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Although it is not explained up to that point in the Qiskit textbook, the quantum toss is in reality applying the Hadamard gate, denoted $H$. In matrix form, this operator looks like: $$H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$$ Now, we express the basis states in column form as follows: $$\begin{gather} |0\rangle = \... 1 A linear operator A:V\to W represents a transformation in some underlying vector space (or more generally, from a vector space to a different one). Let's stick to the case of finite-dimensional spaces for simplicity. Such an operator is not the same as a matrix. A matrix is a way to represent the operator A with respect to a given pair of bases. Given ... 2 TL;DR: Active and passive transformations The dichotomy between the two types of unitary transformations is real and is an example of a division of transformations into active and passive types. This duality is inherent to any use of coordinates and arises from the fact that there is a degree of arbitrariness in the way coordinates are assigned to objects ... 6 OK, honestly I did not follow the later part of your post (where you asked the questions) -- it was too confusing. But I suspect that your confusion arises because you were trying to go between abstract bra-ket notation and matrix notation (which entails choosing some basis to express the operators in). Maybe this will help. Let$$ \hat{\rho} = \sum_i p_i |\...

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To set a context for my understanding of your question let's start with a single qubit. Then, the state can be described by two complex numbers $|\psi\rangle=a|0\rangle+b|1\rangle$ subject to a constraint $|a|^2+|b|^2=1$ and up to an equivalence $|\psi\rangle \sim e^{i\phi} |\psi\rangle$. Absolute values of $a$ and $b$ determine probabilities to obtain ...

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Yes. Intuitively, the set of pure product states has lower dimension than the set of all pure states. Therefore, almost all pure two-qubit states are entangled. Let $\mathcal{F}$ denote the set of all pure states of two qubits and $\mathcal{S}$ denote the set of all pure product states of two qubits. Note that $\mathcal{S}$ can be thought of as the Cartesian ...

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In the specific $|\psi \rangle$ there is symmetry on the order of bits, so we must only show that one bit is entangled and it then follows that they all are. Notice we may write: $$|\psi \rangle = \frac{1}{2}\Big(|0 \rangle \big(|001 \rangle +|010 \rangle +|100 \rangle \big) + |1 \rangle \big(|000 \rangle \big)\Big)=\frac{\sqrt3 |0\rangle |A \rangle + |1\... 1 I was also flummoxed by the apparent puzzle why the value of \mathbb{E}[\langle z|C|0^n\rangle|^2] is 2/2^n instead of 1/2^n, but in my opinion I think the confusion arises from the symbol \mathbb{E} (expectation value) whose meaning is rather ambiguous. Expectation value over what? Case 1: Fix a bit-string z. We ask for the value of \langle z|C|0^... 1 This is a great question! Naively, one might get quite puzzled because it would be natural (although incorrect) to think in the following way: If we first measure the Alice qubit and it turns out 1 (or 0) then we know that the Bob qubit instantaneously collapses to 1 (or 0) and we can confirm this when we measure it. Similarly, if we first measure ... 3 The answer is: no, it is not true that any n exchangeable state is a linear combination of density matrices of states in the symmetric subspace (that is supported on the symmetric subspace). Actually, there are even pure state counterexamples when n=2. Consider the state$$ \rho = |\phi\rangle\langle \phi|, $$where$$ |\phi\rangle = \frac{1}{\sqrt{2}}(|...

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An intuitive way to think about it is that $E[M]=E[X_1 \otimes Z_2]=E[X_1 \otimes \mathbb{1}]E[\mathbb{1} \otimes Z_2]$ If we only think about $E[\mathbb{1} \otimes Z_2]$, it is just the expectation value of $Z_2$ on the second qubit. Consider that our second Qubit in the entangled state $\frac{| 00\rangle + | 11\rangle}{\sqrt{2}}$ is measured to be \frac{+\... 1 Taking the last two terms of last expression you gave, we can do the following \begin{align} M \left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) &= X_1\otimes Z_2\left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{X_1|0\rangle \otimes Z_2|0\rangle+X_1|1\rangle \otimes Z_2|1\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{|1\... 4 We know that \begin{gather} |0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}} \\ |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}} \end{gather} $$Thus, we can rewrite the GHZ state as$$ \begin{align} |GHZ\rangle &= \frac{1}{\sqrt{2}}\left(|0\rangle|00\rangle+|1\rangle|11\rangle\right) \\ &=\frac{1}{2}\left(|+\rangle|00\rangle+|-\rangle|00\rangle+... 4 Qubits are more flexible than bits in a way that's difficult to summarize. But one key difference is that qubits support "phase kickback", and bits have no concept of phase kickback. Here is a puzzle. Fill in the blanks to make these two circuits equal: With bits, this is impossible. There is no single-input single-output process that can reverse ... 2 Calculating\rho_1$Let$N=2^n$denote the dimension of the Hilbert space where$|\psi\rangle$lives. For$i=0,\dots,N-1$, let$V_i$be any unitary that maps$|i\rangle$to$|0\rangle$. The action of$V_i$on other computational basis states$|j\rangle$with$j\ne i$is irrelevant. Exploiting the invariance of the Haar measure to absorb$V_i$into the ... 2 As you are asking specifically for the evaluation of the energy only, I will be brief. I will assume that you have a init_state (a quantum circuit) that produces the the Hartree-Fock wavefunction or any other wavefunction you like to test. I could not find a Qiskit function that provides the energy expectation value of a given wavefunction, given some ... 2$U_f$is defined as$U_f: |x\rangle|y\rangle \rightarrow |x\rangle|y\oplus f(x)\rangle$. Now, let's write the product state of the complete system of two qubits before applying$U_f$. We can do this with tensor products as follows: $$\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes |0\rangle = \frac{1}{\sqrt{2}}(|0\rangle\otimes|0\rangle + |1\rangle\otimes|... 0 On the last few lines at page 30 (Nielsen and Chuang section 1.4.2) is stated: "Suppose f(x) : \{0,1\} \to \{0,1\} is a function with one bit domain and range". So, the value of f(x) is only either 0 or 1 (a single bit), for every x. U_f is just a gate to achieve the transformation y \oplus f(x) in the ancillary qubit. Now consider one ... 0 It took me some time, but I found a more direct approach, alongside my initial idea. It turns out that statevector can indeed be saved multiple times for a given circuit, where one must provide a unique label each time state is saved. The same reasoning should also be valid for, e.g. density_matrix and unitary type of constructs (availability of specific ... 2 Entanglement is analogous to correlation. Correlation over a property simply exists where simultaneous determination is not equivalent to individual determination of a property. What does it mean? Say you have 2 coins - one black and one white. Let's say if you have a black coin you get a +1 colour value whereas with white you get a -1 colour value. If ... 4 It depends on what you want to take as your definition of maximally entangled. But, here's a good one: Given a Bell state, I can convert it into any other two-qubit entangled state using only local operations and classical communication Given that local operations and classical communication cannot increase entanglement, the possibility of performing |\... 22 TL;DR: This is probably going to be disappointing. If a cat enters a superposition and we lose track of the relative phase \phi then there is only one deterministic operation that returns to the |\text{alive}\rangle state: the state preparation channel. In other words, we have to get a new cat. Let us represent the states of the cat on the Bloch sphere ... 3 Source of the problem The purported contradiction arises due to the use of incorrect assumptions for Klein equality$$ S(\rho||\sigma) \ge 0. $$The inequality does not require any particular relationship^1 between the support of \rho and the support of \sigma. However, it does require that \rho and \sigma be states, i.e. unit trace positive ... 0 It is always useful to check the documentation: https://qiskit.org/documentation/_modules/qiskit/visualization/state_visualization.html You can see there the source code for all visualization tools in qiskit.visualization. The function plot_bloch_multivector is somehow factorizing the state and converting the statevector data into products of individual ... 5 There are several, these are the ones I have seen: |+\rangle_y,|-\rangle_y a bit lazy but easy to remember |+i\rangle,|-i\rangle the same as before but you replace the sub index with an imaginary unit i |\circlearrowleft\rangle,|\circlearrowright\rangle this notation is borrowed from light polarization, as you can use photons for light too, circular ... 5 As the other answer mentioned, they are often denoted as$$|+i\rangle= \dfrac{|0\rangle + i|1\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i\end{pmatrix} \ \ \ \textrm{and} \ \ \ |-i\rangle = \dfrac{|0\rangle - i|1\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i\end{pmatrix} $$but sometime you might also see them denoted as ... 2 They are commonly denoted as$$ \left\{ |+i\rangle = \frac{|0\rangle + i|1\rangle}{\sqrt{2}}, |-i\rangle = \frac{|0\rangle - i|1\rangle}{\sqrt{2}}\right\} $$So, you could use |\pm i\rangle = \frac{|0\rangle \pm i|1\rangle}{\sqrt{2}}, which would be very similar to the case of X, i.e., Y|\pm i\rangle = \pm|\pm i\rangle. 1 You can also obtain the states at any point during circuit construction using Statevector, the class from Qiskit's quantum_info module as follows. First, import the Statevector class, from qiskit.quantum_info import Statevector And for your example, the code below will produce all the intermediate states that you want. qc = QuantumCircuit(2) st0 = ... 2 Overview The short answer is no you cannot prepare this state efficiently if you wish to know the values of y_x; but, in fact, your question is a special case of a more general case I prove below which I will outline first as it is applicable to a more general audience. After I will show that your question is indeed a special case of the theorem. Finally, ... 5 The answer is no. Define X=[[0,1,0],[0,0,1],[1,0,0]] Z=[[1,0,0],[0,w,0],[0,0,w^2]], w^3=1 Then the Pauli group is generated by X and Z and is of order 27. With H being your matrix, you can check that H'XH and H'ZH are not in the group. Calculations like this are easy to do in gap The dim=3 counterpart of the Hadamard gate is the 3 dimensional Fourier ... 2 Error correcting codes work the same on entangled qubits as any other qubit. All Alice and Bob have to do is separately encode their qubit into a Shor code. They each run an encoding circuit, apply noise, then run a decoding circuit and apply the corrections it inferred. 2 As you say, the difference is in the global phase. Let me explain using the first of your examples,$$ \left[\begin{array}{cc} \frac{1-i}{2} & \frac{1+i}{2} \end{array}\right]. $$Mathematically, this is the same as$$ \left[\begin{array}{cc} \frac{e^{-i\pi/4}}{\sqrt{2}} & \frac{e^{i\pi/4}}{\sqrt{2}} \end{array}\right]=\left[\begin{array}{cc} e^{-i\... 4 Notice that it is somewhat a coincidence of that particular Bell state and choice of basis. The states$|0\rangle$and$|1\rangle$are in the$z$axis of the Bloch sphere and$|+\rangle$,$|-\rangle$are in the$x$-axis. The state you chose is a sum of products of single states that are the same, and it turns out that the same is true when you convert it to ... 2 One strong element of the intuition is related to the fact that it is maximally entangled. One definition of a pure state$|\psi\rangle$being maximally entangled is that the individual systems have density matrices $$\rho_A=\text{Tr}_B(|\psi\rangle\langle\psi|)=\frac{I}{d}$$ where$d$is the dimension of$A$'s Hilbert space. Now, one thing that the ... 14 The minimum overlap is zero and the maximum overlap is$\frac{1}{d}$. The overlap is a linear function of$\rho$and the set$S$of separable states is convex, so the overlap is both minimized and maximized at extreme points. Extreme points of$S$are the states of the form$^1\rho = \overline\sigma\otimes\tau$. The reason we choose to define$\sigma$as ... 2 Yet another derivation Applying a local unitary$U^A$on the first subsystem of a bipartite maximally entangled state$|\psi^{AB}\rangle$is equivalent to applying a possibly different unitary$V^B$on the second subsystem $$(U^A\otimes I)|\psi^{AB}\rangle = (I\otimes V^B)|\psi^{AB}\rangle\tag1.$$ In the specific case of the Bell state$(|00\rangle+|11\...

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