$\newcommand{\ket}[1]{\left|#1\right>}$ Let $\cal{H}$ be a Hilbert space with dimension $d$. Define an MES (maximally entangled state) $\ket{\Phi} \in \cal{H} \otimes \cal{H}$ to be one with a Schmidt decomposition of the form

$$ \ket{\Phi} = \sum_{i=1}^d \frac{1}{\sqrt{d}} \ket{\alpha_i} \ket{\beta_i}. $$

We also define the standard MES in a basis $(\ket{\phi_i})$ to be the MES

$$ \ket{\Omega_\phi} := \sum_{i=1}^d \frac{1}{\sqrt{d}} \ket{\phi_i} \ket{\phi_i}, $$

so it is an MES which has a Schmidt decomposition where the Schmidt bases are identical. Of course, there is freedom in choosing the Schmidt basis, so there are other decompositions where this is not the case; the important thing is that among those decompositions is the one of the required form.

Here is my question. If $\ket\Phi$ is any MES, then can we guarantee that in some basis $(\psi_i)$, it is the standard MES, $\ket\Phi = \ket{\Omega_\psi}$ -- does the freedom of choice of $\ket\Phi$'s Schmidt decomposition guarantee the existence of such a basis?

The motivation for the question is that I see conflicting definitions for MESs. I think that defining it as a state with Schmidt coefficients $(\frac{1}{\sqrt{d}}, \frac{1}{\sqrt{d}}, ..., \frac{1}{\sqrt{d}})$ is what makes the most sense, as such a state has maximal Schmidt rank and entanglement entropy; thus, it is maximally entangled. However, I sometimes see it defined as being one which, per my definitions above, is equal to the standard MES in some basis. It would be convenient for me conceptually if these two definitions coincided.

Here's how far I got. Work in some basis $(\ket{i})$ with standard MES $\ket\Omega$. We know there exists a unitary $U$ such that

$$ \ket\Phi = (I \otimes U) \ket\Omega. $$

Then we can compute a condition on the vectors $\ket{\psi_i}$ :

\begin{align} \ket\Phi &= \frac{1}{\sqrt{d}} \sum_i \ket{\psi_i}\ket{\psi_i} \\ \frac{1}{\sqrt{d}}\sum_{i, j, k} \psi_{ij} \ \psi_{ik} \ \ket{j}\otimes\ket{k} &= \frac{1}{\sqrt{d}} \sum_{j,k} \ket{j} \otimes U_{jk} \ket{k} \end{align}

and hence, if the components of the $(\ket{\psi_i})$'s are arranged into a matrix $M_{ij} = \psi_{ij}$,

$$ M^T M = U. $$

So the question reduces to: given a unitary $d \times d$ matrix $U$, does there exist a $d \times d$ complex matrix such that $M^T M = U$?


1 Answer 1


The answer is "no".

To this end, note that every $U=M^TM$ has the property that $$ U^T = (M^TM)^T = M^TM = U\ , $$ i.e. $U$ is its own transpose.

However, there are clearly $U$ which are not of this form, such as $$ U = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix} $$ (that is, a $\pi$ rotation about the $y$ axis).

(Note that this behavior cannot be changed by absorbing a global phase into $U$, since the phase remains the same under transposition.)

More generally, for qubits, we can parametrize any $\mathrm{U}(2)$ matrix (up to a global phase, which does not matter here) as \begin{align} U &= \exp(i\, \vec u \cdot \vec \sigma)\ ,\\ M &= \exp(i\, \vec m \cdot \vec \sigma)\ . \end{align} Then, $$ M^T = \exp(i\, \vec m \cdot \vec\sigma^T) = \exp(i\,\vec m'\cdot \vec \sigma)\ , $$ and since $\sigma_{x,z}^T=\sigma_{x,z}$ but $\sigma_{y}^T=-\sigma_y$, we have that $$ \vec m' = (m_x,-m_y,m_z)\ . $$ That is, $U$ can be written as $U=M^TM$ iff the rotation about $u$ can be decomposed into a rotation about $m$, followed by one about $m'$.

As we have seen above, this is not possible for e.g. rotations $\vec u=(0,u_y,0)$ about the $y$ axis.

On the other hand, it is also immediately clear that this is possible for rotations which do not involve the $y$ axis, $\vec u=(u_x,0,u_z)$. In particular, the state $\tfrac{1}{\sqrt{2}}(\vert0\rangle\vert+\rangle+\vert1\rangle\vert-\rangle)$ can be written in such a form.

  • $\begingroup$ Hi there. Could you please elaborate? Do you mean that in fact not all MESs can be expressed as standard (with repeated basis), and if so, could you provide a counter-example to $M^T M = U$? $\endgroup$
    – Movpasd
    Commented Jun 12, 2021 at 13:21
  • $\begingroup$ Yes, I understand this, as mentioned in my original post. My question is, is this freedom sufficient to ensure that every MES can be represented with a repeated basis? $\endgroup$
    – Movpasd
    Commented Jun 13, 2021 at 13:50
  • $\begingroup$ Very clean argument, thank you! $\endgroup$
    – Movpasd
    Commented Jun 13, 2021 at 16:43

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