I am new to quantum computing and reading the book "Introduction to Classical and Quantum Computing", by Wong (link).

I do not understand how to calculate the qubit state for the below question. Can someone please help me understand the answer? Thanks

Excercis 2.11. The following two states are opposing points on the Bloch sphere: $$|a\rangle=\frac{\sqrt 3}{2}|0\rangle+\frac{i}{2}|1\rangle,$$ $$|b\rangle=\frac{i}{2}|0\rangle+\frac{\sqrt 3}{2}|1\rangle.$$ So, we can measure relative to them. Now consider a qubit in the state $$\frac{1}{2}|0\rangle-\frac{\sqrt 3}{2}|1\rangle.$$

(a) Write the qubit's state in terms of $|a\rangle$ and $|b\rangle$.

(b) If you measure the qubit in the basis $\{|a\rangle,|b\rangle\}$, what states would you get and with what probabilities?


2 Answers 2


The point here is that you're going to be able to write $$ \frac12|0\rangle-\frac{\sqrt{3}}{2}|1\rangle=\alpha|a\rangle+\beta|b\rangle. $$ Your job is to find the parameters $\alpha,\beta$. You've got a few strategies open to you. In a simple case like this, you could just expand the states $|a\rangle$ and $|b\rangle$. If you equate the coefficients for $|0\rangle$ and $|1\rangle$, you'll get two simultaneous equations that will let you find $\alpha,\beta$.

However, long-term, I wouldn't recommend that as a way of doing the calculation. Instead, try to think if there's a state $|c\rangle$ you could use to take the inner product with: $$ \langle c|(\frac12|0\rangle-\frac{\sqrt{3}}{2}|1\rangle)=\alpha\langle c|a\rangle+\beta \langle c|b\rangle. $$ How does this help? If you can pick your $|c\rangle$ such that $\langle c|b\rangle=0$, then it immediately isolates the $\alpha$ term for you. Do you know any state with the required property?


a) First rewrite the equations in terms of $\left| 0 \right>$ and $\left| 1 \right>$ as follows (here is how to do so): $$\begin{aligned} \left| 0 \right> &= \frac{\sqrt{3}}{2}\left| a \right> - \frac{i}{2}\left| b \right>\\ \left| 1 \right> &= -\frac{i}{2}\left| a \right> + \frac{\sqrt{3}}{2}\left| b \right>\\ \end{aligned}$$

Then substitute as follows: $$\begin{aligned} \frac{1}{2}\left| 0 \right> - \frac{\sqrt{3}}{2}\left| 1 \right>&=\frac{1}{2}(\frac{\sqrt{3}}{2}\left| a \right> - \frac{i}{2}\left| b \right>) - \frac{\sqrt{3}}{2}(- \frac{i}{2}\left| a \right> + \frac{\sqrt{3}}{2}\left| b \right>)\\ &= \frac{\sqrt{3}}{4}\left| a \right> - \frac{i}{4}\left| b \right> + \frac{\sqrt{3}i}{4}\left| a \right> - \frac{3}{4}\left| b \right>\\ &= \frac{\sqrt{3}(1 + i)}{4}\left| a \right> - \frac{i + 3}{4}\left| b \right> \end{aligned}$$

b) The probability of $\left| a \right>$ and $\left| b \right>$ respectively is: $$\begin{aligned} \left| \frac{\sqrt{3}(1 + i)}{4} \right|^2 &= \left( \frac{\sqrt{3}(1 + i)}{4} \right) \left( \frac{\sqrt{3}(1 - i)}{4} \right) = \frac{3}{8}\\ \left| \frac{i + 3}{4} \right|^2 &= \left( \frac{i + 3}{4} \right) \left( \frac{3 - i}{4} \right) = \frac{5}{8} \end{aligned}$$

It's worth noting that the author explains the process for these types of problems on page 88 and 89 in the book.

  • $\begingroup$ You need some kets on the RHS of your first two equations... $\endgroup$
    – hft
    Feb 28, 2023 at 23:05
  • 1
    $\begingroup$ My mistake! Thanks for catching that @hft. $\endgroup$ Feb 28, 2023 at 23:08

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