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I'm currently going through Introduction to Classical and Quantum Computing, by Thomas Wong, and I'm struggling with an example given on page 88 (point 4).

The author gives the two following states, which form a basis as they are on opposite points of the Block sphere: $$\begin{align} \left| a \right> &= \frac{\sqrt{3}}{2}\left| 0 \right> + \frac{i}{2}\left| 1 \right>\\ \left| b \right> &= \frac{i}{2}\left| 0 \right> + \frac{\sqrt{3}}{2}\left| 1 \right> \end{align}$$

He then writes $\left| 0 \right>$ and $\left| 1 \right>$ in terms of $\left| a \right>$ and $\left| b \right>$ as follows: $$\begin{align} \left| 0 \right> &= \frac{\sqrt{3}}{2}\left| a \right> - \frac{i}{2}\left| b \right>, \,\, \left| 1 \right> = -\frac{i}{2}\left| a \right> + \frac{\sqrt{3}}{2}\left| b \right> \end{align}$$

In the above equations, the author uses in both cases the conjugate of $\frac{i}{2}$. What's the reasoning behind that?

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$$\begin{align} \left| a \right> &= \frac{\sqrt{3}}{2}\left| 0 \right> + \frac{i}{2}\left| 1 \right> \tag{A}\\ \left| b \right> &= \frac{i}{2}\left| 0 \right> + \frac{\sqrt{3}}{2}\left| 1 \right> \tag{B} \end{align}$$

He then writes $\left| 0 \right>$ and $\left| 1 \right>$ in terms of $\left| a \right>$ and $\left| b \right>$ as follows: $$\begin{align} \left| 0 \right> &= \frac{\sqrt{3}}{2}\left| a \right> - \frac{i}{2}\left| b \right>, \,\, \left| 1 \right> = -\frac{i}{2}\left| a \right> + \frac{\sqrt{3}}{2}\left| b \right> \end{align}$$

In the above equations, the author uses in both cases the conjugate of $\frac{i}{2}$. What's the reasoning behind that?

There is no reasoning other than algebra. The rearrangement is literally just algebra. It may also help to recall that $\frac{1}{i} = -i$.

You rearrange the equations (the equations I tagged as "A" and "B" above) so that you have solved for $|0\rangle$ and $|1\rangle$. For example, multiply the equation I have tagged above as "A" by $2/i$ to find: $$ |a\rangle \frac{2}{i} = \frac{\sqrt{3}}{i}|0\rangle + |1\rangle $$ and multiply the equation I have tagged above as "B" by $2/\sqrt{3}$ to find: $$ |b\rangle\frac{2}{\sqrt{3}} = \frac{i}{\sqrt{3}}|0\rangle + |1\rangle $$

Then subtracting the above two equations to solve for $|0\rangle$ gives: $$ \frac{2}{\sqrt{3}}|b\rangle + 2i|a\rangle = \left(\frac{i}{\sqrt{3}}+i\sqrt{3}\right)|0\rangle\;, $$ and rearranging to isolate $|0\rangle$ on one side gives: $$ |0\rangle = \frac{-i}{2}|b\rangle + \frac{\sqrt{3}}{2}|a\rangle\;. $$

You can similarly solve for $|1\rangle$ using algebra.

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    $\begingroup$ That makes sense, thank you for this detailed explanation! $\endgroup$ Feb 28 at 23:14

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