I've pinpointed that the difference between solutions occurs when I apply the $H$, up to there, the solutions match. So, perhaps it is better to phrase my question as: how to apply an $H$ gate to one qubit in a 2-qubit system?

End Edit

I have the following quantum circuit, and I need to calculate the values of the two qubits afterwards.


Now I know the values are:

  • $\lvert \psi_1 \rangle = \frac{1}{\sqrt{2}}(\lvert 0 \rangle+\lvert 1 \rangle)$
  • $\lvert \psi_2 \rangle = \lvert 0 \rangle$

It's pretty simple to see, and I checked them with a simulator just to be sure.

The trouble I'm getting into is when I try to express this as a two qubit system.

As I understand it should be as follows:

$$\lvert \psi\prime \rangle=M\lvert \psi \rangle=\frac{1}{\sqrt{2}}(\lvert 00 \rangle+\lvert 10 \rangle)$$

Where $\lvert \psi\prime \rangle$ is the state after the circuit is complete, $\lvert \psi \rangle = \lvert 00 \rangle$ is the initial state, and $M$ is the matrix representing the combined effect of all the gates of the circuit.

Now, I calculate $M$ as follows:

$$M=\frac{1}{\sqrt{2}}(I\otimes X)\begin{bmatrix}I&0\\0&X\end{bmatrix}(I\otimes X)(H\otimes I)$$

I wrote the $CNOT$ matrix in block matrix form here, and extracted the $\frac{1}{\sqrt{2}}$ from $H$ to the beginning.

$$\begin{align}M&=\frac{1}{\sqrt{2}}\begin{bmatrix}X&0\\0&X\end{bmatrix}\begin{bmatrix}I&0\\0&X\end{bmatrix}\begin{bmatrix}X&0\\0&X\end{bmatrix}\begin{bmatrix}I&I\\I&-I\end{bmatrix} \\ &=\frac{1}{\sqrt{2}}\begin{bmatrix}X&0\\0&I\end{bmatrix}\begin{bmatrix}X&0\\0&X\end{bmatrix}\begin{bmatrix}I&I\\I&-I\end{bmatrix} \\ &=\frac{1}{\sqrt{2}}\begin{bmatrix}I&0\\0&X\end{bmatrix}\begin{bmatrix}I&I\\I&-I\end{bmatrix} \\ &=\frac{1}{\sqrt{2}}\begin{bmatrix}I&I\\X&-X\end{bmatrix}\end{align}$$

Now, plugging $M$ back into the first equation...

$$\begin{align}\lvert\psi\prime\rangle &=\frac{1}{\sqrt{2}}\begin{bmatrix}I&I\\X&-X\end{bmatrix}\lvert\psi\rangle \\ &=\frac{1}{\sqrt{2}}\begin{bmatrix}1&0&1&0\\0&1&0&1\\0&1&0&-1\\1&0&-1&0\end{bmatrix}\begin{bmatrix}1\\0\\0\\0\end{bmatrix}\\&=\frac{1}{\sqrt{2}}\begin{bmatrix}1\\0\\0\\1\end{bmatrix}\\&=\frac{1}{\sqrt{2}}(\lvert 00\rangle + \lvert 11\rangle)\end{align}$$

Which is not the correct solution. Where am I at fault?


1 Answer 1


The problem is in the order of the transformations. The first transformation you apply in the circuit is $I\otimes X$, whereas in the computation of $M$ it is actually the last. Recall that the rightmost operator is the first to act on the state (matrix multiplication is not commutative).

Fixing that we obtain:

$$ M=(H\otimes I)(I\otimes X)CNOT(I\otimes X) = \begin{bmatrix} H & 0 \\ 0 & H \end{bmatrix} \begin{bmatrix} X & 0 \\ 0 & X \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & X \end{bmatrix} \begin{bmatrix} X & 0 \\ 0 & X \end{bmatrix}= \begin{bmatrix} H & 0 \\ 0 & H \end{bmatrix} \begin{bmatrix} X & 0 \\ 0 & X \end{bmatrix} \begin{bmatrix} X & 0 \\ 0 & I \end{bmatrix}= \begin{bmatrix} I & I \\ I & -I \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & X \end{bmatrix}= \begin{bmatrix} I & X \\ I & -X \end{bmatrix} $$

Now applying $M$ on $|00\rangle$ will yield the right result.

  • $\begingroup$ It seems to me that H X I should yield [[ I I ], [ I -I ]] and not the first explicit matrix above? The result shown would be for (I X H)? $\endgroup$
    – Francky_V
    Apr 20, 2020 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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