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Let $f(x,y)$ be a random 2n- to 1-bit function.

Consider the quantum circuit $|x,y,z\rangle \to |x,y,z\oplus f(x,y)\rangle$.

Is the new state entangled in general?

Is it entangled if $x,y$ are $H^{\bigotimes n}|0\rangle$ ?

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  • $\begingroup$ You still haven't specified what $z$ is. $\endgroup$ Apr 30 '19 at 8:19
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If applied on $H^{\otimes n}|0\rangle$ then the result will be entangled if $f$ is not constant also in the simpler situation.

Suppose $x,y$ are $n$-bit strings, $\forall x,y:~ U|x,y\rangle = |x, y \oplus f(x) \rangle$ and $|\phi\rangle = U \cdot H^{\otimes n}|0\rangle\otimes|y\rangle$ for some bit string $y$. Let's calculate $|\phi\rangle\langle\phi|$. $$ |\phi\rangle\langle\phi| = U \cdot \frac{1}{2^{n-1}}\sum_{i=0}^{2^n-1}|i\rangle\otimes|y\rangle \cdot \frac{1}{2^{n-1}}\sum_{j=0}^{2^n-1}\langle j|\otimes\langle y| \cdot U^\dagger = $$ $$ = \frac{1}{2^{n}} \sum_{i=0}^{2^n-1}|i\rangle\otimes|y\oplus f(i)\rangle \cdot \sum_{j=0}^{2^n-1}\langle j|\otimes\langle y \oplus f(j)| = $$ $$ = \frac{1}{2^{n}} \sum_{i=0}^{2^n-1}\sum_{j=0}^{2^n-1}|i\rangle\langle j|\otimes|y\oplus f(i)\rangle\langle y \oplus f(j)| $$ Now we can calculate partial trace of $|\phi\rangle\langle\phi|$ over the first subsystem $$ \text{tr}_1\left(|\phi\rangle\langle\phi|\right) = \text{tr}_1\left(\frac{1}{2^{n}} \sum_{i=0}^{2^n-1}\sum_{j=0}^{2^n-1}|i\rangle\langle j|\otimes|y\oplus f(i)\rangle\langle y \oplus f(j)|\right) = $$ $$ =\frac{1}{2^{n}} \sum_{i=0}^{2^n-1}|y\oplus f(i)\rangle\langle y \oplus f(i)| $$ The result is pure state if and only if $f$ is constant. Hence $|\phi\rangle$ is entagled iff $f$ is not constant.

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  • $\begingroup$ But f is a function of two parameters, not one $\endgroup$ May 2 '19 at 4:55
  • $\begingroup$ You can treat two $n$-bit parameters as one $2n$-bit parameter. $\endgroup$
    – Danylo Y
    May 2 '19 at 7:00
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It depends on the input state.

If f(x, y) is zero for all x, y in the superposition, then the output will equal the input.

This operation is its own inverse, so if you apply it to an unentangled state u and get an entangled state v, that means when you get v as an input state the operation actually disentangles it.

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  • $\begingroup$ I've added the extra condition, please take a look $\endgroup$ Apr 30 '19 at 3:43

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