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I'm currently going through Introduction to Classical and Quantum Computing, by Thomas Wong, and I'm struggling with exercise 2.29 (page 107):

Exercise 2.29. Say $\left| \psi \right> = \alpha\left| 0 \right> + \beta\left| 1 \right>$ is a normalized quantum state, i.e., $|\alpha|^2 + |\beta|^2 = 1$.

(a) Calculate $H\left| \psi \right>$.

Given the two following transformations the Hadamard gate does:

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

I did the following:

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

Though, the answer given at the back of the book (page 360) is (notice the $+$ sign): $$\begin{align} H\left| \psi \right> &= \frac{\alpha + \beta}{\sqrt{2}}\left| 0 \right> + \frac{\alpha + \beta}{\sqrt{2}}\left| 1 \right> \end{align}$$

Why is that? This is a simple exercise, yet I can't seem to find what I'm doing wrong.

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    $\begingroup$ Your result is correct. The answer you quote from the book is wrong and probably a typo. You can quickly check by plugging in $\alpha=0,\beta=1$, in which case that answer results in $H|1\rangle = |+\rangle$. More generally I don't think that is even a valid quantum state. $\endgroup$
    – forky40
    Mar 3 at 21:37
  • $\begingroup$ @forky40 oh, well that explains it... $\endgroup$
    – Maxime Desalle
    Mar 3 at 21:43
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    $\begingroup$ @forky40 I think you still should write this as an answer, should anyone stumble upon the same problem! $\endgroup$
    – Tristan Nemoz
    Mar 3 at 21:45
  • $\begingroup$ Just reached out to the author of the textbook to fix the mistake, hopefully they see my email. $\endgroup$
    – Maxime Desalle
    Mar 3 at 21:49

2 Answers 2

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Your calculation is correct and that is probably a typo in the textbook, as already mentioned in the comments.

Moreover, if you write down your state as $| \psi \rangle = \alpha | 0 \rangle + \beta | 1 \rangle = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$ and the Hadamard gate as $H = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix}$, then computing the result is straightforward: $$ H | \psi \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} \alpha + \beta \\ \alpha - \beta \end{bmatrix} = \frac{\alpha + \beta}{\sqrt{2}} | 0 \rangle + \frac{\alpha - \beta}{\sqrt{2}} | 1 \rangle $$

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Your result is correct. The answer you quote from the book is wrong and probably a typo. You can quickly check by plugging in α=0,β=1, in which case that answer results in $H|1\rangle = |+\rangle$. More generally that is not a valid quantum state - for example, set $\alpha=\beta=1/\sqrt{2}$ and the result will be $\sqrt{2}|+\rangle$.

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