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Let $\newcommand{\ket}[1]{\lvert#1\rangle}\{\ket{u_k}\}_k,\{\ket{v_k}\}_k\subset\mathcal H$ be orthonormal bases in an $N$-dimensional space. It then follows that the state $$\ket\Psi = C\sum_{k=1}^N \ket{u_k}\otimes\ket{v_k}\tag1$$ is maximally entangled (where $C$ is a normalisation constant). Or more generally, it means that $|\Psi\rangle$ has rank $N$ (which if the embedding space is larger does not correspond to a maximally entangled state).

Does the opposite direction hold? In other words, if we know that a state $\ket\Psi$ is maximally entangled and can be written as (1), can we conclude that $\langle u_j|u_k\rangle=\langle v_j|v_k\rangle=\delta_{jk}$? Equivalently, suppose $\ket\Psi$ has the form in (1) with $\{\ket{u_k}\}_k,\{\ket{v_k}\}_k$ not orthogonal. Can $\ket\Psi$ then be maximally entangled?

More generally, suppose $\dim\mathcal H=M$ with $M>N$ (the vectors are not a basis). If $|\Psi\rangle$ is as in (1), can it have rank $N$ even if $\{\ket{u_k}\}_k,\{\ket{v_k}\}_k$ are not orthogonal?

For example, in the simplest case with $M=2$ and $N>M$, the question is whether a state of the form $$\frac{1}{\sqrt{2(1+\Re[\langle u_1|u_2\rangle\langle v_1|v_2\rangle])}} \left(\ket{u_1}\otimes\ket{v_1}+\ket{u_2}\otimes\ket{v_2}\right)$$ can be maximally entangled (or more precisely, have Schmidt coefficients $(1/\sqrt2,1/\sqrt2,0,...,0)$) even if $\langle u_1|u_2\rangle,\langle v_1|v_2\rangle\neq0$.


I'll show here that it is crucial that both bases are non-orthogonal for this to be possible.

If only one of the two sets, say $\{\ket{v_k}\}_k$, is orthonormal, then the matrix of coefficients of $\ket\Psi$, write it with $\Psi$, has the form $\Psi = U \sqrt D V^T$, where $U,V$ are the matrices whose columns equal $\ket{u_k}$ and $\ket{v_k}$, respectively, and $D$ is diagonal. The orthonormality of $\{\ket{v_k}\}_k$ (and thus of $\{\ket{\bar v_k}\}_k$) then implies that $$\Psi\Psi^\dagger = UDU^\dagger,$$ which tells us that the Schmidt coefficients of $\ket\Psi$ majorize the diagonal of $\sqrt{D}$. E.g. if $D$ is a multiple of the identity then $\Psi\Psi^\dagger \simeq UU^\dagger\neq I$, and thus $\ket\Psi$ is not maximally entangled.

This would seem to suggest that, if at least one of the bases is orthonormal, then indeed $\ket\Psi$ is maximally entangled only if the other basis is also orthonormal. But this still leaves open the possibility of it being possible when both bases are not orthonormal.


Here is an example of a pair of non-orthogonal states $|\psi\rangle$ and $|\phi\rangle$ such that $|\psi\psi\rangle+|\phi\phi\rangle$ has rank 2. Define \begin{align} 2\sqrt2 \ket\psi &= \ket1 + (2+i)\ket2 - \ket3 + i \ket4, \\ 2\sqrt2 \ket\phi &= \ket1 + i\ket2 + (1-2i)\ket3 - i \ket4. \end{align} Let $\ket\Psi\equiv (\ket{\psi\psi}+\ket{\phi\phi})/\sqrt{3/2}$. You can then verify that the corresponding matrix of coefficients is $$C = \frac{1}{2\sqrt{6}}\begin{pmatrix} 1 & 1+i & -i & 0 \\ 1+i & 1+2 i & 0 & i \\ -i & 0 & -1-2 i & -1-i \\ 0 & i & -1-i & -1 \end{pmatrix}.$$ As can be readily checked, the only non-vanishing eigenvalue of $C^\dagger C$ is a two-fold degenerate $+1/2$, hence $\ket\Psi$ has rank $2$.

It's possible that this is only possible because the states live in a larger space, i.e. $M=4$ but $N=2$. I'm wondering if there is a good way to understand why this can happen, and if it possible also when $N=M$.

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To my surprise, your statement is not true. Consider the qubit example: $$ |\Psi\rangle=|00\rangle-|++\rangle. $$ This is clearly of the correct form but with non-orthogonal states (in case you're worrying about the negative sign, we put $|u_2\rangle=|+\rangle$ and $|v_2\rangle=-|+\rangle$). However, it is maximally entangled. To see this, we expand it out: $$ |\Psi\rangle=\frac12\left(|00\rangle-|01\rangle-|10\rangle-|11\rangle\right)=\frac{1}{\sqrt{2}}(|0-\rangle+|1+\rangle), $$ which is equivalent under local unitaries (Hadamard on the second qubit) to the Bell state, and hence maximally entangled.

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    $\begingroup$ ah! $|00\rangle+|++\rangle$ was actually the first state I tried this on, and it didn't work. I didn't think a simple sign change would make a difference in this context. Interesting. $\endgroup$ – glS Oct 2 at 10:07
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Some thoughts; I am not $100\%$ sure yet and will revisit:

A bi-partite (pure) state $|\Psi\rangle \in \mathcal{H}_{A} \otimes \mathcal{H}_{B}$ is maximally entangled if the reduced state in either $\mathcal{H}_{A}$ or $\mathcal{H}_{B}$ is the maximally mixed state.

Suppose we have a state \begin{equation} |\hat{\Psi}\rangle = C \sum_{k=1}^{N}|u_{k}\rangle \otimes |v_{k}\rangle, \end{equation} where $\{|u_{k}\rangle\}$ and $\{|v_{k}\rangle\}$ may or may not be orthonormal. Lets assume that they are at least normalized.

Furthermore, let $\{|a_{j}\rangle\}$ be an orthonormal basis for $\mathcal{H}_{A}$. The reduced state $\rho_{B} = \mathrm{tr}_{A}\big[|\hat{\Psi}\rangle\langle \hat{\Psi}|\big]$ on $\mathcal{H}_{B}$ is: \begin{equation} \begin{split} \rho_{B} &= \sum_{j} \big(\langle{a_{j}}|\otimes I \big) |\hat{\Psi}\rangle\langle \hat{\Psi}| \big(|{a_{j}}\rangle\otimes I \big) \\ &= |C|^{2}\sum_{j}\sum_{k}\sum_{k'} \langle a_{j}|u_{k}\rangle \langle u_{k'}|a_{j}\rangle \otimes |v_{k}\rangle\langle v_{k'}| \\ &= |C|^{2}\sum_{k}\sum_{k'} \langle u_{k'}| \Big( \sum_{j}| a_{j}\rangle\langle a_{j} |\Big) |u_{k}\rangle |v_{k}\rangle\langle v_{k'}| \\ &= |C|^{2}\sum_{k}\sum_{k'} \langle u_{k'}| I |u_{k}\rangle |v_{k}\rangle\langle v_{k'}|. \\ \end{split} \end{equation} If we put the $\{|u_{k}\rangle\}$ and $\{|v_{k}\rangle\}$ in matrices $U$ and $V$: \begin{equation} \begin{split} U &= \begin{bmatrix}|u_{1}\rangle & |u_{2}\rangle \cdots |u_{n}\rangle\end{bmatrix} \\ V &= \begin{bmatrix}|v_{1}\rangle & |v_{2}\rangle \cdots |v_{n}\rangle\end{bmatrix} \end{split} \end{equation} we get for: \begin{equation} U^{\dagger}U = \begin{bmatrix} \langle u_{1}|u_{1}\rangle & \cdots & \langle u_{1}|u_{n}\rangle \\ . & \ddots & . \\ \langle u_{n}|u_{1}\rangle & \cdots & \langle u_{n}|u_{n}\rangle \\\end{bmatrix} = I + \hat{D}_{U} \end{equation} for some off-diagonal (Hermitian, likely sparse) matrix $\hat{D}_{U}$. We can do the same for $V$, getting $\hat{D}_{V}$

That allows us to write for $\rho_{B}$:

\begin{split} \rho_{B} &= |C|^{2}\sum_{k}\sum_{k'} \langle u_{k'} |u_{k}\rangle |v_{k}\rangle\langle v_{k'}|\\ &= |C|^{2}\sum_{k}\sum_{k'} |v_{k}\rangle \langle u_{k'} |u_{k}\rangle \langle v_{k'}|\\ &= |C|^{2}VU^{\dagger}UV^{\dagger} \\ \end{split}

The question is now when this is or is not equal to the maximally mixed state i.e. $|C|I$ (note that the normalisation factor is different because we're in a smaller space!). We know, since $\{|u_{k}\rangle\}$ and $\{|v_{k}\rangle\}$ are in fact bases, both $U$ and $V$ are invertible, such that $\rho_{B} = |C|^{2}VU^{\dagger}UV^{\dagger}= |C|I$ implies that: \begin{equation} U^{\dagger} U = V^{-1} (V^{\dagger})^{-1} = (V^{\dagger}V)^{-1} \end{equation} up to some normalisation factor. Lets investigate the three (without loss of generality) options:

  • The bases $\{|u_{k}\rangle\}$ and $\{|v_{k}\rangle\}$ are orthonormal. Then, $U^{\dagger}U = V^{\dagger}V = I$ and the state is maximally entangled, as expected.

  • $\{|u_{k}\rangle\}$ is, but $\{|v_{k}\rangle\}$ isn't orthonormal. Then, $U^{\dagger}U = I \not = V^{\dagger}V$. This shows that if only one of the bases is orthonormal, the state cannot be maximally entangled.

  • Both $\{|u_{k}\rangle\}$ and $\{|v_{k}\rangle\}$ aren't orthonormal. Then, $U^{\dagger}U,V^{\dagger}V \not = I $. The state might still be maximally entangled, as long as $U^{\dagger}U$ is the inverse of $V^{\dagger}V$, or in other words: $|C|U^{\dagger}UV^{\dagger}V = I$. We can say a little more: \begin{equation} \begin{split} I &= |C|U^{\dagger}UV^{\dagger}V\\ &= |C|(I + \hat{D}_{U})(I + \hat{D}_{V}) \\ &= |C|I + |C|\hat{D}_{U} + |C|\hat{D}_{V} + |C|\hat{D}_{U}\hat{D}_{V} \\ \end{split} \end{equation} Both $\hat{D}_{U}$ and $\hat{D}_{V}$ are purely anti-diagonal, which means that their product $\hat{D}_{U}\hat{D}_{V}$ has diagonal elements that are equal to $1-|C|$. The off-diagonal terms need to cancel out; in higher dimensions this could get messy.

Some extra thoughts about the bi-partite case

For a bi-partite state (and thus $\hat{D}_{U}\hat{D}_{V}$ purely diagonal) we get that $\hat{D}_{U} = -\hat{D}_{V}$, which means that

$$\langle u_{1}|u_{2}\rangle = - \langle v_{1}|v_{2}\rangle.$$

Moreover, for the bi-partite state we also get $\frac{1}{2}I = I + \hat{D}_{U}\hat{D}_{V}$ so that

$\hat{D}_{U}\hat{D}_{V} = -\frac{1}{2}I$ which leads to $$\langle u_{1}|u_{2}\rangle \langle v_{2}|v_{1}\rangle = \langle u_{2}|u_{1}\rangle \langle v_{1}|v_{2}\rangle = -\frac{1}{2}.$$

You can combine this to say that $\langle v_{1}|v_{2}\rangle\langle v_{2}|v_{1}\rangle = \langle u_{1}|u_{2}\rangle\langle u_{2}|u_{1}\rangle = \frac{1}{2}$. This puts quite strong constraints on the bases. Essentially, the bases provided by Daftwullies answer seem to be (up to permutations and some extra phases) the only ones.

An example after Daftwullies answer

The relation $U^{\dagger}UV^{\dagger}V = I$ can very well be met even if $U$ and $V$ are not unitary; we have (for the bases Daftwullie came up with):

\begin{equation} \begin{split} U &= \begin{bmatrix} \sqrt{2} & 1 \\ 0 & 1 \end{bmatrix} \\ V &= \begin{bmatrix} \sqrt{2} & -1 \\ 0 & -1 \end{bmatrix} \\ \end{split} \end{equation}

which can be readily checked to obey $U^{\dagger}U V^{\dagger}V = I$:

\begin{equation} \begin{split} U^{\dagger}UV^{\dagger}V &= \begin{bmatrix} \sqrt{2} & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \sqrt{2} & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \sqrt{2} & 0 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} \sqrt{2} & -1 \\ 0 & -1 \end{bmatrix} \\ &= \begin{bmatrix} 2 & \sqrt{2} \\ \sqrt{2} & 2 \end{bmatrix} \begin{bmatrix} 2 & -\sqrt{2} \\ -\sqrt{2} & 2 \end{bmatrix} \\ &\sim I \end{split} \end{equation}

The states also obey the above equations for the inner products in the bi-partite case.

Old incorrect analysis

Now I first wrote the text below, but this is actually incorrect. After the answer of DaftWullie it became clear that there actually are examples of this. My error in the below derivation is that the Frobenius inner product $2\langle VV^{\dagger}, V\hat{D}_{U}V^{\dagger}\rangle$ does not bear a square, and can therefore be negative. That means that the inequality actually does not hold.

I don't actually think that this is possible under the assumption that all states need to be normalized. (which means, if you recall, that $U^{\dagger}U = I + \hat{D}_{U}$ with that identity being, well, the identity).

We already had $U^{\dagger}U = I + \hat{D}_{U}$, so that (using the Frobenius norm): $$||U^{\dagger}U||^{2} = ||I + \hat{D}_{U}||^{2} =||I||^{2} + ||\hat{D}_{U}||^{2} + 2\langle I,\hat{D}_{U}\rangle_{F} = ||I||^{2} + ||\hat{D}_{U}||^{2} \geq ||I||^{2},$$ where $\langle I, \hat{D}_{U}\rangle_{F}$ is the Frobenius inner product, which is zero here because $\hat{D}_{U}$ is by definition traceless. Moreover, the inequality becomes strict when $\hat{D}_{U}$ is non-empty and therefore $||\hat{D}_{U}|| > 0$ (the Frobenius norm is a norm). This is only true if the basis encoded in $U$ is non-orthogonal.

For our relation $\rho_{B} = I = VU^{\dagger}UV^{\dagger} = V(I + > \hat{D}_{U})V^{\dagger} = VV^{\dagger} + V\hat{D}_{U}V^{\dagger}$, we get:

\begin{equation} \begin{split} ||I||^{2} &= ||VU^{\dagger}UV^{\dagger}||^{2} = ||VV^{\dagger} + V\hat{D}_{U}V^{\dagger}||^{2} \\ &= ||VV^{\dagger}||^{2} + ||V\hat{D}_{U}V^{\dagger}||^{2} + 2\langle VV^{\dagger},V\hat{D}_{U}V^{\dagger}\rangle \\ &\geq ||VV^{\dagger}||^{2} + ||V\hat{D}_{U}V^{\dagger}||^{2} \\ & = ||I ||^{2} + || \hat{D}_{V}||^{2} + ||V\hat{D}_{U}V^{\dagger}||^{2} \\ & \geq ||I ||^{2} + ||V\hat{D}_{U}V^{\dagger}||^{2} \\ \end{split} \end{equation}

with the second bound only an equality if $||\hat{D}_{V}|| = 0$, which is not the case if the basis encoded in $V$ is not orthogonal. So in that case we have $||I||^{2} \geq ||I||^{2} + || \hat{D}_{V}||^{2} > + ||V\hat{D}_{U}V^{\dagger}||^{2} > ||I||^{2}$, which is a contradiction. We have to conclude that $\hat{D}_{V}$, and conversely $\hat{D}_{U}$ are zero, which means that the bases must be orthogonal.

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  • $\begingroup$ The completeness relation on the $\{|u_j\rangle\}$ only holds if it's an orthonormal basis, which we are explicitly excluding here, surely? $\endgroup$ – DaftWullie Sep 30 at 9:15
  • $\begingroup$ @DaftWullie I think you 're right. I've closed my answer for now. $\endgroup$ – JSdJ Sep 30 at 10:17
  • $\begingroup$ @DaftWullie I changed the basis over which the trace is performed to some orthonormal basis $\{|a_{j}\rangle\}$ so that the completeness relation holds. $\endgroup$ – JSdJ Sep 30 at 10:28
  • $\begingroup$ the trace in the middle of the expression is acting on a scalar, so it seems quite superfluous. That aside, if I understand you, you are showing that the partial trace does not equal $\sum_k |v_k\rangle\!\langle v_k|$. But what is this telling you? The $|v_k\rangle$ are not necessarily orthogonal either, so you are not showing that the partial trace is not (proportional to) the identity $\endgroup$ – glS Sep 30 at 14:45
  • $\begingroup$ The trace is more or less unnecessary, but it does add a little more clout to re-arranging the inner products so that it 'becomes' and operator. Also, I was working under the assumption that $\{|v_{k}\rangle\}$ actually is an orthonormal basis. Then this shows that the partial trace becomes something which is not (proportional to) the identity, and therefore that the bi-partitie state can not be a maximally entangled state. $\endgroup$ – JSdJ Sep 30 at 15:31

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