What is the conversion factor for qubits (qudits) to bits/bytes in classical information theory/computation theory?
I mean, how can we know how many "bits/bytes" process, e.g., a 60 qubit quantum computer (quamputer), are equivalent to classical bits (dits)?What about memories and velocities in "attaniable" quantum computers versus classical current memories and computer velocities(GHz)?
Quantum information is highly constrained compared to classical information, yet has extra degrees of freedom that in a sense are "worth more" than classical bits. But this is not merely that N qubits can store M bits:
It is sometimes useful to distinguish between the different quantum information resources in terms of (classical) bits, qubits and ebits (entanglement bits). A maximally entangled two qubit state (a Bell state) has 1 ebit: tasks requiring X ebits can be done with X or more Bell states but not fewer.
This extra value can be expressed through "Bennett's laws":
1 qubit $\succeq$ 1 bit (qubits can transmit bits)
1 qubit $\succeq$ 1 ebit (qubits can generate entanglement)
1 qubit + 1 ebit $\succeq$ 2 bits (superdense coding)
1 ebit + 2 bits $\succeq$ 1 qubit (quantum teleportation)
where $\succeq$ means "can do the job of". The no-signalling theorem rules out using one ebit to do the job of one bit, and the no-teleportation theorem rules out using any number of bits to do the job of one qubit.
The Holevo bound implies that amount of classical information that can be retrieved from $n$ qubits is just $n$ classical bits, despite the much larger amount of information in the qubits.
$n$ qubits in a computation allow computing up to $2^n$ function evaluations in parallel... but only $n$ bits, randomly sampled from a probability distribution due to the quantum computation done, can be retrieved at the end. So 60 qubits are at most worth $2^{60}$ operations but will only give you 60 bits of information at the end. They can also perform nonclassical tricks according to Bennett's laws.
As for speed, see the quantum speed limits. Basically, the fastest you can move between distinguishable quantum states is set by the energy you put into the system, not the number of bits.
I think one concept to grasp here is quantum parallelism. GHz refers to how fast a computer does 1 computation (A billionth of a second) and classic computers do 1 computation at a time (or a few in multi-core computers). However in a quantum computer multiple computations can be done at once in parallel.
Ultimately when you input Qubits into a system they are identical to regular bits... and when you measure the Qubits they are the same as bits... the interesting property about Qubits is the algorithms you can perform on them because of the quantum properties (assuming you can bring them back to a form where the measured results become useful bits).
Qubit (Qudit) equivalence with classical bits:
$1 \; qubit=2\; classical\; bits$ and $1 \; qudit=D\; classical\; dits $
$N \; qubits=2^N\; classical\; bits$ and $N \; qudit=D^N\; classical\; dits $
As 1 byte= 8 bits=$2^8\;bits$, or $1\;bit= 2^{-3}\;bytes=0.125\;bytes$, we have also
$N \; qubits=2^{N-3}\; classical\; bytes$ and $N \; qudytes=D^{N-3}\; classical\; dits $
Of course, one could define also the qubyte (8 qubits) and the qudyte avoid the -3 extra exponent and keep parallelism with classical information theory. It is trivial to get the equivalence with kilobytes.
The value of a qubit is, in effect, a complex number. To specify a complex number requires an infinite (and in fact uncountable) number of bits. So in terms of the information that's stored, a qubit is the equivalent of a large infinite number of bits.
You cannot, of course, extract all of that information. As Peter Shor points out in comments, Holevo's Theroem says that you can extract at most one bit.
