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?


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.

enter image description here

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.