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It is known that in a classical computer we can't generate a purely random number by deterministic process. I am taking a course in Quantum Computing, recently I learnt that using state we can generate a deterministic process which will produce random number $|0\rangle$ and $|1 \rangle $. But, I am not aware of the physical implementation and very excited to know about it. I have searched online and got this paper, but this needs knowledge of architecture inside QC I think.

Can anyone explain(from basic please) how we can develop a deterministic process to generate $|0\rangle$ and $|1 \rangle $ random-ly? Also explain please how that process is a deterministic one.

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We start with the following qbit state:

$|\psi\rangle = |1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

Then, we apply the Hadamard gate to that qbit:

$H|\psi\rangle = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix}\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \end{bmatrix}$

The resulting state is also known as the $|-\rangle$ state.

We can measure this state in the $|0\rangle$, $|1\rangle$ basis (also called the computational basis) and it will collapse to $|0\rangle$ or $|1\rangle$ with the following probabilities:

$P[|0\rangle] = |\frac{1}{\sqrt{2}}|^2 = \frac 1 2$

$P[|1\rangle] = |\frac{-1}{\sqrt{2}}|^2 = \frac 1 2$

Assuming that quantum mechanics is indeed fundamentally probabilistic, this gives you a deterministic random number generator.

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  • $\begingroup$ Weird use of terminology, calling quantum indeterminism deterministic. $\endgroup$ – jknappen - Reinstate Monica Sep 10 '18 at 17:57
  • $\begingroup$ The process we follow to generate the random number is deterministic. I thought it best to answer the spirit of OP's question rather than delve into semantics. $\endgroup$ – ahelwer Sep 10 '18 at 18:02
  • $\begingroup$ Yes, from the terminology of automata theory or algorithms, this is perfectly deterministic. But the physical side isn't. $\endgroup$ – jknappen - Reinstate Monica Sep 10 '18 at 18:03

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