50 lines
1.8 KiB
Markdown
50 lines
1.8 KiB
Markdown
The first quantum algorithm we will look at is the Bernstein-Vazirani algorithm.
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Given a secret s $\in \{0,1\}^n$ and the function f : $\{0,1\}^n \rightarrow \{0,1\}$ defined as f(x) = x * s
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\* denotes the inner product of two bitstrings here
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For Bitstrings x and y of length n the inner product x * y is $x_1y_1 + ... + x_ny_n$ mod 2
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The goal is to find the secret s using as few queries of f as possible.
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So as few evaluations of f as possible.
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We will look at a quantum algorithm that will find s with only one evaluation of f.
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![[Pasted image 20260730144713.png]]
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The top wire consists of n qubits in state 0. $\ket{0}^n = \ket{0} \otimes ... \otimes \ket{0}$
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The bottom wire is in state 1.
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Both wires together are in state $\ket{0^n1} = \ket{0}^n \otimes \ket{1}$
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First, we apply the Hadamard gate on all qubits.
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The resulting state is calculated as follows:
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![[Pasted image 20260730145119.png]]
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We are now in superposition between all classical possibilities on the top wire
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and in $\ket{-}$ on the bottom wire.
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Next we apply the unitary $U_f$ on both wires.
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The Unitary is defined as:
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$U_f \ket{x,y} = \ket{x,y \otimes f(x)}$
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This unitary applies the function f to the bottom wire y.
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It can be calculated as follows:
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![[Pasted image 20260730153305.png]]
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And we can rewrite $U_f(\ket{x} \otimes \ket{-})$ as
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![[Pasted image 20260730154444.png]]
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The bottom wire has not changed.
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The top wire now has f(x) encoded in its quantum state.
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The phenomenon that the output of f is encoded as a -1 is called phase kickback.
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We need one final step before measuring.
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Now, we perform one final hadamard gate on the top wire.
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We want the result to be $\ket{s} \otimes \ket{-}$
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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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![[Pasted image 20260730155153.png]]
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