5
$\begingroup$

Suppose I have the entangled state

$$|\psi\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |110\rangle).$$

If i want to factor the non-entangled parts of this state out, I can easily write that down as

$$|\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) \otimes |0\rangle.$$

Suppose now, however, that I have a state

$$|\phi\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |101\rangle).$$

In $|\phi\rangle$, qubits $1$ and $3$ are entangled as opposed to qubits $1$ and $2$ in $|\psi\rangle$. Is there a notation for factoring out qubit $2$ in this state that is elegant like for $|\psi\rangle$?

$\endgroup$
0
5
$\begingroup$

I might say:

$$\vert\phi\rangle=\frac{1}{\sqrt 2}(\vert 0_A0_B0_C\rangle +\vert 1_A0_B1_C\rangle)= \frac{1}{\sqrt 2}(\vert 0_A0_C\rangle +\vert 1_A1_C\rangle)\otimes \vert 0_B\rangle,$$

because I think that might be intuitively understandable without much parsing. It also might help to label each qubit if I have to give the qubits to Alice, Bob, and Charlie later on.

Another option might be to consider adding wildcards. For example I might say:

$$\vert\phi\rangle=\frac{1}{\sqrt 2}(\vert 000\rangle +\vert 101\rangle)= \frac{1}{\sqrt 2}(\vert 0.0\rangle +\vert 1.1\rangle)\otimes \vert .0.\rangle,$$

but there I’d have to define my convention appropriately.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.