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authorPrefetch2024-07-17 10:01:43 +0200
committerPrefetch2024-07-17 10:01:43 +0200
commit075683cdf4588fe16f41d9f7b46b9720b42b2553 (patch)
treef200b341b35e9c89aa1030a2f6da94cd0e1e958b /source/know/concept/deutsch-jozsa-algorithm
parentc4d597e8d695eb145755464cffbf88a68fd0c88a (diff)
Improve knowledge base
Diffstat (limited to 'source/know/concept/deutsch-jozsa-algorithm')
-rw-r--r--source/know/concept/deutsch-jozsa-algorithm/index.md22
1 files changed, 11 insertions, 11 deletions
diff --git a/source/know/concept/deutsch-jozsa-algorithm/index.md b/source/know/concept/deutsch-jozsa-algorithm/index.md
index 44b06ad..223877a 100644
--- a/source/know/concept/deutsch-jozsa-algorithm/index.md
+++ b/source/know/concept/deutsch-jozsa-algorithm/index.md
@@ -72,8 +72,8 @@ $$\begin{aligned}
+ \frac{1}{2} \Ket{1} \Big( \Ket{0 \oplus f(1)} - \Ket{1 \oplus f(1)} \Big)
\end{aligned}$$
-The parenthesized superpositions can be reduced.
-Assuming that $$f(b) = 0$$, we notice:
+The parenthesized superpositions can be reduced:
+let us suppose that $$f(b) = 0$$, then:
$$\begin{aligned}
\Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
@@ -91,7 +91,7 @@ $$\begin{aligned}
\end{aligned}$$
We can thus combine both cases, $$f(b) = 0$$ or $$f(b) = 1$$,
-into the following single expression:
+into the following expression:
$$\begin{aligned}
\Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
@@ -106,8 +106,8 @@ $$\begin{aligned}
\frac{1}{2} \Big( (-1)^{f(0)} \Ket{0} + (-1)^{f(1)} \Ket{1} \Big) \Big( \Ket{0} - \Ket{1} \Big)
\end{aligned}$$
-The second qubit in state $$\Ket{-}$$ is garbage; it is no longer of interest.
-The first qubit is given by:
+The second qubit in state $$\Ket{-}$$ is garbage (i.e. no longer of interest).
+The first qubit is:
$$\begin{aligned}
\frac{1}{\sqrt{2}} \Big( (-1)^{f(0)} \Ket{0} + (-1)^{f(1)} \Ket{1} \Big)
@@ -126,8 +126,8 @@ $$\begin{aligned}
\end{aligned}$$
Depending on whether $$f$$ is constant or balanced,
-the mearurement outcome of this state will be $$\Ket{0}$$ or $$\Ket{1}$$
-with 100\% probability. We have solved the problem!
+the measurement outcome of this state will be $$\Ket{0}$$ or $$\Ket{1}$$
+with 100% probability. We have solved the problem!
Note that we only consulted the oracle (i.e. applied $$U_f$$) once.
A classical computer would need to query it twice,
@@ -146,7 +146,7 @@ This algorithm is then implemented by the following quantum circuit:
alt="Deutsch-Jozsa circuit" %}
There are $$N$$ qubits in initial state $$\Ket{0}$$, and one in $$\Ket{1}$$.
-For clarity, the oracle $$U_f$$ works like so:
+The oracle $$U_f$$ performs this action:
$$\begin{aligned}
\Ket{x_1} \Ket{x_2} \cdots \Ket{x_N} \Ket{y}
@@ -167,7 +167,7 @@ $$\begin{aligned}
Where $$\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$$ denotes a classical binary state.
For example, if $$x = 5 = 2^0 + 2^2$$ in the summation,
then $$\Ket{x} = \Ket{1} \Ket{0} \Ket{1} \Ket{0}^{\otimes N-3}$$
-(from least to most significant).
+(from least to most significant digit).
We give this state to the oracle,
and, by the same logic as for the Deutsch algorithm,
@@ -217,8 +217,8 @@ we only need to measure the $$N$$ qubits once;
$$f$$ is constant if and only if all are zero.
The Deutsch-Jozsa algorithm needs only one oracle query to give an error-free result,
-whereas a classical computer needs $$2^{N-1} + 1$$ queries in the worst case;
-a revolutionary discovery.
+whereas a classical computer needs $$2^{N-1} + 1$$ queries in the worst case.
+A revolutionary discovery!
## References