# Tag Info

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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 ...

6

Given $\rho$ and a fixed ensemble $\{ |\psi_i \rangle \}$ it might not be possible to write $\rho$ as $\sum_i p_i |\psi_i \rangle \langle \psi_i |$. For example, let $| + \rangle = \frac{1}{\sqrt{2}} (| 0 \rangle + | 1 \rangle )$. Then the state $|+\rangle \langle + |$ cannot be expressed as a convex combination in the ensemble \{ | 0 \rangle, |1\rangle \... 5 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)^{\... 5 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 ... 5 "I understand that quantum physics supports the concept that the probability of a qubit collapsing into (say) 1, can be negative or positive [...] If that is wrong, please say so; I understand that to be fundamental." It is wrong. The most general form of a qubit is: c_0|0\rangle + c_1|1\rangle,\tag{1} $$and the probability of collapsing into ... 4 Since the classical algorithm samples "uniformly and independently 𝑘 times from the search space", equation (5) should be, P(X_i=1, X_j=1)= P(X_i=1)P(X_j=1)=\frac{M^2}{N^2} instead. If you substitute (5) with this, you would arrive at the book's standard deviation. 4 In the following, I'll show the evaluation of the probability densities of the transition probabilities: |\langle \psi | z\rangle^2 and their pairwise independence. I didn't work out the full mutual independence. The n-qubit pure states span the complex projective space CP^{N-1} with N=2^n. Pure n-qubit states can be parametrized almost everywhere ... 4 Just to note that the complex number also specify a quantum phase of a qubit. In this particular case$$ \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 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)$and three other quantities. Further, the sum in the first equation ranges over a single index suggesting that all matrices under the trace are diagonal. In fact, since we are working with a composite system this also suggests ... 3 The paper does not specify the exact algorithm or class of distributions$\mathcal{D}$for which such algorithm fails to refute XQUATH, and some classes of distributions$\mathcal{D}$do not satisfy XQUATH, so some additional assumptions about$\mathcal{D}$are needed. In this answer we will try to follow the idea of the paper and try to show that it fails ... 3 The authors are certainly thinking about finite frames. In this case, your statement is correct, since the number of elements in every spanning set is at least the vector space dimension. As glS already pointed out, the frames could be infinite: for countable frames, the summation is then meant as convergence in the$L^2$-sense (i.e. w.r.t. Hilbert-Schmidt ... 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)}...

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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{split} \|P-...

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$U-I$ 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 ...

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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

It seems like you are trying to apply a Hadamadard gate to the state $\frac{\sqrt{3}}{2}|0\rangle + \frac{1}{2}|1\rangle$. This would go as follows: \begin{align} H\left[\frac{\sqrt{3}}{2}|0\rangle + \frac{1}{2}|1\rangle\right] &= \frac{\sqrt{3}}{2}H|0\rangle + \frac{1}{2}H|1\rangle \\ &= \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} \left( |0\... 3 The problem that led to an important role of IQP circuits is Boson Sampling. Boson sampling algorithm has to sample from a distribution based on the permanent of some complex matrix. Computing complex matrices permanents is #P-hard, meaning it is at least NP-Complete (or harder, not proven yet.) The same problem is efficiently solvable by a sub-universal IQP ... 3 Let \mathcal{H}_A denote the Hilbert space of qubits in partition A and similarly for \mathcal{H}_\bar{A}. Define the operator P:=Q|0^n\rangle\langle 0^n|Q^\dagger and write its Schmidt decomposition P = \sum_a R_a\otimes Q_a\tag1 $$where R_a are operators on \mathcal{H}_A and Q_a are operators on \mathcal{H}_\bar{A}. See for example the ... 2 Let  p_i = |\langle i | \phi \rangle|^2 \sim Dir(a_1, .., a_{2^n}) = Dir(1, .., 1)  and  m_i  the occurences of outcome  |i\rangle  on samples z_1, .. z_k. Since the Dirichlet distribution is the conjugate prior of the categorical (see here), meaning  \bf{p}  | Z, (1, .. 1),  \bf{m}   \sim Dir(2^n,  \bf{m} + 1) and using the formula for the ... 2 So probability of the second qubit being in state |1\rangle is the probability of the 5 qubit system being in a state that has |1\rangle as the second qubit. So among all the 32 states, find the ones that have |1\rangle in the second qubit, which will be half of them, for example |01100\rangle and |11111\rangle. Add up the corresponding ... 2 You'll place the phase within the CRz gate. The approach you've taken essentially argues that:$$ 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 \alpha coefficient 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 ... 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 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 ...

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