Can we prepare a state regarding a transformation in quantum computing that seems to generate another equal amplitude superposition state when applying a Hadamard gate?

Specifically, I observed that starting with an equal complex amplitude superposition of all possible state and applying a Hadamard gate resulted in another equal amplitude superposition state with different phases.

$$\frac{1}{\sqrt{2^n}}\sum_{x=0}^{2^n-1}a_{x}\left|x\right>\xrightarrow{\mathbb H^{\otimes n}}\frac{1}{\sqrt{2^n}}\sum_{y=0}^{2^n-1}b_{y}\left|y\right>$$

I would like to discuss how can we prepare the initial state and explore its potential applications and implications for quantum algorithms.

  • 1
    $\begingroup$ Assuming $a,b$ are imaginary numbers, this relation doesn't hold for $|+\rangle$, i.e. $H|+\rangle = H(|0\rangle + |1\rangle)/\sqrt{2} = |0\rangle$. $\endgroup$
    – forky40
    Commented Apr 20 at 15:55
  • $\begingroup$ @forky40 , I've tested my state up to n=8 (N = 256). Since my PC is somewhat outdated, it can't simulate more than that. However, I have a doubt; maybe for some n, it may fail to give the required equal superposition. I'd like to know the benefits of such a state, as someone might be looking for this kind of state for a specific purpose. That's my intention. $\endgroup$
    – Aman
    Commented Apr 20 at 16:25
  • 2
    $\begingroup$ a and b are just global phases, since it's the same term for all x. This means the left term and the right term are both an equal superposition of all basis states on n qubits. Applying Hadamard gates to each qubit in such a state yields an all-zero state, so the equality doesn't hold. $\endgroup$ Commented Apr 20 at 19:26
  • $\begingroup$ A state that is an equal superposition of all basis states is a well-known thing in quantum algorithms - Deutsch-Jozsa and Grover are just two examples that come to mind - but it's also not challenging to prepare, just apply Hadamards to an all-zero basis state. $\endgroup$ Commented Apr 20 at 19:27
  • $\begingroup$ @MariiaMykhailova, I am so sorry my mistake when writing the equations a and b are not global phases, for each components they have their own phases $\endgroup$
    – Aman
    Commented Apr 21 at 11:20


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