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 ...
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 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 &...
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\...
3 votes

What is $HTHTH\left| 0 \right>$?

I quickly computed in Mathematica. You are correct. The book is incorrect.
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\\ \...
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(\...
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 \...
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 ...
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 | ...
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} & -\...
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 ...
  • 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 ...
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 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

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