Summation over set notation

Question

I am reading some lecture notes on quantum computing and ran into the following expression that I cannot seem to understand:

For example, if we apply the Hadamard gate $H$ to each bit in a register of $n$ zeroes, we obtain $\frac{1}{\sqrt{2}} \sum_{j \in \{0,1\}^n} | j \rangle$, which is a superposition of all $n$-bit strings. More generally, if we apply $H^{\otimes n}$ to an initial state $| i \rangle$, with $i \in \{0,1\}^n$, we obtain \begin{equation} \tag{2.1} H^{\otimes n} | i \rangle = \frac{1}{\sqrt{2^n}} \sum_{j \in \{0,1\}^n} (-1)^{i \cdot j} | j \rangle, \end{equation} where $i \cdot j = \sum_{k=1}^n i_k j_k$ denotes the inner product of the $n$-bit strings $i, j \in \{0, 1\}^n$. For example: $$ H^{\otimes 2} | 01 \rangle = \frac{1}{\sqrt{2}} ( | 0 \rangle + | 1 \rangle ) \otimes \frac{1}{\sqrt{2}} ( | 0 \rangle - | 1 \rangle ) = \frac{1}{2} \sum_{j \in \{0,1\}^2} (-1)^{01 \cdot j} | j \rangle. $$

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In both of the summations, a set is implied but I do not see how it would work if it were expanded. Any help would be appreciated. Thank you.

Answer 1

$\{0,1\}^n$ denotes the set of all bit strings of length $n$. So, for instance, with $n=3$ and $i = 011$, we have $$ \begin{align*} H ^{\otimes 3}|i \rangle &= \frac{1}{\sqrt{2^3}} \sum_{j \in \{0,1\}^3} (-1)^{i \cdot j} \\ & = \frac{1}{\sqrt{2^3}}\left[(-1)^{(000)\cdot i}|000 \rangle + (-1)^{(001) \cdot i}|001 \rangle + \cdots + (-1)^{(111) \cdot i}|111 \rangle\right] \\ & = \frac{1}{\sqrt{2^3}}\left[|000 \rangle -|001 \rangle -|010 \rangle + |011 \rangle + |100 \rangle - |101 \rangle - |110 \rangle + |111 \rangle\right] \end{align*} $$

Answer 2

The sum in (2.1) is over all length-$n$ vectors with each entry $0$ or $1$. The symbol $i$ is also such a vector, so the dot product $i\cdot j$ in $(-1)^{i\cdot j}$ is the sum of values of $k$ for which $i_k=j_k=1$. In the unnumbered display-line equation below, the same thing happens with the special case $n=2,\,i_1=0,\,i_2=1$.