How instantaneous is state preparation in a quantum register, if all possible superpositions are to be initialized equally?

Question

Before the start of a quantum algorithm qubits need to be initialized into a quantum register. How fast can a quantum register of length $n$ be initialized in a way that all possible superpositions of the $N=2^n$ basis states exist with equal amplitudes.
I ask because I want to find out if there's a standard procedure to put a quantum register in an equal superposition of all possible states simultaneously. I imagine if there isn't a way to do this at once then it would take much longer to prepare say a quantum register of $N=2^{1024}$ basis states than that with $N=2^{256}$ basis states in an equal superposition.

KAJ226 · Accepted Answer

We know that giving a  single qubit  starting in the state $|0rangle$, which is a state one can initialize very fast with high fidelity, then we can put it in the superposition state $|psi rangle = dfrac{|0rangle + |1rangle}{sqrt{2}}$ by applying a Hadamard gate. That is,
$$H |0rangle = dfrac{|0rangle + |1rangle}{sqrt{2}}$$
And we can do for $N$-qubit by applying each Hadmard gate to each qubit individually. That is,
begin{align} overbrace{H|0rangle otimes H|0rangle otimes cdots otimes H|0rangle}^{n  textrm{times}} &= overbrace{ bigg( dfrac{|0rangle +|1rangle}{sqrt{2} } bigg)otimes  bigg( dfrac{|0rangle +|1rangle}{sqrt{2} } bigg) otimes cdots otimes bigg( dfrac{|0rangle +|1rangle}{sqrt{2} } bigg) }^{n  textrm{times}} 
&= dfrac{1}{sqrt{2^{n}}}big( overbrace{ |00cdots0rangle + |00cdots1rangle + cdots + |11cdots 1rangle }^{2^n  textrm{terms} } big)
&= dfrac{1}{sqrt{2^n}}sum_{i=0}^{2^n-1} |irangle
end{align}
You can do these $N$ operations of applying Hadamard gate to each qubit in parallel as applying a Hadamard to qubit 1 does not effect the state of qubit 2 and etc. You can see this parallelization in term of quantum circuit as well:

All these Hadamard gates can be and will be execute at the same time on the quantum processor.
Furthermore, each Hadamard gate can be executed pretty quick. In fact, if you execute a circuit with a Hadamard and do a measurement, the measurement process takes much much longer than the execution of Hadamard gate. See the pulse schedule for this particular circuit below: (The pink rectangular boxes indicate the measurement process)

Martin Vesely · Answer

You can prepare equal superposition by application of Hadamard gate on each qubit. The result will be state
$$
|qrangle=frac{1}{sqrt{2^n}}sum_{i=0}^{2^n}|irangle,
$$
i.e. the desired equally distributed superposition.
As Hadamard gate belongs to Clifford group, it acts quickly. It can be even simulated on classical computer in polynomial time. But the actual duration of Hadamard gate depends on particular physical realization of a quantum processor. Also the instantaneous depends on physical properties of a quantum processor.
But overall, preparation of equal superposition is quick in comparison with preparation of general state in which case you generally need non-Clifford gates and depth of a preparation circuit increases exponentially in number of qubits.

How instantaneous is state preparation in a quantum register, if all possible superpositions are to be initialized equally?

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