# Derivation for the result of performing the Hadamard transform on $|0\rangle^{\otimes n}$ being $2^{-n/2}\sum_x|x\rangle$

It's said that the result of performing the Hadamard transform on n qubits initially in the all |0> state is

$$\frac{1}{\sqrt{2^n}}\sum_x|x\rangle$$

where the sum is over all possible values of x.

I'm confused about what is meant by "the sum is over all possible values of x", what is x? I also wonder why the normalization can be generalized to $$\frac{1}{\sqrt{2^n}}$$? What's the proof and intuition for this?

$$H |0\rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$$
And we can do for $$N$$-qubit by applying each Hadmard gate to each qubit individually.
\begin{align} \overbrace{H|0\rangle \otimes H|0\rangle \otimes \cdots \otimes H|0\rangle}^{n \ \textrm{times}} &= \overbrace{ \bigg( \dfrac{|0\rangle +|1\rangle}{\sqrt{2} } \bigg)\otimes \bigg( \dfrac{|0\rangle +|1\rangle}{\sqrt{2} } \bigg) \otimes \cdots \otimes \bigg( \dfrac{|0\rangle +|1\rangle}{\sqrt{2} } \bigg) }^{n \ \textrm{times}} \\ &= \dfrac{1}{\sqrt{2^{n}}}\big( \overbrace{ |00\cdots0\rangle + |00\cdots1\rangle + \cdots + |11\cdots 1\rangle }^{2^n \ \textrm{terms} } \big)\\ &= \dfrac{1}{\sqrt{2^n}}\sum_{x=0}^{2^n-1} |x\rangle \end{align}