Quantum Computing Stack Exchange is a question and answer site for engineers, scientists, programmers, and computing professionals interested in quantum computing. It only takes a minute to sign up.
Sign up to join this community
So I am reading this book and I found this worked out example which I didn't get. If you understand these bits of pieces, can you please let me know.
Question:
Answer:
H over here? I thought it would be just H * Cnot?
If $\left|\psi\right> = \alpha \left|0\right> + \beta \left|1\right>$, then $\alpha = \left<0|\psi\right>$ and $\beta = \left<1|\psi\right>$. A 1-qubit unitary operator can be written as $U = a\left|0\right>\left<0\right| + b\left|0\right>\left<1\right| + c\left|1\right>\left<0\right| + d\left|1\right>\left<1\right|$ and this will correspond to a matrix $\begin{pmatrix}a & b\\c & d\end{pmatrix}$: to see this, notice that through the projection form $U \left|\psi\right> = (a \left<0|\psi\right> + b \left<1|\psi\right>)\left|0\right> + (c\left<0|\psi\right> + d\left<1|\psi\right>)\left|1\right>$. To get the values of the matrix out of $U$ and bras and kets of the computational basis states, notice $U\left|0\right> = a\left|0\right> + c\left|1\right>$ and $U\left|1\right> = b\left|0\right> + d\left|1\right>$: from this, the general matrix form can be given as $U = \begin{pmatrix}\left<0\right|U\left|0\right> & \left<0\right|U\left|1\right> \\\left<1\right|U\left|0\right> & \left<1\right|U\left|1\right> \end{pmatrix}$: this gives the answer to your second question when $X$ is said operator.
For the other question, if $T = \begin{pmatrix}\left<+|0\right> & \left<+|1\right> \\\left<-|0\right> & \left<-|1\right> \end{pmatrix}$, then an operator $U$'s form in the new basis is given by $U^T = \begin{pmatrix}\left<+\right|U\left|+\right> & \left<+\right|U\left|-\right> \\\left<-\right|U\left|+\right> & \left<-\right|U\left|-\right> \end{pmatrix}$, and it represents the operation's matrix when $\left|+\right>$ and $\left|-\right>$ are treated as $\begin{pmatrix}1 \\0\end{pmatrix}$ and $\begin{pmatrix}0 \\ 1\end{pmatrix}$ instead of $\left|0\right>$ and $\left|1\right>$. Then, $U^T = T U T^{-1}$ since $T^{-1}$ turns the representation in terms of the new basis back to its representation in the computational basis, $U$ runs the operator in the computational basis, and then $T$ converts it back to the new basis. As stated in the text, $H$ is its own inverse, so $H U_{NOT}H^{-1} = HU_{NOT}H$.
the operation of U's form on the original basis is performed this is just the "TUT'" U part right? What are the roles of those T's?
$\endgroup$
Dec 23, 2020 at 21:02
then an operator U's form in the new basis is given by UT Why is it U^T? Seriously I am not getting it. You have wrote a great answer but idk why I am not getting it,
$\endgroup$
Dec 24, 2020 at 8:25
Joseph's answer explains a lot of this already but I just want to give a much simpler answer in case others have the same question:
"Why is there two H over here? I thought it would be just H * Cnot?"
To put the $X$ matrix into the $H$ basis, simply calculate $H^{-1}XH$ as you can see here. Since $H^{-1} = H$, the expression is as given in the answer you found in that book :)
"Where from do we get these representations?"
In the basis containing $|0\rangle$ and $|1\rangle$, each of the 4 elements of the $X$ matrix will be given by $\langle 0| X |0\rangle$, $\langle 0| X |1\rangle$ , $\langle 1| X |0\rangle$, $\langle 1| X |1\rangle$ int he exact positions shown in the screenshot you gave us. If you do the matrix multiplication of a row vector, times a matrix, times a column vector (from left-to-right), combined with knowledge of how "Dirac notation works", you'll see why this is the case :).
