6
$\begingroup$

Let's say we have the following quantum circuit:

enter image description here

Let's say we input the state $|00\rangle$ . Both of the $H$ gates produce the output $1/\sqrt{2}$, but which one of the following $2$ vectors is the input of $\operatorname{CNOT}$ gate: $1: [1/\sqrt{2}, \ 1/\sqrt{2}, \ 1/\sqrt{2}, \ 1/\sqrt{2}]$ or $2: [1/2, \ 1/2, \ 1/2, \ 1/2]$? Also, after applying a $2$ qubit gate (in this case $\operatorname{CNOT}$) how do we find out what to input in the following single qubit gates (in this case two $H$ gates)? Note that I gave this example for simplicity, a general answer will also be accepted.

$\endgroup$
4
$\begingroup$

Step 1 (application of two hadamard gates):

$$|0\rangle_A \otimes |0\rangle_B \to \left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_A\otimes\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_B$$

This is equivalent to the state vector: $\begin{bmatrix}1/2\\1/2\\1/2\\1/2\end{bmatrix}$, which will act as the input for your CNOT gate.

Step 2: (application of CNOT)

Let's remind ourselves what the CNOT gate does:

enter image description here

So clearly, $$\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_A\otimes\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_B = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)$$ $$\to \frac{1}{2}(|00\rangle + |01\rangle + |11\rangle + |10\rangle) = \left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_A\otimes\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_B$$

Step 3: (re-application of two Hadamard gates)

$$\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_A\otimes\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_B \to |0\rangle_A \otimes |0\rangle_B$$

Here, we used that fact that the Hadamard gate maps the state $\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$ to the state $|0\rangle$, for each qubit. You basically need to input the vector $\begin{bmatrix}1/\sqrt{2}\\1/\sqrt{2}\end{bmatrix}$ into the two Hadamard gates at the end. Find the state transformation on each qubit. From there you can construct the state vector for the final 2-qubit state. The final answer will be $\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}$. Clearly, we end up with the same 2-qubit state which we started with.

Exercise: Carry out steps $2$ and $3$ using the matrix notation. Also, check this related answer.

$\endgroup$

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.