

A Quantum System

A system in the library is a finite-dimensional complex Hilbert space.^✦ A state $\rho$ of a system $S$ is a positive operator of unit trace on the Hilbert space of $S$.^✦ A pure state $\psi$ of $S$ is a unit vector of the Hilbert space of $S$, with density operator $|\psi\rangle\langle\psi|$.^✦ Two systems $A$ and $B$ combine into the system $A \otimes B$, whose Hilbert space is the tensor product, and a state $\rho$ of $A$ and a state $\sigma$ of $B$ combine into the product state $\rho \otimes \sigma$ of $A \otimes B$. A unitary $U$ of a system sends a state $\rho$ to $U\rho U^{\dagger}$; conjugation by a unitary preserves positivity and the trace, so the result is again a state of the same system.^✦

If $\rho$ is a state of $A \otimes B$, Alice sees only the reduced state, the partial trace over $B$. The reduced state $\operatorname{Tr}_{2}(\rho)$ of $\rho$ on $A$ is the partial trace of $\rho$ over the second factor, and is a state of $A$.^✦ For the shared pure state, Alice’s reduced state is $\operatorname{Tr}_{2}(|\psi\rangle\langle\psi|)$. Its non-zero eigenvalues are the squared Schmidt coefficients $\lambda_\psi$; the rest are zeros.

For two states $\rho$ and $\sigma$, on the same system or not, $\rho \cong \sigma$ means that their eigenvalue lists are $\cong$ in the exercise’s sense.^✦ The spectrum of $\sigma \otimes \tau$ consists of all products of an eigenvalue of $\sigma$ with one of $\tau$. The book’s $\lambda_\varphi \otimes x$ is therefore the spectrum of $\operatorname{Tr}_{2}(|\varphi\rangle\langle\varphi|) \otimes \tau$ for a state $\tau$ with spectrum $x$, and $x$ is a probability vector because $\tau$ is a state.

A Local Operation

The library takes the most general thing one party can do to a system $S$ to be a dilated channel $\mathcal{E}_{U,\omega}$, in the form of Chapter 8: attach an ancilla $C$ in a pure state $|\omega\rangle$, apply a unitary $U$ of $S \otimes C$ to system and ancilla together, and trace the ancilla out again as the second factor, so that a state $\sigma$ of $S$ goes to $$\mathcal{E}_{U,\omega}(\sigma) = \operatorname{Tr}_{2}\bigl(U\,(\sigma \otimes |\omega\rangle\langle\omega|)\,U^{\dagger}\bigr) .$$(1)^✦

Bob’s channel was already defined in Chapter 11: a dilated channel on the second factor of a state $\rho$ of $A \otimes B$, which we write $(\mathrm{id} \otimes \mathcal{E}_{U,\omega})(\rho)$. In full, with $\alpha_{A,B,C}$ the regrouping of $A \otimes (B \otimes C)$ as $(A \otimes B) \otimes C$, $$(\mathrm{id} \otimes \mathcal{E}_{U,\omega})(\rho) = \operatorname{Tr}_{2}\Bigl(\alpha_{A,B,C}\bigl((\mathrm{id} \otimes U)\,\alpha_{A,B,C}^{-1}(\rho \otimes |\omega\rangle\langle\omega|)\,(\mathrm{id} \otimes U)^{\dagger}\bigr)\Bigr) ,$$(2)^✦ In (2), $\mathrm{id}$ is the identity unitary of $A$, the one with operator $1$.^✦ The operator $\mathrm{id} \otimes U$ is the product unitary of the identity on $A$ with $U$ on $B \otimes C$: the tensor product of the two operators, acting on $A \otimes (B \otimes C)$, and again a unitary.^✦

Alice’s channel is defined from Bob’s by symmetry. Swap the two factors, run Bob’s construction, swap back: $$(\mathcal{E}_{U,\omega} \otimes \mathrm{id})(\rho) = \mathrm{swap}_{B,A}\bigl((\mathrm{id} \otimes \mathcal{E}_{U,\omega})(\mathrm{swap}_{A,B}(\rho))\bigr) ,$$(3)^✦

Both parties may act, Alice on her side and Bob on his. With Alice’s data $(U_A, \omega_A)$ on $A \otimes C_A$ and Bob’s $(U_B, \omega_B)$ on $B \otimes C_B$, Bob’s channel is applied first, then Alice’s: $$(\mathcal{E}_{U_A,\omega_A} \otimes \mathcal{E}_{U_B,\omega_B})(\rho) = (\mathcal{E}_{U_A,\omega_A} \otimes \mathrm{id})\bigl((\mathrm{id} \otimes \mathcal{E}_{U_B,\omega_B})(\rho)\bigr) .$$(4)^✦

In (4) the two unitaries are fixed in advance: that is the exercise’s “no communication”.

We write $\psi \rightsquigarrow \varphi$, $\psi$ converts to $\varphi$ without communication, when some choice of ancillas, unitaries and pure ancilla states sends $|\psi\rangle\langle\psi|$ to $|\varphi\rangle\langle\varphi|$, $$\psi \rightsquigarrow \varphi \iff \exists\, C_A, C_B, U_A, U_B, \omega_A, \omega_B:\ |\varphi\rangle\langle\varphi| = (\mathcal{E}_{U_A,\omega_A} \otimes \mathcal{E}_{U_B,\omega_B})(|\psi\rangle\langle\psi|) .$$(5)^✦ The conversion is deterministic: reaching the target only with some probability does not count.

The Claim

What Is Checked

The claim is stated for pure states $\psi$ and $\varphi$ of $A \otimes B$, over the conversion relation (5), the partial trace over Bob’s factor, the product of a reduced state with a state $\tau$ of some system $C$, and the relation $\cong$ of equal nonzero spectra.

The library proves that the reduced state, the product state and the regrouped state are states.