New answers tagged

1

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 |\...


1

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 ...


7

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 ...


1

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}}(|...


1

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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