I know that $QFT_n|0\rangle$ is equivalent to $H_n|0\rangle$ (mathematical proof).

And it is also easy to prove that $QFT_1$ is equivalent to $H_1$ (applied to one QuBit).

From looking at the circuit below it seems clear to me that the gates should also be equivalent if $|x_1\rangle$ is in any state and all other QuBits are $|0\rangle$. This should be true because none of the controlled $R$ gates are applied to $|x_1\rangle$.

I do not know how to prove this mathematically. Can anyone provide an elegant proof?

from wikipedia 3

  • $\begingroup$ No my question is slightly different. I ask for a prove that x_1 qubit can be in any state, and qft is still equivalent to walsh. $\endgroup$ – birneee Aug 1 at 11:39
  • $\begingroup$ "This should be true because none of the controlled R gates are applied" is a mathematical proof (if you replace "should be" by "is"). $\endgroup$ – Norbert Schuch Aug 2 at 12:58

I think your explanation based on the circuit is perfectly adequate.

For a more rigorous "proof", why not simply take the output of the circuit? Substitute in $x_i=0$ for all $i$ and see that all the outputs are $(|0\rangle+|1\rangle)/\sqrt{2}$ for $i\neq 1$ and $(|0\rangle+(-1)^{x_1}|1\rangle)/\sqrt{2}$ for $i=1$, exactly as it would be for the Hadamard transform.

Alternatively, simply repeat the proof in the answer you cite for two inputs: $|0\rangle^{\otimes n}$ and $|1\rangle|0\rangle^{\otimes(n-1)}$. If it works for those two, by linearity it must work for any input on the first qubit.

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