Express any bipartite quantum state as evolution of the maximally entangled state

Quantum Computing is part of Stack Overflow’s open communities: specialist spaces where curiosity is welcome, knowledge is shared freely, and the best answers rise to the top. Turn scattered docs and code into trusted AI context. Explore the new Stack Internal platform. Assume we have two Hilbert Spaces HAHA\mathcal{H}_{A} and HBHB\mathcal{H}_{B} of dimensions dA≥dBdA≥dBd_{A}\ge d_{B}, equipped with respective orthonormal bases {|iA⟩}{|iA⟩}\{\lvert i_{A} \rangle\} and {|iB⟩}{|iB⟩}\{\lvert i_{B} \rangle\}. Consider a random bipartite state |ψ⟩∈HA⊗HB|ψ⟩∈HA⊗HB\lvert \psi \rangle\in \mathcal{H}_{A}\otimes\mathcal{H}_{B} whose general form is : |ψ⟩=∑i=1dA∑j=1dBcij|iA⟩|jB⟩|ψ⟩=∑i=1dA∑j=1dBcij|iA⟩|jB⟩\begin{equation} \label{eq:21} \lvert \psi \rangle = \sum_{i=1}^{d_{A}}\sum_{j=1}^{d_{B}}\,c_{ij}\,\lvert i_{A}\rangle\lvert j_{B} \rangle \end{equation} where C=[cij]C=[cij]C=[c_{ij}] is a matrix of shape dA×dBdA×dBd_{A}\times d_{B}. I want to demonstrate that one can find a local operation VVV so that the state |ψ⟩|ψ⟩\lvert \psi \rangle can be written as a transformation of the maximally bipartite entangled state |Ω⟩|Ω⟩\lvert \Omega \rangle on HB⊗HBHB⊗HB\mathcal{H}_{B}\otimes \mathcal{H}_{B} : |ψ⟩=(V⊗I)|Ω⟩,|Ω⟩=1d−−√∑i=1d|iB⟩|iB⟩|ψ⟩=(V⊗I)|Ω⟩,|Ω⟩=1d∑i=1d|iB⟩|iB⟩\begin{equation} \label{eq:22} \lvert \psi\rangle\, = \, (V\otimes I)\,\lvert \Omega \rangle\;,\quad \lvert \Omega \rangle = \frac{1}{\sqrt{d}} \sum_{i=1}^{d}\lvert i_{B}\rangle\lvert i_{B} \rangle \end{equation} with d=min{dA,dB}=dBd=min{dA,dB}=dBd = \min\{d_{A},d_{B}\} = d_{B}. Assume we have two unitaries UAUAU_{A} and UBUBU_{B} acting respectively in HAHA\mathcal{H}_{A} and HBHB\mathcal{H}_{B}, I will use in the sequel the property : UA⊗UB|ψ⟩=∑i=1dA∑j=1dB[UAC(UB)T]ij|iA⟩|jB⟩UA⊗UB|ψ⟩=∑i=1dA∑j=1dB[UAC(UB)T]ij|iA⟩|jB⟩\begin{equation} \label{eq:23} U_{A}\otimes U_{B}\,\lvert \psi \rangle = \sum_{i=1}^{d_{A}}\sum_{j=1}^{d_{B}}\, [U_{A}\, C\, (U_{B})^{T}]_{ij}\,\lvert i_{A}\rangle\lvert j_{B} \rangle \end{equation} Recall now the Singular Value Decomposition of CCC : C=UAΣU†B,Σ=diag(σ1,…,σd),d=min{dA,dB}C=UAΣUB†,Σ=diag(σ1,…,σd),d=min{dA,dB}\begin{equation} \label{eq:24} C = U_{A}\,\Sigma\,U_{B}^{\dagger}\;,\quad \Sigma = \,\text{diag}\,(\sigma_{1},\dots,\sigma_{d})\;,\quad d= \min\{d_{A}, d_{B}\} \end{equation} where : Note that the previous decomposition can be turned into : CUB=ΣUA⇔∀k∈{1,…,d},C|uBk⟩=σk|uAk⟩CUB=ΣUA⇔∀k∈{1,…,d},C|ukB⟩=σk|ukA⟩\begin{equation} \label{eq:26} C\, U_{B} = \Sigma\, U_{A}\quad\Leftrightarrow\quad \forall\,k\in\{1,\dots,d\}\;,\quad C\lvert u_{k}^{B} \rangle = \sigma_{k} \lvert u_{k}^{A} \rangle \end{equation} Then one can write : U†A⊗UTB|ψ⟩=∑i=1dA∑j=1dB[U†ACUB]ij|iA⟩|jB⟩⟩=∑i=1dA∑j=1dBΣij|iA⟩|jB⟩=∑i=1dσi|iA⟩|iB⟩UA†⊗UBT|ψ⟩=∑i=1dA∑j=1dB[UA†CUB]ij|iA⟩|jB⟩⟩=∑i=1dA∑j=1dBΣij|iA⟩|jB⟩=∑i=1dσi|iA⟩|iB⟩\begin{equation} \label{eq:27} U_{A}^{\dagger}\otimes U_{B}^{T}\,\lvert \psi \rangle = \sum_{i=1}^{d_{A}}\sum_{j=1}^{d_{B}}\, [U_{A}^{\dagger}\, C\, U_{B}]_{ij}\,\lvert i_{A}\rangle\lvert j_{B} \rangle \rangle = \sum_{i=1}^{d_{A}}\sum_{j=1}^{d_{B}}\, \Sigma_{ij}\,\lvert i_{A}\rangle\lvert j_{B} \rangle = \sum_{i=1}^{d}\sigma_{i}\lvert i_{A}\rangle\lvert i_{B} \rangle \end{equation} and in view that the inverse of U†A⊗UTBUA†⊗UBTU_{A}^{\dagger}\otimes U_{B}^{T} is UA⊗U⋆BUA⊗UB⋆U_{A}\otimes U_{B}^{\star}, we can continue into : |ψ⟩=UA⊗U⋆B(∑i=1dσi|iA⟩|iB⟩)=(I⊗U⋆B)(UA⊗I)(∑i=1dσi|iA⟩|iB⟩)=(I⊗U⋆B)(∑i=1dσi(UA|iA⟩)|iB⟩)=(I⊗U⋆B)(∑i=1dσi|uAi⟩|iB⟩)=(I⊗U⋆B)(∑i=1dC|uBi⟩|iB⟩)=(I⊗U⋆B)(C⊗I)(∑i=1d|uBi⟩|iB⟩)=(C⊗U⋆B)(∑i=1dUB|iB⟩|iB⟩)=(C⊗U⋆B)(UB⊗I)(∑i=1d|iB⟩|iB⟩)=(CUB⊗U⋆B)(∑i=1d|iB⟩|iB⟩)=d−−√(CUB⊗U⋆B)|Ω⟩|ψ⟩=UA⊗UB⋆(∑i=1dσi|iA⟩|iB⟩)=(I⊗UB⋆)(UA⊗I)(∑i=1dσi|iA⟩|iB⟩)=(I⊗UB⋆)(∑i=1dσi(UA|iA⟩)|iB⟩)=(I⊗UB⋆)(∑i=1dσi|uiA⟩|iB⟩)=(I⊗UB⋆)(∑i=1dC|uiB⟩|iB⟩)=(I⊗UB⋆)(C⊗I)(∑i=1d|uiB⟩|iB⟩)=(C⊗UB⋆)(∑i=1dUB|iB⟩|iB⟩)=(C⊗UB⋆)(UB⊗I)(∑i=1d|iB⟩|iB⟩)=(CUB⊗UB⋆)(∑i=1d|iB⟩|iB⟩)=d(CUB⊗UB⋆)|Ω⟩\begin{equation} \label{eq:28} \begin{split} \lvert \psi \rangle &= U_{A}\otimes U_{B}^{\star}\biggl(\sum_{i=1}^{d}\sigma_{i}\lvert i_{A}\rangle\lvert i_{B} \rangle \biggr) = (I\otimes U_{B}^{\star})\,(U_{A}\otimes I)\biggl(\sum_{i=1}^{d}\sigma_{i}\lvert i_{A}\rangle\lvert i_{B} \rangle \biggr) = (I\otimes U_{B}^{\star})\,\biggl(\sum_{i=1}^{d}\sigma_{i}(U_{A}\lvert i_{A}\rangle)\lvert i_{B} \rangle \biggr)\\ & = (I\otimes U_{B}^{\star})\,\biggl(\sum_{i=1}^{d}\sigma_{i}\lvert u_{i}^{A}\rangle\lvert i_{B} \rangle \biggr) = (I\otimes U_{B}^{\star})\,\biggl(\sum_{i=1}^{d}C \lvert u_{i}^{B}\rangle\lvert i_{B}\rangle\biggr) = (I\otimes U_{B}^{\star})\,(C\otimes I)\, \biggl(\sum_{i=1}^{d} \lvert u_{i}^{B}\rangle\lvert i_{B} \rangle \biggr) \\ & = (C\otimes U_{B}^{\star})\, \biggl(\sum_{i=1}^{d} U_{B}\lvert i_{B}\rangle \lvert i_{B} \rangle \biggr) = (C\otimes U_{B}^{\star})\,(U_{B}\otimes I)\,\biggl(\sum_{i=1}^{d} \lvert i_{B}\rangle\lvert i_{B} \rangle \biggr) = (CU_{B}\otimes U_{B}^{\star})\,\biggl(\sum_{i=1}^{d} \lvert i_{B}\rangle\lvert i_{B} \rangle \biggr)\\ &= \sqrt{d}\,(CU_{B}\otimes U_{B}^{\star})\,\lvert \Omega \rangle \end{split} \end{equation} By using the previous property one finally gets : |ψ⟩=d−−√(CUB⊗I)(I⊗U⋆B)|Ω⟩=d−−√(CUB⊗I)(U†B⊗I)|Ω⟩=d−−√(CUBU†B⊗I)|Ω⟩=d−−√(C⊗I)|Ω⟩|ψ⟩=d(CUB⊗I)(I⊗UB⋆)|Ω⟩=d(CUB⊗I)(UB†⊗I)|Ω⟩=d(CUBUB†⊗I)|Ω⟩=d(C⊗I)|Ω⟩\begin{equation} \label{eq:30} \begin{split} \lvert \psi \rangle &= \sqrt{d}\,(CU_{B}\otimes I)\,(I\otimes U_{B}^{\star})\,\lvert \Omega \rangle = \sqrt{d}\,(CU_{B}\otimes I)\,(U_{B}^{\dagger}\otimes I)\,\lvert \Omega \rangle \\ & =\sqrt{d}\,(CU_{B}U_{B}^{\dagger}\otimes I)\,\lvert \Omega \rangle = \sqrt{d}\,(C\otimes I)\,\lvert \Omega \rangle \end{split} \end{equation} Furthermore, I wonder about the meaning of this result : the local operation d−−√CdC\sqrt{d}\,C is not necessarily a unitary, which means that this transformation is not physical. Then, is this result in general a simple mathematical reformulation ? I haven't checked the minutiae of your derivation, but conceptually it is correct. You can certainly do something like that. In words: if the Schmidt coefficients are {λi}{λi}\{\lambda_i\}, you apply an operator C=∑iλi|i⟩⟨i|C=∑iλi|i⟩⟨i|C=\sum_i\lambda_i|i\rangle\langle i| to the maximally entangled state, and then apply the unitaries that convert to the Schmidt basis. It's not true that CCC is unphysical. It is a measurement (provided you complete it with other measurement operators). It just tells you that this particular conversion process is probabilistic. If you get the correct measurement outcome, the transformation has succeeded, otherwise you've got some junk. (In fact, there is a way to make it such that the other measurement outcomes can be transformed into the correct result. If interested, look up Nielsen's majorization criteria.) Thanks for contributing an answer to Quantum Computing Stack Exchange! 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