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Given two separate states, I'd like to entangle them to a specific configuration.

Example 1: $$|+\rangle \otimes |+\rangle \rightarrow \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$

Example 2: $$(a_1|0\rangle + b_1|1\rangle) \otimes (a_2|0\rangle + b_2|1\rangle) \rightarrow a_1 a_2|00\rangle + b_1 b_2|11\rangle$$.

By intuition, I am lead to believe this map is not possible, as it would mean bringing to 0 the configurations $a_1 b_2|01\rangle$ and $a_2 b_1|10\rangle$.

However, Grover's oracle and diffusion may be used to achieve the correct mapping. But all applications I found start from balanced superposition and I don't know whether and how I could adapt the algorithm to make it work in my scenario.

Any hint?

Initial amplitudes $a_1,b_1,a_2,b_2$ can be positive if necessary.

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  • $\begingroup$ What would be the image of $|00\rangle$? Both coefficients should be $0$, so that it's not possible to normalize it. $\endgroup$ Commented Apr 10 at 11:23

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Insert an extra qubit in the $|0\rangle$ state.

Apply two controlled-nots, one controlled off each qubit and targetting the extra qubit.

Measure the extra qubit. If you get outcome $|0\rangle$, you succeeded.

If you have a method for preparing the initial state in either example, and prior knowledge of what the weights are, you can turn this into a Grover search (but if you know that much, you can probably build the state directly, far more efficiently).

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  • $\begingroup$ Thank you. I was thinking of Grover search to avoid the case of failure in measurement. Do I need to have prior knowledge of the weights to use Grover? $\endgroup$ Commented Apr 10 at 11:32
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    $\begingroup$ you need to know the probability of getting the 0 answer. There are versions that work even if the relevant amplitude isn't known, but it slows things down. I would expect (without justification) that it would be much slower than the non-unitary version for such a small problem. $\endgroup$ Commented Apr 10 at 12:07
  • $\begingroup$ Could you please give me a reference? $\endgroup$ Commented Apr 10 at 12:09
  • $\begingroup$ start here: en.wikipedia.org/wiki/… $\endgroup$ Commented Apr 10 at 12:39
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Given two separate states, I'd like to entangle them to a specific configuration.

Example 1: $$|+\rangle \otimes |+\rangle \rightarrow \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$

This is possible to do with a 2-qubit unitary transformation. The transformation is $$ U_{\text{Ex. 1}}=(\text{CX})(I\otimes H)\;, $$ which, in matrix form, looks like $$ \bar U_{\text{Ex. 1}} = \frac{1}{\sqrt{2}}\left(\begin{matrix} 1&1&0&0\\ 1&-1&0&0\\ 0&0&1&-1\\ 0&0&1&1 \end{matrix}\right) $$

Example 2: $$(a_1|0\rangle + b_1|1\rangle) \otimes (a_2|0\rangle + b_2|1\rangle) \rightarrow a_1 a_2|00\rangle + b_1 b_2|11\rangle\;.$$

This is not possible to do with a 2-qubit unitary transformation, for general $a_1$, $b_1$, $a_2$, and $b_2$.

It's not possible in general because you have specified that the coefficient of $|00\rangle$ must end up as $a_1 a_2$ and the coefficient of $|11\rangle$ must end up as $b_1 b_2$. For example, if you tried to specify that in the previous example, you would need $\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} = \frac{1}{2}$ to also end up being equal to $\frac{1}{\sqrt{2}}$, which is not possible.

To prove that the transformation of Example 2 is not possible in general, start by assuming that it is possible that a 2-qubit unitary $U$ could make the transformation and then go looking for trouble.

We assume there exists a 2-qubit unitary $U$ such that: $$ U(a_1|0\rangle + b_1|1\rangle) \otimes (a_2|0\rangle + b_2|1\rangle) \stackrel{?}{=} a_1 a_2|00\rangle + b_1 b_2|11\rangle\;. $$

We then take the inner product of each side of the equation with itself, assuming $U$ is unitary. This gives: $$ 1=|a_1|^2|a_2|^2 + |a_1|^2|b_2|^2+|b_1|^2|a_2|^2+|b_1|^2|b_2|^2 \stackrel{?}{=} |a_1|^2|a_2|^2 + |b_1|^2|b_2|^2 = 1\;, $$ where the first and last equalities follow because we assumed that initial state was normalized to $1$ and that $U$ is unitary.

But this means that $$ |a_1|^2|b_2|^2+|b_1|^2|a_2|^2 = 0\;, $$ which is not true in general.

By intuition, I am lead to believe this map is not possible, as it would mean bringing to 0 the configurations $a_1 b_2|01\rangle$ and $a_2 b_1|10\rangle$.

Yes, it's not possible (in general) via a 2-qubit unitary transformation.

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