A lot of quantum algorithms start from the uniform superposition state, and then do some finagling to transform that state into the one they cared about. I was wondering if there is any advantage from starting from another "easy to prepare" state (i.e. O(1) circuit depth), that has a closer total variational distance from the state you wanted to prepare?

For example, if I wish to prepare a Gaussian distribution on N states, starting from the uniform superposition, one would would have to "move" amplitudes from the states closer to 0 and N, onto the states near "the middle" states around N/2.

So to me, a natural question to ask is, does this task become easier if you start from a better initial state that's also very efficiently preparable?

Is there a classification of what states can be prepared with only a constant circuit depth without ancilla?

  • 1
    $\begingroup$ If that better initial state is efficiently preparable (e.g. from all zeros), wouldn't you just make that preparation the first part of your circuit? $\endgroup$
    – DaftWullie
    Commented Sep 22, 2023 at 9:07
  • $\begingroup$ Yes, that’s exactly the point. I just don’t know when that is the case! $\endgroup$
    – Cuhrazatee
    Commented Sep 22, 2023 at 17:53

1 Answer 1


Suppose you have a real function $f$ such that the state you want to prepare is: $$\newcommand\ket[1]{\left|#1\right\rangle}\ket{\psi_f}=\sum_xf(x)\ket{x}$$ (up to normalisation). We also assume $f$ to be differentiable.

Zylberman and Debbasch proved that every such state on $n$ qubits can be prepared with fidelity at least $1-\varepsilon$ in depth $\mathcal{O}\left(\frac{1}{\sqrt{\varepsilon}}\right)$ with probability of success $\Theta(\varepsilon)$. So this, already accounts for the Gaussian distribution for instance.

The authors claim that their scheme can be implemented for complex states and even for non-differentiable functions, though I'm not sure that they claim the efficiency of this scheme carries on to these cases.


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