5
$\text{Tr}(AB)$ is always real and non-negative if $A,B$ are positive semi-definite hermitian matrices.
To see this note that $A = UDU^\dagger$, for some unitary $U$ and diagonal matrix $D$ with $d_{ii} \ge 0$.
Then $\text{Tr}(AB) = \text{Tr}(UDU^\dagger B) = \text{Tr}(DU^\dagger B U).$
But $B^\prime = U^\dagger B U$ is also positive semi-definite and ...
4
In[13]:= H = 1/Sqrt[2]*{{1, 1}, {1, -1}};
T = {{1, 0}, {0, Exp[I*Pi/4]}};
CNOT = {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}};
KroneckerProduct[IdentityMatrix[2], T].CNOT.
KroneckerProduct[T,ConjugateTranspose[T]].CNOT // MatrixForm
{"1", "0", "0", "0"},
{"0", "1", "0", "0"},
{"0", "0", "1", "0"},
{"0", "0", "0", "I"}
So that red boxed ...
4
When using a simulator, it doesn't really matter what kind of qubit you refer to. You can even mix-and-match the types. The type of qubit only becomes relevant when you intend to run on a device, because devices have qubits at specific locations.
For example, if you wanted to run on Bristlecone, you would limit yourself to GridQubit instances that actually ...
4
I think this question makes two assumptions:
the system does not evolve intentionally (i.e., no gates or measurements are performed on it), and
there is no noise that would cause the system to evolve in an unintended way.
Of course, if you change the state of the system by applying some gates, or if the system is noisy, this will not be the case.
3
The Solovay-Kitaev algorithm is not practical. It is very useful theoretically because it proves that once you have a "dense" set of quantum gates (i.e. a set with which you can approximate any other quantum gate) you can approximate up to an arbitrary precision and quickly any quantum gate.
In practice, the Solovay-Kitaev works as follow:
Fill the space ...
3
A density matrix $\rho$ on two qubits has 16 complex amplitudes (although not all are free variables due to constraints from normalization and Hermeticity), so the City plot is showing those amplitudes as well. The $|00\rangle\langle 11|$ and $|11\rangle\langle 00|$ amplitudes shown are not going to directly impact your measurement if you were to measure in ...
2
There is no general exact formula for $N(\epsilon, d)$ and some special cases (for example SIC-POVM) is an area of active research.
However there is a Welch bound that gives $\epsilon^2 \ge \frac{n-d}{d(n-1)}, n=N(\epsilon, d)$ and hence bounds $N(\epsilon, d)$ from above.
2
This seems like it should be a known mathematical property of Hilbert spaces, but I can't immediately lay my hand on any such result. In lieu of that, this is very far from an answer to your question, but it perhaps indicates the difficulty of (some of) what you're asking...
First, perhaps we can clarify your problem statement. I assume you mean
$$
|\langle ...
2
Short answer. The trace distance between two states more or less determines how distinguishable they are by any operational means. A trace distance of 0 means that they are indistinguishable (because they're equal); a trace distance of 2 indicates that they can be perfectly distinguished in principle.
Long answer. We will show how, from the objective of ...
2
One way to understand the trace distance is to notice that it equals the (classical) trace distance (also referred to as Kolmogorov distance, see this post for some information about it) maximised over all possible POVMs on the states.
To see this, start from the following expression for the trace distance:
$$D(\rho,\sigma)\equiv\frac{1}{2}\mathrm{Tr}|\rho-\...
2
So probability of the second qubit being in state $|1\rangle$ is the probability of the 5 qubit system being in a state that has $|1\rangle$ as the second qubit.
So among all the 32 states, find the ones that have $|1\rangle$ in the second qubit, which will be half of them, for example $|01100\rangle$ and $|11111\rangle$. Add up the corresponding ...
2
Forget that you know that non-orthogonal states cannot be distinguished, as this is what you're trying to prove. It's like a proof by contradiction.
The question is talking about a hypothetical device that could, if it existed, distinguish between $|\psi\rangle$ and $|\phi\rangle$. Assume it exists, and prove that it gives you cloning (I know which state I ...
2
No, $|\psi\rangle$ and $|\phi\rangle$ are non-orthogonal by assumption, they can't be effectively (whatever this means) orthogonal. The device just gives us the answer $\phi$ or $\psi$ (you can think it returns label and destroys the input state), so we can prepare $|\psi\rangle$ or $|\phi\rangle$ separately (or many copies of it). It is that trivial.
1
On one hand, the oracle can use an arbitrary subset of the register qubits to decide whether it marks the state as the solution to the encoded problem. For example, if you're looking for any 4-qubit state with the first 2 qubits in state $|11\rangle$, you can do a CCNOT with those qubits as controls and the marked qubit as target to implement this condition -...
1
Doing a 5 to 2-qubit reduction is a little tedious, but we can illustrate how it works with a simpler example.
Let's take a 3-qubit state $\lvert\psi_{ABC}\rangle$ and boil it down to $\lvert\psi_{BC}\rangle$, say
$$\lvert\psi_{ABC}\rangle = \alpha\lvert 001\rangle + \beta\lvert 101\rangle + \gamma\lvert 011\rangle.$$
First of all, the density operator ...
1
I know that I should write states using the creation operator and $|00⟩$, and then let $U$ act on creation operators.
This is exactly right.
Writing the state in terms of creation operators gives $$(\alpha|0\rangle+\beta|1\rangle+\gamma|2\rangle)|10\rangle = \left(\alpha + \beta a_1^\dagger + \gamma a_1^\dagger a_1^\dagger\right)a_2^\dagger\left|000\right&...
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