In its report Quantum Decade, IBM claims that we need 512 classical bits to represent two-qubits entangled state (see graphic on pg. 2 in the report). This seems a little bit odd to me. Lets have a general two-qubits quantum state $|\psi\rangle = \alpha|00\rangle + \beta|01\rangle + \gamma|10\rangle + \delta|11\rangle$ where probability amplitudes are complex numbers. Assume that the amplitudes are expressed with two real values each, i.e. 8 real numbers in total. The only way how to come to 512 bits is using 64 bits real numbers (double precission) on a classical computer.

I understand that the report is intended for managers rather than experts in computers sciences and quantum computing but still this seems a little bit misleading as the number of classical bits required depends on a precission of a quantum computer simulation (for single precission we would need only 256 bits), moreover, the memory requirements hold for any two-qubits state, not only entangled ones.

On top of that, IBM also claims in the report that

Entanglement means the combined state of the qubits contains more information than the qubits do independently.

If we have for example Bell state and measure both qubits, we are left with two classical bits. So, there is the same amount of information as in two classical bits. Only difference is that in entangled state we can infer state of other qubits from measuring only few in the state but in the end we have same amount of information.

Is my understanding right and the report is more marketing material or am I missing something important?


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