An n-qubit pure state has 2n computational-basis amplitudes, so a dense full state-vector simulation stores 2n complex values. In contrast, n classical bits hold one definite bitstring at a time, selected from 2n possible bitstrings.
complex64 normally uses 8 bytes per complex amplitude; complex128 normally uses 16. The result excludes simulator, operating-system, gate-buffer, communication, and GPU overhead.
| Question | n classical bits | n-qubit pure state | n-qubit density matrix |
|---|---|---|---|
| Stored state | One definite n-bit string | 2n complex amplitudes, subject to normalization | 4n complex matrix elements, with physical constraints |
| Basis/configuration count | 2n possible bitstrings | 2n computational-basis vectors | Acts on a 2n-dimensional basis |
| Readout | The stored n-bit value | One sampled n-bit outcome per standard measurement | One sampled n-bit outcome per standard measurement |
| Dense classical storage | n bits for the current configuration | 2n ร bytes per complex amplitude | 4n ร bytes per complex element |
Accessible doubling table; complex128 values (16 bytes), binary units.
| Qubits | Basis amplitudes (2n) | Dense vector memory |
|---|---|---|
| 1 | 2 | 32 B |
| 2 | 4 | 64 B |
| 3 | 8 | 128 B |
| 10 | 1,024 | 16 KiB |
| 20 | 1,048,576 | 16 MiB |
| 30 | 1,073,741,824 | 16 GiB |
| 40 | 1,099,511,627,776 | 16 TiB |
| 50 | 1,125,899,906,842,624 | 16 PiB |
Variables: n = qubit count; A = number of pure-state amplitudes; M = dense-state bytes; b = bytes per complex value; R = available bytes. Every additional pure-state qubit doubles M; every additional density-matrix qubit multiplies M by four.
Amplitudes = 210 = 1,024.
Memory = 1,024 ร 16 B = 16,384 B = 16 KiB.
Amplitudes = 230 = 1,073,741,824.
Memory = 230 ร 16 B = 234 B = 16 GiB.
Memory = 240 ร 16 B = 244 B = 16 TiB.
One extra qubit would require 32 TiB.
Memory = 250 ร 16 B = 254 B = 16 PiB.
For comparison, 45 qubits require 512 TiB.
nmax = โlog2(16 GiB รท 16 B)โ = โlog2(230)โ = 30 qubits.
At an 80% usable setting, the calculator reports 29 as the safer planning maximum.
Practical warning: these exact values cover only the dense vector or matrix. A machine with exactly the theoretical amount of RAM generally cannot devote every byte to it.
No. A qubit has no fixed classical-bit equivalent. One qubit has two computational-basis states, but measurement returns one classical bit, and describing an arbitrary qubit state requires amplitudes rather than a fixed number of ordinary bits.
The computational-basis dimension is 2n: 1 qubit has 2 basis states, 2 have 4, 10 have 1,024, and 50 have 1,125,899,906,842,624. A pure state has one complex amplitude per basis state.
Each qubit has two computational-basis labels, 0 and 1. Combining n two-way choices gives 2 multiplied by itself n times, or 2n basis labels. Superposition changes the amplitudes, not the number of basis labels.
No. A standard measurement produces one n-bit outcome per shot. Quantum algorithms use interference to change outcome probabilities; exponential state-space dimension alone does not guarantee a speedup.
A dense 30-qubit pure state vector with complex128 amplitudes requires exactly 16 GiB for the vector: 230 ร 16 bytes. A real simulator generally needs additional memory.
The estimate covers only the dense quantum-state data. The simulator, operating system, gates, work buffers, copies, communication, and GPU allocations also consume memory, so reserving capacity is prudent.
Sometimes. Tensor-network methods can use much less memory for circuits with limited entanglement or favorable structure, but cost depends on circuit connectivity and entanglement and may grow quickly for generic highly entangled circuits.
A physical qubit is a hardware device. A logical qubit is an error-corrected encoded unit built from multiple physical qubits. This calculator's n is the number of simulated qubits in the mathematical state, not a hardware-overhead estimate.
