Fix states and coefficients

Author: casellimarco

intro.Rmd

@@ -475,7 +475,7 @@ Now we apply $U_f$ using the first register as input and the ancilla register as
 $$\left(\frac{1}{\sqrt{2^n}}\sum_{x\in\{0,1\}^n}(-1)^{f(x)}\ket{x} \right)\ket{-}$$
 Now we apply $n$ Hadamard gates to the $n$ qubits in the first registers. Recalling lemma \@ref(lem:hadamard-on-bitstring), this gives the state
 
-$$\left(\frac{1}{2^n} \sum_{x\in\{0,1\}^n}(-1)^{f(x)} \sum_{j \in \{0,1\}^n }(-1)^{xj} \ket{j} \right) \ket{+} = \left(\frac{1}{2^n} \sum_{x\in\{0,1\}^n}\sum_{j \in \{0,1\}^n} (-1)^{f(x)+ xj} \ket{j} \right)\ket{+}$$
+$$\left(\frac{1}{2^n} \sum_{x\in\{0,1\}^n}(-1)^{f(x)} \sum_{j \in \{0,1\}^n }(-1)^{xj} \ket{j} \right) \ket{-} = \left(\frac{1}{2^n} \sum_{x\in\{0,1\}^n}\sum_{j \in \{0,1\}^n} (-1)^{f(x)+ xj} \ket{j} \right)\ket{-}$$
 
 In this state, note that the normalization factor has changed from $\frac{1}{\sqrt{2^n}}$ to $\frac{1}{2^n}$, and recall that $(-1)^{xj}$ is read as $(-1)^{ \sum_{p} x_pj_p \text{mod}2 }$.
 The key idea of the proof of this algorithm lies in asking the right question to the previous state: what is the probability of measuring the state $\ket{0}^n$ in the first register? The answer to this question will conclude the proof of this theorem. Before looking at the probability, observe that the amplitude of the state $\ket{j=0}$ we will see that it is just $\frac{1}{2^n}\sum_{x}(-1)^{f(x)}$, as $x^Tj=0$ if $j=0_1\dots 0_n$, for all $x$. Then,
@@ -518,7 +518,7 @@ The algorithm follows exactly the same steps as the Deutsch-Josza algorithm. The
 
 
 \begin{equation}
-\left(\frac{1}{2^n} \sum_{x\in\{0,1\}^n}(-1)^{f(x)}  \ket{x} \right) \ket{+} = \left(\frac{1}{2^n} \sum_{x\in\{0,1\}^n} (-1)^{a^T x}\ket{x} \right)\ket{+}
+\left(\frac{1}{\sqrt{2^n}} \sum_{x\in\{0,1\}^n}(-1)^{f(x)}  \ket{x} \right) \ket{-} = \left(\frac{1}{2^n} \sum_{x\in\{0,1\}^n} (-1)^{a^T x}\ket{x} \right)\ket{-}
 \end{equation}
 
 Now we resort again to Lemma \@ref(lem:hadamard-on-bitstring), and we use the fact that the Hadamard it is also a self-adjoint operator (i.e. it is the inverse of itself: $H^2 = I$). Thus applying $n$ Hadamard gates to the first register leads to the state $\ket{a}$ deterministically.