D-Wave Computing Power

Computer Science Level 2

As of 2016 D-Wave's newest device can supposedly manipulate about 1000 quantum states simultaneously. About how many classical bits does this correspond to?

10^{300}

10^9

1000

10^{100}

2 solutions

Matt DeCross
Jan 20, 2016

To represent $n$ quantum states requires $2^n$ classical bits. If D-Wave's devices really manipulate $1000$ quantum states, this would contain the information of $2^{1000} \approx 10^{300}$ classical bits.

This assumes solely linear relationships be confined linearly to each successive binary. While this may be an actual mechanical constraint, it was undefined in the problem. Consideration of full field effects provides for more robust answers.

DrA Trimpi - 3 years, 4 months ago

The above comment does not parse for me. The problem is just a rough estimate of the classical information content of a qubit, which of course faces innumerable technical challenges in implementation. One way of thinking about the exponential scaling comes from the amount of classical bits versus qubits needed to simulate a quantum system of n particles.

Matt DeCross - 3 years, 3 months ago

I have a doubt. n electrons can have 2^n states. So you need 2^n classical bits. If a system can have 2^n states then you need 2^n classical bits. So if the device can manipulate 1000 states, then doesn't that mean you just need 1000 classical bits?

Rohit John - 11 months, 2 weeks ago

Yeah, this is kind of an issue in the wording. I should have been more careful to specify 1000 qubits rather than 1000 states. Anyway, the relevant wiki (Quantum Computing) and the problem drastically needs updating since many developments have occurred since 2016 including a much improved-understanding of what D-Wave was actually doing.

Matt DeCross - 3 months, 1 week ago

Niraj Venkat
Feb 27, 2021

Each qubit resides in a complex vector space $\mathbb{C}^2$ which is spanned by two basis states $|0\rangle$ and $|1\rangle$ . The number of states represented by 1000 qubits is thus $2^{1000}$ .

The options are in base-10, so to find the corresponding exponent $x$ is simply a change of base:

$2^{1000} = {10}^x \Rightarrow x = \frac{1000}{\log_2{10}} = 301.03 \approx 300$

D-Wave Computing Power

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