# Violation of the Normalization Constraint?

Say we have two qubits $|a\rangle$ and $|b\rangle$ both initialized to $|0\rangle$. We then apply the rotation gate $R_{x}(\frac{\pi}{2})$ of matrix representation

$\left( \begin{array}{} \frac{1}{\sqrt2} & \frac{-i}{\sqrt2} \\ \frac{-i}{\sqrt2} & \frac{1}{\sqrt2} \end{array} \right)$

to get $|a\rangle = |b\rangle = \left( \begin{array}{} \frac{1}{\sqrt2}\\ \frac{-i}{\sqrt2}\end{array} \right)$

$|a\rangle|b\rangle$ is then $\left( \begin{array}{} \frac{1}{2}\\ \frac{-i}{2}\\ \frac{-i}{2}\\ \frac{1}{2}\end{array} \right)$

A controlled rotation gate $R_{x}(\frac{\pi}{2})$ of matrix representation

$\left( \begin{array}{} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 &\frac{1}{\sqrt2} & \frac{-i}{\sqrt2} \\ 0 & 0 &\frac{-i}{\sqrt2} & \frac{1}{\sqrt2} \end{array} \right)$

is then applied to $|a\rangle|b\rangle$ to get

$|a\rangle|b\rangle = \left(\begin{array}{} \frac{1}{2}\\ \frac{-i}{2}\\ \frac{-i}{\sqrt2}\\ \frac{1}{\sqrt2}\end{array} \right)$

But the sum of the squares of the amplitudes is $\frac{3}{2}$, violating the normalization constraint. Where did I go wrong?

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## 1 Answer

You made a mistake in calculation. In the last matrix multiplication, when you multiplied the last row with the column vector the result should be 0. I suspect that you accidently took (-i)^2=1 instead of (-i)^2=-1

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