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I read in a book that (https://hub.packtpub.com/quantum-expert-robert-sutor-explains-the-basics-of-quantum-computing/)

160 qubits (quantum bits) could hold $2^{160} \approx1.46\times 10^{48}$ bits while the qubits were involved in computation."

How does this calculation come about?

The context of the statement is that a caffine molecule would require $10^{48}$ bits to be represented by a classical computer. However a quantum computer would require 160 qubits and is thus well suited for such representation.

If I look at this question on Quora, a 512 bit computer (which I suppose are real) would give a largest 155 digit number (https://www.quora.com/How-many-digits-are-in-a-512-bit-number). Isn't that big enough to represent atoms, molecules etc.?

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If you have $n$ bits you can combine them in $2^n$ different bit string (this come from combinatorics). Now take $n$ qubits. As any qubit can in superposition of two state, i.e. 0 and 1, $n$ qubits can be in superposition representing all $2^n$ possible bit strings.

The notion that $n$ qubits can hold $2^n$ classical bits is unfortunately misleading because when you measure the qubits, they will collapse to one particular state. This means that information content of $n$ qubits is $n$ classical bits.

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  • $\begingroup$ Thanks. I am unfortunately still unable to comprehend why a classical computer needs 10^48 bits while a quantum computer will need only 160 bits. Would you know something about that? $\endgroup$ Jan 25 at 11:11
  • $\begingroup$ This has to do with molecular simulation algorithms, I am no expert in this area, but Quantum Computers allow for easy molecular simulation, like they do Hamiltonian Simulation (it is the same thing), the Quantum computer is not 'magically more powerful' than a classical one, but because of the computational constructs available in Quantum Computing, the algorithms for molecular simulation scale linearly, and for classical computers they scale exponentially. $\endgroup$ Jan 25 at 19:19
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    $\begingroup$ @ManuChadha: Number of all bits combinations needed for the simulation is $2^{160} \approx 10^{48}$, so you need as many memory places. However, on quantum computer you can save all these combination into superposition state of 160 qubits. $\endgroup$ Jan 26 at 9:35
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    $\begingroup$ A bit represents a state. On its own, 1 or 0. 2 possible states. 2 bits, 00, 01, 10, 11. 4 states. 3 bits, 000, 001, ..., 111. 8 states. The number of possible states doubles every time you add a bit. Classically, each single state needs to be modeled explicitly by its own bit string (e.g. 1010110). In a quantum computer we can have a superposition of qubits, say, 3 qubits that models all of the (8) possible bit strings. This is done without having to explicitly assign 8 bits to each bit string. Instead, we have a probability (amplitude) associated with each bitstring modeled by just 3 qubits. $\endgroup$
    – Greenstick
    Jan 26 at 19:00
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    $\begingroup$ @ManuChadha: The problem is not in the bus width but memory size. $10^{48}$ bits is $10^{36}$ Tb (terra bits) and this is impossible to achieve. However, with quantum computer you need only 160 qubits in superposition to represent the simulation input. Of course, in the end you are left with only 160 classical bits representing the result (i.e. the optimum) but this is the case for classical computer too. $\endgroup$ Jan 27 at 13:54

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