7
votes
Accepted
Parametrization of a two-qubit state
TL;DR: The logic behind the equation $|\psi\rangle=\cos\frac{\theta}{2}|0\rangle+e^{i\phi}\sin\frac{\theta}{2}|1\rangle$ can be iterated to obtain a real parameterization for states of arbitrary ...
- 18.2k
5
votes
Accepted
HHL Algorithm: How to compute the signs of the solution vector
Let us denote:
$$|x\rangle=\sum_ix_i|i\rangle$$
If it was possible to learn the sign of each $x_i$, then you would have a way to distinguish $|x\rangle$ and $-|x\rangle$. Since these states only ...
- 4,278
5
votes
Accepted
How to derive the state of a qubit after a partial measurement?
Your expression of the final state and your computation of the probability of measuring $0$ are both correct. We thus have to find the expression of $\left|\psi_f\right\rangle$ once the first qubit ...
- 4,278
4
votes
Accepted
Why is the conjugate being used when rewriting a qubit state in another basis?
$$\begin{align}
\left| a \right> &= \frac{\sqrt{3}}{2}\left| 0 \right> + \frac{i}{2}\left| 1 \right> \tag{A}\\
\left| b \right> &= \frac{i}{2}\left| 0 \right> + \frac{\sqrt{3}}{...
- 606
4
votes
Accepted
What is $HTHTH\left| 0 \right>$?
...
$$\frac{1}{2\sqrt{2}}[(1 - i + 2e^{i\pi/4})\left| 0 \right> + (1 + i)\left| 1 \right>]$$
...
However, the answer given at the back of the book (page 360) is (notice the $e^{i\pi/4}$ as ...
- 606
4
votes
Why does “discarding” a qubit register, in the Hidden Subgroup Problem, give a randomly chosen coset $|x+H\rangle$?
(Made CW as an expansion of some other other answers)
Understanding the answer to why quantum algorithms often ignore the second register is a bit of a pons asinorum test toward figuring out more of ...
4
votes
Accepted
Why does “discarding” a qubit register, in the Hidden Subgroup Problem, give a randomly chosen coset $|x+H\rangle$?
Let's start from the state (ignoring normalization)
$$
|\psi\rangle = \sum_{g \in G} | x\rangle |f(x)\rangle.
$$
There are two registers here: the "data" register that stores $x$ and the &...
- 658
3
votes
Why does “discarding” a qubit register, in the Hidden Subgroup Problem, give a randomly chosen coset $|x+H\rangle$?
That description is a bit imprecise. "Discard" really means "measure this register and discard the result". It's impossible to just "discard" registers in quantum ...
- 1,802
3
votes
Why for every state there is always a measurement that has a deterministic outcome?
How can we say this is true of all quantum states and gates – "Whatever state our quantum system is in, there is always a measurement that has a deterministic outcome.” Can someone explain this ...
3
votes
Accepted
Adding phases of two qubits
You've asked for a $2$-qubit quantum circuit, thus we are looking for a $2$-qubit unitary that performs the desired transformation. Note that if we take $\theta=0$, then $\frac{1}{\sqrt{2}}\left(|0\...
- 4,278
3
votes
What is $HTHTH\left| 0 \right>$?
I quickly computed in Mathematica. You are correct. The book is incorrect.
- 658
3
votes
Accepted
Where am I going wrong in my understanding of qubit associativity?
Looks like a simple error in the left-associative calculation:
$$
\left(
\begin{pmatrix}
0\\
1\\
\end{pmatrix}
\otimes
\begin{pmatrix}
1\\
0\\
\end{pmatrix}
\right)
\otimes
\begin{pmatrix}
0\\
1\\
\...
- 303
3
votes
Accepted
What are the possible $\beta$ in a qubit state $\frac{e^{i\pi/8}}{\sqrt5}|0\rangle+\beta|1\rangle$?
The exercise asks for a possible value for $\beta$, which you've given. So your answer is correct.
The fact that you get an answer different from the textbook comes from your first equation:
$$\left(\...
- 4,278
3
votes
Accepted
Fidelity concentration bound for random stabilizer states
No such bound holds for general $|\Phi\rangle$. The set $\mathcal{S}_n$ of $n$-qubit stabilizer states is finite, so $m=\min_{|\psi\rangle\in\mathcal{S}_n} |\langle\Phi|\psi\rangle|^2$ is well-defined ...
- 18.2k
2
votes
Calculate qubit state in terms of two states that are opposite points on Bloch sphere
a) First rewrite the equations in terms of $\left| 0 \right>$ and $\left| 1 \right>$ as follows (here is how to do so):
$$\begin{aligned}
\left| 0 \right> &= \frac{\sqrt{3}}{2}\left| a \...
- 167
2
votes
Quantum hardness of XQUATH conjecture
Usually, the complexity to compute output probabilities from quantum circuit sampling (considering that depth and number of gates is in poly($n$) obviously) depend on the number of samples required to ...
- 21
2
votes
Accepted
What is $H\left| \psi \right>$ with $\left| \psi \right> = \alpha\left| 0 \right> + \beta\left| 1 \right>$?
Your calculation is correct and that is probably a typo in the textbook, as already mentioned in the comments.
Moreover, if you write down your state as $| \psi \rangle = \alpha | 0 \rangle + \beta | ...
- 1,161
2
votes
What is $H\left| \psi \right>$ with $\left| \psi \right> = \alpha\left| 0 \right> + \beta\left| 1 \right>$?
Your result is correct. The answer you quote from the book is wrong and probably a typo. You can quickly check by plugging in α=0,β=1, in which case that answer results in $H|1\rangle = |+\rangle$. ...
- 5,610
2
votes
Accepted
Is it possible to use quantum state to store and read information without destorying it?
Reading information from a quantum system is possible only via a measurement, which mathematically can be described as applying a Hermitian operator (also called "observable") on the Hilbert ...
- 1,574
2
votes
How to derive the state of a qubit after a partial measurement?
@TristanNemoz's answer is perfect. I just want to emphasize an important point regarding the effect of the measurement of the ancillary qubit on the main register in the Hadamard test, which is also ...
- 787
2
votes
Upper bound on trace distance of subsystems based on full system
In general I would imagine not much. For simplicity consider the case where $p_1 = p_2$. Then the worst case would be that for all $(i,j) \neq (a,b)$ we have $\rho_{ij} = \sigma_{ij}$. Thus the entire ...
- 4,516
2
votes
Does applying an operator to a state have nothing to do with taking a measurement?
Both quantum gates and measurements can be represented as matrices. Every quantum gate corresponds to a unitary matrix, and every measurement corresponds to a Hermitian matrix. Some matrices happen to ...
2
votes
Recover local systems from composite systems
I think what you're looking for is the partial trace. Generally speaking, given two finite-dimensional Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, if $\text{L}(\mathcal H_1)$ and $\text{L}(\...
2
votes
How do I set a specific amplitude?
You can use RY-gate for this task.
$$\begin{align}\begin{aligned}\newcommand{\th}{\frac{\theta}{2}}\\\begin{split}RY(\theta) = \exp\left(-i \th Y\right) =
\begin{pmatrix}
\cos{\th} & -\...
- 7,109
1
vote
Accepted
Is there another parameterization of a qutrit 3-level system, besides Gell-Mann?
First, note that the direct answer to the question "Is there another parameterization of a qutrit 3-level system, besides Gell-Mann?" is a pretty obvious yes. There isn't anything ...
glS♦
- 21.2k
1
vote
Is there another parameterization of a qutrit 3-level system, besides Gell-Mann?
Updated answer after DaftWullie's comment:
I think I over-complicated; if $U$ is a random unitary in $SU(3)$, $U\vert0\rangle$ is good enough to get a random pure state, so you could get a random pure ...
- 11
1
vote
Distributive property of Ket over addition
If what the book meant was |a> + |b>; I believe it would just be element-wise addition. i.e. |a>+|b> = \begin{bmatrix}a_1+b_1\\a_2+b_2\end{bmatrix}
As well, to be more explicit, $$|0> + ...
1
vote
Accepted
How to measure an unknown state produced by a source of qubits?
I thought I'd elaborate a little on @Chris E's answer with some details about what kind of precision we can expect in our approximation as we increase the number of measurements we make. I'll be ...
1
vote
Accepted
Quantum algorithm to add zeroes between two halves of a vector
Yes, should be possible.
Using your notation, the state your starting with (on n+1 qubits) is:
$$|0,\psi_1,0 \rangle + |1, \psi_2, 0 \rangle $$
You can apply n-1 swap gates in the following order:
...
- 1,601
1
vote
Sorting two numbers using quantum computing
In a quantum circuit, a sorting step looks like this:
Or, in code:
let cmp = a > b
if cmp:
swap a, b
The main tricky thing here is that you can't discard ...
- 28.7k
Only top scored, non community-wiki answers of a minimum length are eligible