7
Completeness ensures that Cauchy sequences have limits. This guarantees the existence of objects defined via limits of such sequences. These include derivatives, exponentials, trigonometric functions etc. Perhaps the most spectacular consequence of dropping the completeness assumption is the inability to describe state evolution.
State evolution
By the ...
5
There is a physical meaning to the complex amplitude.
The phase of the state affects what happing to the state.
For example, consider photon qubit, which might be located in a superposition of $2$ optic fibers (fiber $A$ is $|0\rangle$, fiber $B$ is $|1\rangle$) they might be in the same phase, they might be mirrors (phase $-1 / \pi / 180^\circ$) or phase $...
4
Following up on a comment in @gIS's answer, there is a particular sense in which the computational power of algorithms involving only real amplitudes is very much equivalent, up to a small overhead, as the power of algorithms involving both real and complex amplitudes.
I say this because it is known that Toffoli gates and Hadamard gates are sufficient for ...
4
I'm assuming you are asking specifically about the need for complex numbers in the context of a quantum algorithm written as a decomposition in terms of quantum gates.
If you are instead asking about the need for complex numbers more in general in quantum mechanics, the answer would be a bit different, depend on what precisely you mean with "need", ...
2
This is basically up to you: which elements are you transposing? If you're talking about transposing just the third system, then you'd be talking about
$$
|abc\rangle\langle xyz|\mapsto |abz\rangle\langle xyc|
$$
but you could do this on any of the three individual subsystems, or any of the three pairs of subsystems. Of course, if you're talking about doing ...
1
Yes, the reasoning is correct.
In fact, it can be generalized beyond pure states. By definition, every mixed quantum state $\rho$ is a positive semidefinite operator with unit trace. Since every positive semidefinite operator is Hermitian, we may interpret $\rho$ as an observable.
In this case, the expectation of observable $\rho$ in state $\sigma$
$$
\...
1
I'll consider vector spaces over $\mathbb R$ or $\mathbb C$.
In finite dimensions, an inner product space is automatically complete, and thus a Hilbert space, so no issues there.
More generally, one important feature of a Hilbert space $V$ is the isomorphism between $V$ and its (continuous) dual. This duality is what allows to work seamlessly in bra-ket ...
1
Explicit indices
The difficulty here arises from making indices implicit in tensor product expressions. For example, the unitary $U$ corresponding to controlled-NOT gate on qubits $1$ and $2$ and identity on qubit $3$ is often written down as
$$
U=\text{CNOT}\otimes I\tag1
$$
but a similar unitary $U'$ corresponding to controlled-NOT on qubits $1$ and $3$ ...
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