What is $HTHTH\left| 0 \right>$?

Question

I'm currently going through Introduction to Classical and Quantum Computing, by Thomas Wong, and I'm struggling with exercise 2.33 (page 108):

Exercise 2.33. Answer the following:

(a) Calculate $HTHTH\left| 0 \right>$.

Given the two following transformations the Hadamard gate does:

$$\begin{align} H\left| 0 \right> &= \frac{1}{\sqrt{2}}(\left| 0 \right> + \left| 1 \right>)\\ H\left| 1 \right> &= \frac{1}{\sqrt{2}}(\left| 0 \right> - \left| 1 \right>) \end{align}$$

And the two following transformations the T gate does:

$$\begin{align} T\left| 0 \right> &= \left| 0 \right>\\ T\left| 1 \right> &= e^{i\pi/4}\left| 1 \right> \end{align}$$

I did the following:

$$\begin{align} HTHTH\left| 0 \right> &= HTHT\frac{1}{\sqrt{2}}(\left| 0 \right> + \left| 1 \right>) &\text{(apply H gate)}\\ &= HTH\frac{1}{\sqrt{2}}(\left| 0 \right> + e^{i\pi/4}\left| 1 \right>) &\text{(apply T gate)}\\ &= HT\frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}}(\left| 0 \right> + \left| 1 \right>) + e^{i\pi/4}\frac{1}{\sqrt{2}}(\left| 0 \right> - \left| 1 \right>)) &\text{(apply H gate)}\\ &= H\frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}}(\left| 0 \right> + e^{i\pi/4}\left| 1 \right>) + e^{i\pi/4}\frac{1}{\sqrt{2}}(\left| 0 \right> - e^{i\pi/4}\left| 1 \right>)) &\text{(apply T gate)}\\ &= \frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}}(\left| + \right> + e^{i\pi/4}\left| - \right>) + e^{i\pi/4}\frac{1}{\sqrt{2}}(\left| + \right> - e^{i\pi/4}\left| - \right>)) &\text{(apply H gate)}\\ &= \frac{1 + e^{i\pi/4}}{2}\left| + \right> + \frac{e^{i\pi/4} - i}{2}\left| - \right> &\text{(simplify)}\\ &= \frac{1 + e^{i\pi/4}}{2}\frac{1}{\sqrt{2}}(\left| 0 \right> + \left| 1 \right>) + \frac{e^{i\pi/4} - i}{2}\frac{1}{\sqrt{2}}(\left| 0 \right> - \left| 1 \right>) &\text{(change base)}\\ &= \frac{1 + e^{i\pi/4}}{2\sqrt{2}}\left| 0 \right> + \frac{1 + e^{i\pi/4}}{2\sqrt{2}}\left| 1 \right> + \frac{e^{i\pi/4} - i}{2\sqrt{2}}\left| 0 \right> - \frac{e^{i\pi/4} - i}{2\sqrt{2}}\left| 1 \right> &\text{(distribute)}\\ &= \frac{1 + 2e^{i\pi/4} - i}{2\sqrt{2}}\left| 0 \right> + \frac{1 + i}{2\sqrt{2}}\left| 1 \right>&\text{(simplify)}\\ &= \frac{1}{2\sqrt{2}}[(1 - i + 2e^{i\pi/4})\left| 0 \right> + (1 + i)\left| 1 \right>]&\text{(simplify)}\\ \end{align}$$

However, the answer given at the back of the book (page 360) is (notice the $e^{i\pi/4}$ as opposed to $2e^{i\pi/4}$): $$\begin{align} HTHTH\left| 0 \right> &= \frac{1}{2\sqrt{2}}[(1 - i + e^{i\pi/4})\left| 0 \right> + (1 + i)\left| 1 \right>]\\ \end{align}$$

What am I doing wrong?

Answer 1

...

$$\frac{1}{2\sqrt{2}}[(1 - i + 2e^{i\pi/4})\left| 0 \right> + (1 + i)\left| 1 \right>]$$ ...

However, the answer given at the back of the book (page 360) is (notice the $e^{i\pi/4}$ as opposed to $2e^{i\pi/4}$): $$\begin{align} HTHTH\left| 0 \right> &= \frac{1}{2\sqrt{2}}[(1 - i + e^{i\pi/4})\left| 0 \right> + (1 + i)\left| 1 \right>]\\ \end{align}$$

What am I doing wrong?

You seem to have done nothing wrong.

For notational convenience, I denote $T$ by $$ T\equiv \left(\begin{matrix}1 & 0 \\ 0 & t\end{matrix}\right)\;, $$ where $t=e^{i\pi/4}$.

Then, by matrix multiplication: $$ HTH = \frac{1}{2}\left(\begin{matrix}1+t & 1-t \\ 1-t & 1+t\end{matrix}\right) $$ and $$ THTH = \frac{1}{2}\left(\begin{matrix}1+t & 1-t \\ t-t^2 & t+t^2\end{matrix}\right) $$ and $$ HTHTH = \frac{1}{2\sqrt{2}}\left(\begin{matrix}1+2t-t^2 & 1+t^2 \\ 1+t^2 & 1-2t-t^2\end{matrix}\right) $$ and so $$ HTHTH |0\rangle = \frac{1}{2\sqrt{2}}\left( (1+2t-t^2)|0\rangle +(1+t^2)|1\rangle \right)\;, $$ where, setting $t=e^{i\pi/4}$ and using $t^2 = i$, we see that: $$ HTHTH |0\rangle = \frac{1}{2\sqrt{2}}\left( (1+2e^{i\pi/4}-i)|0\rangle +(1+i)|1\rangle \right)\;. $$

Answer 2

I quickly computed in Mathematica. You are correct. The book is incorrect.