2
$\begingroup$

Say I have the two qubit system $\frac{1}{\sqrt{2}}\begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}$.

I have two 2x2 unitary gates, one is a rotation by $\theta$ radians and the other is a rotation by $-\theta$ radians. How would I apply one of these rotations to one qubit and the other to the other qubit? Can I do this with a single 4x4 gate?

$\endgroup$
2
$\begingroup$

Suppose your $2 \times 2$ unitary gate is an $RY$ rotation. Then you can create your $4 \times 4$ gate as follow: $$U = RY(\theta)\otimes RY(-\theta)$$

Now, when apply $U$ to your state, you simply act $RY(\theta)$ to the first qubit and $RY(-\theta)$ to the second qubit.

To see this more explicitly, first note that your state $|\psi \rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} = \dfrac{|01\rangle + |10\rangle}{\sqrt{2}}$ can be represented in a quantum circuit as:

enter image description here

Then now applying $U$ is equivalent to adding appropriate $RY$ rotation to the appropriate qubit, which is:

enter image description here

$\endgroup$
4
  • $\begingroup$ So what you're saying is I would generate the 4x4 and then multiply it with my state vector. $\endgroup$
    – MLNewbie
    Mar 18 at 20:28
  • $\begingroup$ If you want to do it explicitly by hand then yes. What I am trying to write is that in your question, it seems to me like you try to implement certain one qubit gate says $U_1$ on qubit and another one qubit gate says $U_2$ to the other qubit. If that is the case then your $4 \times 4$ unitary gate is $U = U_1 \otimes U_2$. In my answer, I took $U_1$ and $U_2$ to be $RY$ rotations... $\endgroup$
    – KAJ226
    Mar 18 at 21:27
  • $\begingroup$ That makes sense, thank you for the help! $\endgroup$
    – MLNewbie
    Mar 18 at 22:01
  • $\begingroup$ No problem. Glad I was able to help. :) $\endgroup$
    – KAJ226
    Mar 18 at 22:21

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.