Prove that QFT and Walsh-Hadamard gates give the same output when acting on $\lvert x\rangle\lvert 0\rangle$ [duplicate]

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?

• 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. – birneee Aug 1 at 11:39
• "This should be true because none of the controlled R gates are applied" is a mathematical proof (if you replace "should be" by "is"). – Norbert Schuch Aug 2 at 12:58

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.