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In chapter 7, equation 7.6 says CNOT works as follows:

CNOT: $\frac{1}{\sqrt{2}} (|0\rangle + |1\rangle )\otimes |x\rangle \rightarrow \frac{1}{\sqrt{2}} (|0\rangle + (-1)^x |1\rangle ) \otimes |x\rangle$, where it acts trivially if the target is $x=0$ state, and it flips the control if the target is the $x=1$ state.

I've looked at a few other resources about CNOT and this is the first time I encountered the $(-1)^x$ term.

Could someone explain to me where that term comes from?

Given that the matrix representation of CNOT is $$ \begin{pmatrix} 1 & 0 &0 &0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix}$$ I don't see how that $(-1)^x$ came about.

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Expanding on Jalex

Look at what happens on the possible terms.

\begin{eqnarray*} \mid 0 \rangle \otimes \mid + \rangle &\to& \mid 0 \rangle \otimes \mid + \rangle\\ \mid 0 \rangle \otimes \mid - \rangle &\to& \mid 0 \rangle \otimes \mid - \rangle\\ \mid 1 \rangle \otimes \mid + \rangle &\to& \mid 1 \rangle \otimes \mid + \rangle\\ \mid 1 \rangle \otimes \mid - \rangle &\to& (-1) \mid 1 \rangle \otimes \mid - \rangle\\ \end{eqnarray*}

where the first 2 are unchanged because the control is $0$ so nothing happens. The third is unchanged because NOT applied to $\mid + \rangle$ just gives back $\mid + \rangle$. The last is the only one with change because NOT applied to $\mid - \rangle$ gives $(-1) \mid - \rangle$.

We can summarize these possibilities by knowing that $\mid + \rangle$ goes with $x=0$ and $\mid - \rangle$ with $x=1$ as:

\begin{eqnarray*} \mid 0 \rangle \otimes \mid x \rangle &\to& \mid 0 \rangle \otimes \mid x \rangle\\ \mid 1 \rangle \otimes \mid x \rangle &\to& (-1)^x \mid 1 \rangle \otimes \mid x \rangle\\ \end{eqnarray*}

The first two become the first one above. And third and fourth, the second above.

Now add the two together along with a $\frac{1}{\sqrt{2}}$ prefactor to give

$$ \frac{1}{\sqrt{2}} ( \mid 0 \rangle + \mid 1 \rangle ) \otimes \mid x \rangle \to \frac{1}{\sqrt{2}} ( \mid 0 \rangle + (-1)^x \mid 1 \rangle ) \otimes \mid x \rangle $$

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Here Preskill is using a physics convention that the states $|x\rangle$ are the eigenstates of the $X$ operator. So $|x\rangle$ with $x=0$ actually means $|+\rangle$ and with $x=1$ actually means $|-\rangle$.

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