diff options
Diffstat (limited to 'source/know/concept/deutsch-jozsa-algorithm/index.md')
| -rw-r--r-- | source/know/concept/deutsch-jozsa-algorithm/index.md | 22 |
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 |