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I'm self-studying quantum computing and have gone through the Wikipedia article on the Berstein-Vazirani algorithm and believe that I understand it. I'm looking to verify my understanding and hope that someone could help me.

If I have an 8 bit secret of all 0s, would the oracle just be nothing? Alternatively, if it was all 1s, would my oracle CNOT each input with the output qubit?

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Yes, this is because the initial state will be starting as $|\psi_{init} \rangle = |000\cdots0\rangle = |0\rangle^{\otimes 8} $. Then you apply a layer of Hadamard gates, follow by the Oracle operation, then another layer of Hadamard gates, then finally the state of the 8 qubits of interest. Because the secret bitstring you have in mind are all $0$s, the Oracle would be nothing since this will allow the two layer of the Hadamard gates cancel each other out and keeping the state as $|\psi \rangle = |0\rangle^{\otimes 8}$. Now, upon measurement, you will extract out this very state... which is the secret bitstring you are interested in finding.

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If your secret bitstring is all 1's then you would want your final state to be $|\psi\rangle = |1\rangle^{\otimes 8}$. To do this, we can use something called phase kickback. The idea is simple but clever. First notice the following:

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This tells us, that if we want all $1$s then we will apply CNOT gates to each of the 8 qubits in your state to the ancilla qubit (which is in the state $\dfrac{|0\rangle - |1\rangle}{\sqrt{2}}$, which is created by apply $X$ gate follow by Hadamard gate) between the two layers of the Hadamard gates. Thus you will have something like this:

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Measuring this will return you the state of all 1's.

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With the Quantum Oracle $|x \rangle \xrightarrow{f_s} (-1)^{s\cdot x} |x \rangle$ (between the Hadamard sandwich) for the BV algorithm, with ($n$-bit) input secret bits $s$, here is what it does:

$|00\dots 0\rangle \xrightarrow{H^{\otimes n}} \frac{1}{\sqrt{2^n}} \sum\limits_{x\in \{0,1\}^n} |x\rangle \xrightarrow{f_s} \frac{1}{\sqrt{2^n}} \sum\limits_{x\in \{0,1\}^n} (-1)^{s\cdot x}|x\rangle \xrightarrow{H^{\otimes n}} |s\rangle$

When $s=00\ldots0$, $(-1)^{s.x}=(-1)^0=1$, s.t. the orale $f_s$ become an identity operator, as can be seen from above:

$\frac{1}{\sqrt{2^n}} \sum\limits_{x\in \{0,1\}^n} |x\rangle \xrightarrow{f_\textbf{0}} \frac{1}{\sqrt{2^n}} \sum\limits_{x\in \{0,1\}^n} (-1)^{\textbf{0}\cdot x}|x\rangle$,

i.e.,

$\frac{1}{\sqrt{2^n}} \sum\limits_{x\in \{0,1\}^n} |x\rangle \xrightarrow{f_\textbf{0}} \frac{1}{\sqrt{2^n}} \sum\limits_{x\in \{0,1\}^n} |x\rangle$,

so that the oracle is an identity mapping (transformation) and it has to do nothing.

We need to have a $CNOT$ gate whenever there is a $1$ bit in the secret bits in the Oracle, in order to introduce phase-kickback at the query bits, but don't have to do anything where the secret bit is $0$, which can be seen from below.

  • $|0\rangle \xrightarrow{H} |+\rangle$

  • phase kickback: $|+ \rangle \otimes |- \rangle \xrightarrow{CNOT} |- \rangle \otimes |-\rangle$, with $|+\rangle$ as the control qubit

  • $|-\rangle \xrightarrow{H} |1\rangle$

Hence, when $s=111\ldots 1$, we have the oracle as

$\frac{1}{\sqrt{2^n}} \sum\limits_{x\in \{0,1\}^n} |x\rangle \xrightarrow{f_\textbf{1}} \frac{1}{\sqrt{2^n}} \sum\limits_{x\in \{0,1\}^n} (-1)^{\textbf{1}\cdot x}|x\rangle$,

and we need to have $CNOT$ gates for all the secret bits in the oracle.

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