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## Hot answers tagged probability

7

Measurement postulate The statement you are asking about is a postulate of quantum mechanics, so it cannot be mathematically derived from other facts in the theory. Instead, it is justified by its agreement with observation in the sense that it allows us to mathematically describe every measurement we can perform in practice. Intuitive interpretation The ...

4

Just to note that the complex number also specify a quantum phase of a qubit. In this particular case $$\frac{1+i}{2} = \frac{1}{\sqrt{2}}\mathrm{e}^{i\frac{\pi}{4}},$$ so the relative phase is $\frac{\pi}{4}$. You can rewrite your qubit state as $$|q\rangle = \frac{1}{\sqrt{2}}(|0\rangle + \mathrm{e}^{i\frac{\pi}{4}} |1\rangle).$$ Now, you easily see ...

4

You forgot to take the absolute value. The Born rule for computing measurement outcome probability from the state vector amplitudes says that the probability is the square of the magnitude of the amplitude. In your case, we get $$\left(A_1\right)^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}$$ $$\left(A_2\right)^2 = \left|\frac{i+1}{2}\right|^2 = \... 4 This only holds if the two distributions are independent. In this case$$ \begin{aligned} H_{\beta}(p \times q) &= \frac{1}{1-\beta} \log\left( \sum_{i,j}(p(i) q(j))^{\beta} \right) \\ &= \frac{1}{1-\beta} \log\left( \left(\sum_{i}p(i)^{\beta}\right) \left(\sum_jq(j)^{\beta}\right) \right) \\ &= \frac{1}{1-\beta} \left(\log \left(\sum_{i}p(i)^{\...

4

Since the classical algorithm samples "uniformly and independently $𝑘$ times from the search space", equation $(5)$ should be, $P(X_i=1, X_j=1)= P(X_i=1)P(X_j=1)=\frac{M^2}{N^2}$ instead. If you substitute $(5)$ with this, you would arrive at the book's standard deviation.

4

In the following, I'll show the evaluation of the probability densities of the transition probabilities: $|\langle \psi | z\rangle^2$ and their pairwise independence. I didn't work out the full mutual independence. The $n$-qubit pure states span the complex projective space $CP^{N-1}$ with $N=2^n$. Pure $n$-qubit states can be parametrized almost everywhere ...

4

These are classical conditional probabilities used extensively in Bayesian probability. Suppose Alice can prepare any of a number of pure states $|\psi_i\rangle$. She chooses $i$ randomly from distribution $p(i)$, prepares a system in state $|\psi_i\rangle$ and gives it to Bob without telling him the random choice $i$. Suppose also that Bob knows the ...

4

First, recall that $\mathrm{tr} A = \sum_i \langle i|A|i \rangle$. Each equation is then a sum where all terms are products of $P(z)$ and three other quantities. Further, the sum in the first equation ranges over a single index suggesting that all matrices under the trace are diagonal. In fact, since we are working with a composite system this also suggests ...

3

Note that if $p_{i}q_{i} = 0\,\,\forall i$, then for all $i$ either $p_{i} = 0$, $q_{i} = 0$, or both are $0$. Divide $\{i\} = \{1,\ldots,N\}$ into those $i$ for which these three different things happen: $\{N_{p_{i}}\} = \{i|p_{i} = 0\}$, $\{N_{q_{i}}\} = \{i|q_{i} = 0\}$, $\{N_{pq_{i}}\} = \{i|p_{i} = q_{i} = 0\}$. Then \begin{equation} \begin{split} \|P-...

3

The issue that easily leads to confusion is the dual role played by output bitstring probability. It enters the computation of the average in two ways. On one hand, it determines how often one sees different bitstrings. On the other hand, it determines the contribution that each bitstring makes towards the average. In mathematical terms, the output bitstring ...

3

As far as I'm aware there isn't much of a meaningful connection. The corresponding entropy for $D_{\max}$ is the min-entropy (written $H_{\min}$ or $H_{\infty}$). It measures a sort of `worst case' uncertainty whereas the Shannon or von Neumann entropies measure an average uncertainty. To answer your first question: the quantum relative entropies or ...

3

First of all, that does not imply anything for shorter (constant/logarithmic) depths. Moreover, the 2-design property does not imply that the outcome distribution is the same as for Haar-random unitaries, but only the first and second moment is. In the mentioned paper, they consider anti-concentration of the outcome distribution. To show this feature, a ...

3

The optimal probability of guessing correctly is $$\frac12 + \frac12 \Big\|\frac23 \rho_0 - \frac13 \rho_1 \Big\|_1$$ where $\| X \|_1 = \mathrm{Tr}[\sqrt{X^* X}]$ is the Schatten 1-norm. This success probability is achieved by the POVM with operator $$E_0 = \Pi_{[\tfrac23 \rho_0 - \tfrac13 \rho_1]_+} \qquad E_1=I-E_0$$ where $[X]_+$ denotes the positive ...

3

Given a description of $U_x$ you can efficiently find a decription of $\textbf{B}$ and $\phi$ by iterating over the gates of $U_x$ and adding one "free" variable every time you have a Hadamard gate and imposing the constraint of the Toffoli gate. Regarding the approximation, you can estime $f, g$ to additive error $\frac{2^h}{\text{poly}(h)}... 3 The authors are certainly thinking about finite frames. In this case, your statement is correct, since the number of elements in every spanning set is at least the vector space dimension. As glS already pointed out, the frames could be infinite: for countable frames, the summation is then meant as convergence in the$L^2$-sense (i.e. w.r.t. Hilbert-Schmidt ... 2 The statement made in the research paper is right. The initialization of the$\psi$state is more than one qubit. For n qubits, it's state is in$2^n$dimensions. Coming back to your question, compute your state with the operator$G+A$, i.e.$\frac{1}{2}(G+A) |\psi\rangle=|\phi\rangle$, for some state$|\phi\rangle$. Now the probability of$|\phi\rangle$is$...

2

\begin{align} CNOT |01\rangle &= CNOT \big( |0\rangle \otimes |1\rangle \big) \\ &= \big( |0\rangle \langle 0| \otimes I + |1\rangle \langle 1| \otimes X \big) \big( |0 \rangle \otimes 1\rangle \big) \\ &= \big( |0\rangle \langle 0| \otimes I \big)|0\rangle\otimes|1\rangle + |1\rangle \langle 1| \otimes X \big) |0 \rangle \otimes |1\rangle \\ &...

2

I think there might be a confusion between a Boolean formula (which may be easy to evaluate) and the number of solutions or a difference in the number solutions of the Boolean formula (which may be very difficult to evaluate). You're calling $f$ and $g$ #P-functions, but really you're interested in $\#(0)$ and $\#(1)$ which are values that count the number ...

2

A measurement can be associated with a decomposition of the identity $I$ into a sum of projectors. For example, the measurement of a qubit in the $| + \rangle ,| - \rangle$ basis is associated with the decomposition $$I = P_+ + P_- = |+\rangle\langle+| ~+~ |-\rangle\langle-|.$$ Then the probabilities are $\text{Tr}(P_+|\phi\rangle\langle\phi|) = |\langle\... 2 The two facts are connected in that they both arise as a result of rotational invariance of the Haar measure. We will derive them in the case of large$n$since this is when the Porter-Thomas distribution takes the exponential form given in the question. Also, this case admits an intuitive proof backed by a geometric picture. For small$n$, Porter-Thomas ... 2 Let$ p_i = |\langle i | \phi \rangle|^2 \sim Dir(a_1, .., a_{2^n}) = Dir(1, .., 1) $and$ m_i $the occurences of outcome$ |i\rangle $on samples$z_1, .. z_k$. Since the Dirichlet distribution is the conjugate prior of the categorical (see here), meaning$ \bf{p} | Z, (1, .. 1), \bf{m}  \sim Dir(2^n, \bf{m} + 1$) and using the formula for the ... 2$U-I$is a normal matrix so$||U-I||_{op}$is its eigenvalue with the largest magnitude. The eigenvalue equation for this matrix is $$(U-I)|\psi\rangle=\lambda|\psi\rangle,$$ so $$|\lambda|^2=(\cos\phi-1)^2+\sin^2\phi=4\sin^2\frac{\phi}{2}\Rightarrow |\lambda|=2\left|\sin\frac{\phi}{2}\right|,$$ where$e^{i\phi}$is some eigenvalue of$U$. Now, the ... 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 There is no a way in Qiskit to get the results in that format. So, you will have to go for the pure Python way: counts = {'0000': 66, '0001': 71, '0010': 68, '0011': 70, '0100': 77, '0101': 64, '0110': 64, '0111': 51, '1000': 52, '1001': 67, '1010': 43, '1011': 64, '1100': 61, '1101': 59, '1110': 73, '1111': 74} shots = 1024 ... 2 You'll place the phase within the CRz gate. The approach you've taken essentially argues that: $$e^{it H_3} \approx e^{it \alpha X_1 \otimes Y_2} e^{it \beta Z_1 \otimes Z_2}$$ So, when you're applying the Rz gate, you can select the$it \alphacoefficient to align with the necessary phase (likely you'll use\theta = -2 t \alpha$, depending on the ... 2 Ok let's break this down. Firstly the success probability for QPE and QAE are defined slightly differently. With QPE there are two error bounds$|\hat{x} - x| < \epsilon$to consider$\epsilon < \frac{1}{2^{n+1}}$with probability$ \geq \frac{4}{\pi}$, or$\epsilon < \frac{1}{2^{n}}$with probability$ \geq \frac{8}{\pi}$. We can use the results of ... 1 If we have the state$|\psi \rangle = a_{00}|00\rangle +a_{01}|01\rangle +a_{10}|10\rangle +a_{11}|11\rangle $then the probability of the second qubit being in the state$|1\rangle$is the probability of the state$|\psi \rangle$having$|1\rangle$on the second qubit. In this case, it is from the states$|01\rangle$and$|11\rangle$. So The probability of ... 1 The covariance matrix is a function of the expectation values of powers of position and momentum associated to some state in a continuous-variable system.$\$ \mathbf{\sigma} = \begin{pmatrix} \langle\hat{x}^2\rangle & \frac{1}{2}\langle \{ \hat{x} ,\hat{p}\} \rangle \\ \frac{1}{2}\langle \{ \hat{x} ,\hat{p}\}\rangle & \langle\hat{p}^2\rangle\end{...

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