- Input, output, program and memory are qbits.
- Any computation (step) can be represented by a unitary transformation of the computer as a whole.
- Any computation is reversible. Because of the unitarity of the quantum evolution operator, a deterministic computation can be performed by a quantum computer if and only if it is reversible, i.e., if the program does not involve "deletion" of information or "many-to-one" operations. Only one-to-one operations are allowed. Compared to classical irreversible computation, this may result in a space and time overheads. Furthermore, no "one-to-many" operations are allowed. Thus, unless classical, qbits cannot be copied.
- Unless classical,qbits are context-dependent. That is, their value may depend on the method by which they have been inferred, and on the co-measured qbits.
- Measurements may be carried out on any qbit at any stage of the computation. But, unless classical, a qbit cannot be measured by a single experiment with arbitrary accuracy. The computation process and the measurement have to be repeated in order to obtain sufficient statistics. Any such single measurement will yield merely a "click" on some counter, from which information about the qbit state must be inferred. Thereby, any single measurement is indeterminate and coherence is destroyed. Therefore, it seems more proper to realize that there is no such operational concept of "a single qbit." Because of complementarity, single qbits cannot be determined precisely. What is henceforth called "determination" or "measurement" of a qbit is, in effect, the observation of a successive number of such qbits, one after the other, from "similar" computation processes (same preparation, same evolution). By performing these measurements on "similar" qbits, one can "determine" this qbit within an epsilon-neighborhood only. The parameter epsilon depends on the number of successive measurements made.
- Quantum parallelism: during a computation (step), a quantum computer proceeds down all coherent paths at once. If managed properly, this may give rise to speedups.
- Any subroutine must not leave around any qbits beyond it's computed answer, because the computational paths with different residual information can no longer interfere.

Shor's paper

Svozil's review paper