1.9 KiB
The first quantum algorithm we will look at is the Bernstein-Vazirani algorithm.
Given a secret s \in \{0,1\}^n and the function f : \{0,1\}^n \rightarrow \{0,1\} defined as f(x) = x * s
* denotes the inner product of two bitstrings here
For Bitstrings x and y of length n the inner product x * y is x_1y_1 + ... + x_ny_n mod 2
The goal is to find the secret s using as few queries of f as possible.
So as few evaluations of f as possible.
We will look at a quantum algorithm that will find s with only one evaluation of f.
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The top wire consists of n qubits in state 0. \ket{0}^n = \ket{0} \otimes ... \otimes \ket{0}
The bottom wire is in state 1.
Both wires together are in state \ket{0^n1} = \ket{0}^n \otimes \ket{1}
First, we apply the Hadamard gate on all qubits. The resulting state is calculated as follows:
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We are now in superposition between all classical possibilities on the top wire
and in \ket{-} on the bottom wire.
Next we apply the unitary U_f on both wires.
The Unitary is defined as:
U_f \ket{x,y} = \ket{x,y \otimes f(x)}
This unitary applies the function f to the bottom wire y.
It can be calculated as follows:
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And we can rewrite U_f(\ket{x} \otimes \ket{-}) as
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The bottom wire has not changed. The top wire now has f(x) encoded in its quantum state. The phenomenon that the output of f is encoded as a -1 is called phase kickback. We need one final step before measuring.
Now, we perform one final hadamard gate on the top wire.
We want the result to be \ket{s} \otimes \ket{-}
To check, we calculate (H^{\otimes n})^\dagger \ket{s} \otimes \ket{-} and check if it is equal to \psi_2 :
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Because HH^\dagger = I we now can measure s on the top wire.