I'm trying to work through a self-made exercise, which may be ill formed as a question. Any general advice in dealing with these types of problems is also much appreciated!

I'm looking at a quantum gate $U_f$ for a function $f$, that has the effect $$\sum_x \alpha_x \vert x\rangle\vert 0\rangle \mapsto \sum_x \alpha_x\vert x\rangle\vert f(x)\rangle. $$ This will in most cases be an entangled state: for instance, if $f(x) = x$, then I get what looks like a Bell state.

I want to consider a case where the first register is already maximally entangled with a third party, Eve.

  • One way to proceed is to write the first register as a mixed state which I obtain after tracing out Eve's part. The trouble now is, when we consider the action of the gate, that the gate entangles the two registers. I have no idea how to sort out the entanglement between Eve and the first register and the new entanglement between the first and second registers.

  • Alternatively, if I don't trace out Eve's register and instead implement the gate $\mathbb 1\otimes U_f$, then I'm still not sure what the outcome is. Before the gate, I have $$\sum_x \vert x \rangle_E\vert x\rangle\vert 0\rangle. $$ (I have marked Eve's register for clarity.) After the gate, I could naively write $$\sum_x \vert x\rangle_E\vert x\rangle\vert f(x)\rangle, $$ but this looks dubious to me. Particularly, this looks like Eve is now entangled with the second register but that seems wrong.

I'm not sure how entanglement monogamy fits in but I suspect my guess for the state isn't compatible with it. Can anyone clarify what's going on for me?

  • $\begingroup$ As a side note, $f(x) = x$ is not implementable in this case as this would contradict the non-cloning theorem. $\endgroup$ Nov 27, 2018 at 20:47
  • 4
    $\begingroup$ @Nelimee that is meant to only be the definition for the basis states. Well half of them at least. Not a cloning. $\endgroup$
    – AHusain
    Nov 27, 2018 at 20:59
  • $\begingroup$ @user1936752: I hope you don't mind that I've edited your question somewhat to make it a little more concise. $\endgroup$ Nov 27, 2018 at 22:19
  • $\begingroup$ Thank you - your edits have made it a much clearer question! $\endgroup$ Nov 28, 2018 at 1:10

1 Answer 1


In fact, what you suggest is absolutely right. If you start in the state $$ \sum_x|x\rangle_E|x\rangle|0\rangle, $$ then you will get the state $$ \sum_x|x\rangle_E|x\rangle|f(x)\rangle. $$

Monogamy of entanglement is a slightly different issue that's not really a concern here. What is says (loosely), is that if Alice and Eve share a maximally entangled state, then Alice and Bob cannot share any entanglement at all. But then, there's a tradeoff. As you reduce the amount of entanglement between Alice and Eve, you can correspondingly increase the amount of entanglement between Alice and Bob. The state that you've produced is entangled (a particular type of multipartite entanglement called GHZ), and there's entanglement between all 3 parties. But it sits within the bounds of what's allowed.

To be explicit, there are several different types of entanglement measure to which you can apply monogamy arguments. Probably the most common is the tangle, but I'm going to compute a different one (because it's something I did in one of my papers, which means I remember how to do it). Let's assume we have the state $$ |\Psi\rangle=\frac{1}{\sqrt{n}}\sum_{x=0}^{n-1}|x\rangle_E|x\rangle_A|x\rangle_B. $$ We're going to calculate something called the singlet fraction between Alice and Eve, $p_{AE}$. By symmetry (exchange the labels of E and B), Alice and Bob will share the same singlet fraction $p_{BE}=p_{AE}$.

The monogamy relation that they're supposed to satisfy (according to the paper) is $$ p_{AE}+p_{AB}\leq\frac{n-1}{n}+\frac{1}{n+1}\left(\sqrt{p_{AE}}+\sqrt{p_{AB}}\right)^2, $$ so for $p_{AE}=p_{AB}$, we're expecting to find that $$ p\leq\frac{n+1}{2n}. $$ So long as it does that, there's no problem.

What is the single monogamy? $$ p_{AE}=\max_{U,V}\text{Tr}(|B\rangle\langle B|_{AE}\otimes \mathbb{I}_BV_E\otimes U_A\otimes \mathbb{I}_B|\Psi\rangle\langle\Psi|V_E^\dagger\otimes U_A^\dagger\otimes\mathbb{I}_B). $$ where $|B\rangle=\frac{1}{\sqrt{n}}\sum_x|x\rangle|x\rangle$ is the Bell state. You can certainly verify that if $U=V=\mathbb{I}$, then $p_{AE}=1/n$, and you can prove that this is optimal with just a little more work. Since $$ \frac{1}{n}\leq\frac{n+1}{2n}, $$ we're happy. Indeed, this particular state is nowhere near saturating this monogamy relation for $n\geq 2$.


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