$\newcommand{\Ket}[1]{\left|#1\right>}$In Nielsen's seminal paper on entanglement transformations (https://arxiv.org/abs/quant-ph/9811053), he gives a converse proof for the entanglement transformation theorem presented (starting at eq.10).

In the protocol presented as part of the converse proof, he states, for example, that if $\Ket{\psi'} \sim \Ket{\psi''}$, then $\Ket{\psi''} = (U_A\otimes U_B) \Ket{\psi'}$ for some local unitary transformations $U_A$ and $U_B$. Here, $\Ket{x} \sim \Ket{y}$ iff $\rho_x$ and $\rho_y$ have the same spectrum of eigenvalues. My question is, how does one construct such local unitary transformations? For instance, what would the explicit form of these unitary transformations be for transforming the state in eq.10 to that of eq.12?

$$ \Ket{\psi'} = \sqrt{\alpha_+} \Ket{00} + \sqrt{\alpha_-} \Ket{11} \\ \Ket{\psi''} = \frac{1}{\sqrt{2}} \Big( \Ket{00} + \Ket{1} \big[ \cos(\gamma)\Ket{0} + \sin(\gamma)\Ket{1} \big] \Big ) $$

Note that $\alpha_{\pm} = (1\pm\cos(\gamma))/2$. Basically, I am trying to work through the protocol for the 2D case as presented in the converse proof, but am getting stuck at the explicit form of the unitary transformations. Any help would be much appreciated!

  • 2
    $\begingroup$ posts should contain single, laser-focused questions. The second part of this post seems like a different enough question that you might be better off asking it on a separate post. Regarding the first part, you might notice that the two states have the same Schimdt coefficients. You can choose the unitaries as those that implement the change of basis between the singular vectors of one state to those of the other $\endgroup$
    – glS
    Aug 1, 2021 at 16:29

2 Answers 2


As pointed out in the comments by @gIS, the form of the desired unitary transformation follows from the Schmidt decomposition of the two states. To see this, write down the Schmidt decomposition of $|\psi'\rangle$ and $|\psi''\rangle$

$$ \begin{align} |\psi'\rangle = \sum_i\lambda_i|i'_A\rangle|i'_B\rangle \\ |\psi''\rangle = \sum_i\lambda_i|i''_A\rangle|i''_B\rangle \end{align}\tag1 $$

where $|i'_A\rangle$ and $|i''_A\rangle$ are orthonormal bases for qubit $A$, $|i'_B\rangle$ and $|i''_B\rangle$ are orthonormal bases for qubit $B$ and we use the fact that $|\psi'\rangle \sim |\psi''\rangle$ to justify denoting their shared Schmidt coefficient as $\lambda_i$. The above equations make it clear that we can transform $|\psi'\rangle$ to $|\psi''\rangle$ by mapping $|i'_A\rangle \to |i''_A\rangle$ and $|i'_B\rangle \to |i''_B\rangle$. In other words, if

$$ \begin{align} U_A = \sum_i|i''_A\rangle\langle i'_A| \\ U_B = \sum_i|i''_B\rangle\langle i'_B| \end{align}\tag2 $$


$$ |\psi''\rangle = (U_A\otimes U_B)|\psi'\rangle.\tag3 $$

and we see that $(2)$ exhibits the desired unitary transformations.

In the specific case of

$$ |\psi'\rangle = \cos\frac{\gamma}{2}|00\rangle + \sin\frac{\gamma}{2}|11\rangle \\ |\psi''\rangle = \frac{1}{\sqrt{2}}\left[|00\rangle + |1\rangle(\cos\gamma|0\rangle + \sin\gamma|1\rangle)\right]\tag4 $$

we can immediately identify the bases $|i'_A\rangle$ and $|i'_B\rangle$ as the computational bases on $A$ and $B$ and the Schmidt coefficients as $\lambda_0=\cos\frac{\gamma}2$ and $\lambda_1=\sin\frac{\gamma}2$. In order to find $|i''_A\rangle$ and $|i''_B\rangle$ we solve the equation

$$ \frac{1}{\sqrt{2}} \begin{bmatrix}1\\0\\ \cos\gamma\\ \sin\gamma\end{bmatrix} = \cos\frac{\gamma}{2}\, \begin{bmatrix}x_{00}\\ x_{01}\\ x_{10}\\ x_{11}\end{bmatrix} + \sin\frac{\gamma}{2}\, \begin{bmatrix}y_{00}\\ y_{01}\\ y_{10}\\ y_{11}\end{bmatrix}\tag5 $$

under the constraints that the unknown states are orthonormal product states. Recalling that $\sin^2\alpha+\cos^2\alpha=1$, $\cos2\alpha=\cos^2\alpha-\sin^2\alpha$ and $\sin2\alpha = 2\sin\alpha\cos\alpha$, we have

$$ \frac{1}{\sqrt{2}} \begin{bmatrix}1\\0\\ \cos\gamma\\ \sin\gamma\end{bmatrix} = \frac{1}{\sqrt{2}}\cos\frac{\gamma}{2}\, \begin{bmatrix}\cos\frac{\gamma}{2}\\ \mp\sin\frac{\gamma}{2}\\ \cos\frac{\gamma}{2}\\ \sin\frac{\gamma}{2}\end{bmatrix} + \frac{1}{\sqrt{2}}\sin\frac{\gamma}{2}\, \begin{bmatrix}\sin\frac{\gamma}{2}\\ \pm\cos\frac{\gamma}{2} \\ -\sin\frac{\gamma}{2}\\ \cos\frac{\gamma}{2}\end{bmatrix}\tag6 $$

which becomes

$$ \frac{1}{\sqrt{2}} \begin{bmatrix}1\\0\\ \cos\gamma\\ \sin\gamma\end{bmatrix} = \frac{1}{\sqrt{2}}\cos\frac{\gamma}{2}\, \begin{bmatrix}\cos\frac{\gamma}{2}\\ \sin\frac{\gamma}{2}\\ \cos\frac{\gamma}{2}\\ \sin\frac{\gamma}{2}\end{bmatrix} + \frac{1}{\sqrt{2}}\sin\frac{\gamma}{2}\, \begin{bmatrix}\sin\frac{\gamma}{2}\\ -\cos\frac{\gamma}{2} \\ -\sin\frac{\gamma}{2}\\ \cos\frac{\gamma}{2}\end{bmatrix}\tag7 $$

when we account for the requirement that the two states on the right hand side be product states. Finally, we can rewrite the last equation as

$$ \frac{1}{\sqrt{2}} \begin{bmatrix}1\\0\\ \cos\gamma\\ \sin\gamma\end{bmatrix} = \frac{1}{\sqrt{2}}\cos\frac{\gamma}{2}\, \begin{bmatrix}1\\1\end{bmatrix} \otimes \begin{bmatrix}\cos\frac{\gamma}{2}\\ \sin\frac{\gamma}{2}\end{bmatrix} + \frac{1}{\sqrt{2}}\sin\frac{\gamma}{2}\, \begin{bmatrix}1\\-1\end{bmatrix} \otimes \begin{bmatrix}\sin\frac{\gamma}{2}\\ -\cos\frac{\gamma}{2}\end{bmatrix}.\tag8 $$

Thus, we find

$$ \begin{align} |0''_A\rangle &= |+\rangle = H|0\rangle \\ |1''_A\rangle &= |-\rangle = H|1\rangle \\ |0''_B\rangle &= R_y(\gamma)|0\rangle \\ |1''_B\rangle &= -R_y(\gamma) |1\rangle \end{align}\tag9 $$

which means that

$$ |\psi''\rangle = \left(H \otimes R_y(\gamma) Z\right)|\psi'\rangle.\tag{10} $$

  • $\begingroup$ $\newcommand{\Ket}[1]{\left|#1\right>}$Very elegant solution, thanks @Adam, @glS! I'm still not sure about the next step of the protocol: they implement a POVM ($\{M_1, M_2\}$), resulting in two possible states, $\Ket{\psi_1^{'''}}$ or $\Ket{\psi_2^{'''}}$. Thus Alice applies either $M_1$ or $M_2$ to the state $\Ket{\psi^{''}}$, and gets either measurement 1 or 2, which results in the aforementioned states. If it's state 2 that's obtained, Alice and Bob can use LOCC to transform to state 1, because $\Ket{\psi_1^{'''}} \sim \Ket{\psi_2^{'''}}$; if it's state 1, they do nothing. Is this correct? $\endgroup$ Aug 2, 2021 at 7:07
  • 1
    $\begingroup$ This sounds mostly right to me. Note that a POVM doesn't fix the post-measurement state, so I think you want $\{M_1, M_2\}$ to be general measurement operators (see 2.2.3 in Nielsen & Chuang). $\endgroup$ Aug 2, 2021 at 15:02

I'll give some more detail on how to do this in the general case.

Say you are given two pure states, $\newcommand{\ket}[1]{\lvert #1\rangle}\ket\psi$ and $\ket\phi$, which have the same amount of entanglement, meaning they have the same set of Schmidt coefficients. You are given these states in some fixed orthonormal basis for the underlying bipartite space, i.e. you know the coefficients $\psi_{ij}$ and $\phi_{ij}$ such that $$\ket\psi=\sum_{ij}\psi_{ij} \ket{i,j}, \qquad \ket\phi=\sum_{ij}\phi_{ij} \ket{i,j}.$$ Their having the same Schmidt coefficients means that the matrices $\tilde\psi,\tilde\phi$ have the same singular values. Here, $\tilde\psi,\tilde\phi$ are the matrices whose matrix elements are $\psi_{ij}$ and $\phi_{ij}$, respectively.

Computing the SVD of these matrices, you get expressions of the form $$\tilde\psi = \sum_k \sqrt{s_k}|\psi_k^L\rangle\!\langle\psi_k^R|, \qquad \tilde\phi = \sum_k \sqrt{s_k}|\phi_k^L\rangle\!\langle\phi_k^R|,$$ where $\sqrt{s_k}\ge0$ are the (shared) singular values (the square root is not necessary, you could define $s_k$ as directly the singular values of the matrices), and $\ket{\psi_k^{L(R)}}$ and $\ket{\phi_k^{L(R)}}$ are the left (right) singular vectors of $\tilde\psi$ and $\tilde\phi$, respectively.

In practice, singular values and singular vectors of a matrix $A$ are found computing the spectral decomposition of $A^\dagger A$ and $AA^\dagger$.

Going back to bra-ket notation, the above decompositions correspond to $$ \ket\psi=\sum_k \sqrt{s_k}\ket{\psi_k^L}\otimes\ket{\psi_k^R}, \qquad \ket\phi=\sum_k \sqrt{s_k}\ket{\phi_k^L}\otimes\ket{\phi_k^R}. $$ Once you have these decompositions, to find unitaries $U_A,U_B$ such that $(U_A\otimes U_B)\ket{\psi}=\ket\phi$ you only need to define them as those such that $$U_A = \sum_k |\phi_k^L\rangle\!\langle\psi_k^L|, \qquad U_B = \sum_k |\phi_k^R\rangle\!\langle\psi_k^R|.$$

  • $\begingroup$ $\newcommand{\Ket}[1]{\left|#1\right>}$ Thank you very much, I got it! Just to clarify: would it be correct to say that $\tilde{\psi}^{\dagger}\tilde{\psi} \Ket{\psi_k^R} = s_k\Ket{\psi_k^R}$, and $\tilde{\psi}\tilde{\psi}^{\dagger} \Ket{\psi_k^L} = s_k\Ket{\psi_k^L}$. That is, the L(R) singular vectors are eigenvectors of $\tilde{\psi}\tilde{\psi}^{\dagger} $ ($\tilde{\psi}^{\dagger} \tilde{\psi}$)? (Ref: math.stackexchange.com/questions/3982195/…) $\endgroup$ Aug 5, 2021 at 0:15
  • 1
    $\begingroup$ yes that's right $\endgroup$
    – glS
    Aug 5, 2021 at 9:36

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