No. of bits in 160 qubits computer [duplicate]

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.?

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.
• @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. Jan 26 '21 at 9:35
• @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. Jan 27 '21 at 13:54