Encoding arbitrary quantum gates using qubits

Question

Given an arbitrary 3-qubit state $\sum_{xyz} c_{xyz}|xyz\rangle$, is there a circuit (possibly with measurement) that creates the state $\sum_{xy} c_{xyy}|x\rangle$, up to a normalization constant?

As a weaker question, is the above circuit possible, given that the initial state can be decomposed as a tensor product of the first two qubit and the third? This would allow an encoding of a quantum gate with 2 qubits.

I see that measuring the last two qubits in the Bell basis has a chance of succeeding. But for 3 out of 4 cases, although I know I failed, I cannot correct for it as in the quantum teleporting case, because the gates needed for correction is unknown. Is this task impossible? How can I prove it?

Answer 1

We know that every quantum operation can be described by a completely positive trace-preserving (CPTP) linear map acting on density matrices. You want a map that sends

$$ \sum_{xyz,\tilde x \tilde y\tilde z} c_{xyz}c_{\tilde x\tilde y\tilde z}^*|xyz\rangle\langle \tilde x\tilde y\tilde z|\rightarrow \sum_{xy,\tilde x \tilde y} c_{xyy}c_{\tilde x\tilde y\tilde y}^*|xyy\rangle\langle \tilde x\tilde y\tilde y| $$

Unfortunately, as I've written it, the map isn't even trace-preserving! We can fix that by adding a normalization constant:

$$ \sum_{xyz,\tilde x \tilde y\tilde z} c_{xyz}c_{\tilde x\tilde y\tilde z}^*|xyz\rangle\langle \tilde x\tilde y\tilde z|\rightarrow \frac{1}{\sum_{xy}|c_{xyy}|^2}\sum_{xy,\tilde x \tilde y} c_{xyy}c_{\tilde x\tilde y\tilde y}^*|xyy\rangle\langle \tilde x\tilde y\tilde y| $$

Now we have a trace-preserving operation. But it's no longer linear! Overall, it's not possible to have this operation be both linear and trace-preserving, so it is not a valid quantum operation.