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?


1 Answer 1


First note that

$$H |0\rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$$

And we can do for $N$-qubit by applying each Hadmard gate to each qubit individually.

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Mathematically, this is equivalent to

\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}


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