# What unitary gate produces these quantum states from the computational basis?

Suppose that we have one-qubit unitary $$U$$ that maps $$\left| 0 \right> \longmapsto \frac{1}{\sqrt{2}} \left| 0 \right> + {\frac{1+i}{2}} \left| 1\right>$$ and $$\left| 1 \right> \longmapsto {\frac{1-i}{2}} \left| 0 \right> - \frac{1}{\sqrt{2}} \left| 1\right>$$ What is $$U$$?

• the matrix whose columns are the amplitudes of the two output states
– glS
Jan 9 '20 at 11:14

Firstly simply rewrite probability amplitudes of returned states as columns of a matrix: $$U = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1-i}{2} \\ \frac{1+i}{2} & -\frac{1}{\sqrt{2}} \end{pmatrix}$$ Now do some algebra $$U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & \frac{1-i}{\sqrt{2}} \\ \frac{1+i}{\sqrt{2}} & -1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & \mathrm{e}^{-i\frac{\pi}{4}} \\ \mathrm{e}^{i\frac{\pi}{4}} & -1 \end{pmatrix}$$

There is a quantum gate called $$\mathrm{U2}$$: $$\mathrm{U2}(\phi,\lambda)= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -\mathrm{e}^{i\lambda} \\ \mathrm{e}^{i\phi} & \mathrm{e}^{i(\phi+\lambda)} \end{pmatrix}$$

Setting $$\phi=\frac{\pi}{4}$$ and $$\lambda = \frac{3}{4}\pi$$ you have a resut since $$\phi+\lambda =\pi$$, so $$\mathrm{e}^{i(\phi+\lambda)} = \mathrm{e}^{i\pi} = -1$$ and $$-\mathrm{e}^{i\lambda}=-\mathrm{e}^{i\frac{3}{4}\pi} = -\frac{-1+i}{\sqrt{2}}$$.

Conclusion: $$U=\mathrm{U2}\big(\frac{\pi}{4},\frac{3}{4}\pi\big)$$

• I like how you started from the beginning and never skipped a step, well done! Jan 8 '20 at 21:05

Just to expand on the detail of why writing out the columns works:

Start by writing the action of the unitary: \begin{align*} U|0\rangle=\frac{1}{\sqrt{2}}|0\rangle+\frac{1+i}{2}|1\rangle \\ U|1\rangle=\frac{1-i}{2}|0\rangle-\frac{1}{\sqrt{2}}|1\rangle \end{align*} Before proceeding, it's always worth checking that both sides are correctly normalised. In this case, they are.

Now take the inner product of each equation with $$\langle 0|$$: $$\langle 0|U|0\rangle=\frac{1}{\sqrt{2}}\qquad\langle 0|U|1\rangle=\frac{1-i}{2}$$ Similarly, using $$\langle 1|$$, you get $$\langle 1|U|0\rangle=\frac{1+i}{2}\qquad\langle 1|U|1\rangle=-\frac{1}{\sqrt{2}}.$$ So, these identify all four matrix elements, which you can just insert: $$U=\left(\begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1-i}{2} \\ \frac{1+i}{2} & -\frac{1}{\sqrt{2}} \end{array}\right).$$ (I should say that I always get muddled between the two off-diagonal elements. So I have to stop and think about, for example, $$\langle 0|U|1\rangle$$, and which element is selected by doing the inner product $$\left(\begin{array}{cc}1 & 0\end{array}\right)U\left(\begin{array}{c} 0 \\ 1 \end{array}\right)$$: top row, right-hand column.)

Don't forget to check that your answer is reasonable by verifying $$UU^\dagger=I$$.