New answers tagged quantum-state
2
votes
Accepted
Representation of maximally entangled states of $2n$ qubits with Pauli matrices?
Once you've shown it for $n=1$ it follows for any $n$. This is because you can rearrange the expression for a $2n$-qubit maximally entangled state between a pair of subsystems $A$ and $B$ into a ...
2
votes
How to implement the Circuit of Steane's code for Quantum Error Correction?
Following this, we perform an algebraic verification of the 7-bit encoding and decoding:
Step 0:
$$(a|0\rangle + b|1\rangle)|000000\rangle $$
$$=a|0000000\rangle + b|1000000\rangle $$
Step 1:
$$a|...
1
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How to estimate the qubit frequency from a two-frequency Ramsey experiment?
When you drive a qubit off-resonantly, the Rabi rate, i.e. how quickly you're driving the transition, is given by square root of the sum of the square of the qubit frequency and the square of the ...
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Why are we only interested in the linear combination $|\phi\rangle = a |0\rangle + b |1\rangle$?
A qubit is defined as living in a 2-dimensional Hilbert space. This means one cannot define more than 2 linearly independent basis states. Since $|0\rangle$ and $|1\rangle$ are linearly independent (...
0
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Can we say what happens to phases after a reset?
The state you're starting with is:
$$|\psi\rangle=\frac{1}{3\sqrt{2}}(|00\rangle+2\mathrm{i}|01\rangle-3\mathrm{i}|10\rangle-2|11\rangle)$$
If you post-select on the state $|0\rangle\langle0|$, ...
2
votes
Accepted
Confusion regarding the definition of state of a qubit
The matrix $\phi$ you give is a density matrix representation of a qubit which is different to the statevector form you may be used to.
As you say the state of a qubit can be represented by a ...
0
votes
Accepted
Quantum algorithms and binomial distributions and probabilities
If you have a number $n$ of unentangled qubits and measure these qubits in the computational basis, then indeed the measurement statistics and the counts of the number of times you measure a spin-up ...
2
votes
Accepted
Can you distinguish between $|0\rangle, |1\rangle$, and $\frac{1}{\sqrt 2} (|0\rangle + |1\rangle)$?
If two quantum states $|\psi\rangle$ and $|\phi\rangle$ are such that $\langle\phi|\psi\rangle \neq 0$ (i.e. they are not orthogonal), then it will not be possible to determine which was given with ...
1
vote
Accepted
How is the depth of a circuit creating "Constant size vector states" $O(\log b)$
As mentioned in this answer, it is possible to perform each succession of rotation in parallel.
Let us suppose that we're at depth $k$ in our tree. If we were to implement the successive rotations as ...
1
vote
Accepted
Nonsensical projection of initialized qubits in Qiskit
It seems the Aer backend is not compatible with initializing using integers in this version, here's a small change to make it work: You can use a string to tell the circuit to initialize in the $|1\...
1
vote
Accepted
Is there a polynomial time mapping from 'Grover States' to an orthonormal set of vectors
You have an exponential amount of qubits and You want to be able to apply some gate to each one of them. We know an algorithm is efficient if it can be written as polynomially many universal gates. ...
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What is the conditional min-entropy of a pure bipartite state?
Reading again the paper you linked, I think the way the authors were thinking about the result was of showing this via the relations between conditional min- and max-entropies, see discussion at the ...

glS♦
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3
votes
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What is the conditional min-entropy of a pure bipartite state?
I'll use an equivalent definition of the min-entropy
$$
\begin{aligned}
H_{\min}(A|B) = - \log_2 \min& \quad \lambda \\
\mathrm{s.t.}& \quad \rho_{AB} \leq \lambda I_A \otimes \sigma_B \\
&...
1
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What is the conditional min-entropy of a pure bipartite state?
Remember that the max relative entropy satisfies, in general,
$$D_{\rm max}(\rho\|\sigma) = \log \inf\{\eta\ge0:\,\, \rho\le \eta \sigma\}.$$
In particular,
$$D_{\rm max}(\rho\|I\otimes \sigma)=\log \...

glS♦
- 21.7k
0
votes
Is a quantum state with coefficient 0 still there?
Do not confuse unitary operator and quantum operator in general.
Some quantum operators can transform a qubit state into $0$. For example, the Hamiltonian $\mathcal{H}$ is an operator, you can take ...
4
votes
Transformation of 0 state to superposition of 0 and + state with only using single qubit gates
I believe the following will work...
How did I come up with this? I realised that the state you were after, $|0\rangle+|+\rangle$ is a +1 eigenstate of the Hadamard gate. There's a standard circuit ...
1
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Is a quantum state with coefficient 0 still there?
The state after measuring 0 is $|0\rangle$ times an arbitrary nonzero complex number. In other words, it's the ray in Hilbert space that includes $|0\rangle$.
The number assigned to the measurement ...
2
votes
Transformation of 0 state to superposition of 0 and + state with only using single qubit gates
As given, that is a rather unusual and misleading way to describe a quantum state since $|0\rangle$ and $|+\rangle$ are not orthogonal. As a consequence the coefficients of the two, when multiplied by ...
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