New answers tagged quantum-state
1
The whole point of an EPR pair is that you cannot write (without losing some information) "This is what Alice has" and "This is what Bob has". Partial descriptions can be given using reduced density matrices.
However, if you want to identify which bits Alice and Bob each have, we often use a notation like
$$
(|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B)/\...
2
Right, they are quite similar. The Holevo bound is a bound on the amount of accessible information between your quantum system and your classical system. The I(X;B) object written in the HSW theorem wikipedia page is actually this bound, while the $\chi$ there is the Holevo rate, or product state capacity. What HSW showed was that if you took many copies of ...
8
For any controlled-$U$, if the input state is $|+\rangle|\phi\rangle$ where $|\phi\rangle$ is not an eigenstate of $U$, then the output state is entangled. This immediately deals with trivial cases such as $U=I$ because in that case all states $|\phi\rangle$ are eigenstates, and so it is not entangling. For any other $U$, there is an input state that is ...
3
Two antipodal states in the Bloch sphere (note that $0 \leq \theta \leq \pi$):
\begin{equation}
|\psi \rangle = \cos \frac{\theta}{2} |0 \rangle + e^{i\varphi}\sin \frac{\theta}{2} |1 \rangle
\\
|\psi^\perp \rangle = \cos \frac{\pi - \theta}{2} |0 \rangle + e^{i\varphi + \pi}\sin \frac{\pi - \theta}{2} |1 \rangle = \sin \frac{\theta}{2} |0 \rangle - e^{...
2
For the way round that you've got your inequalities, I don't think there's much that can be said. To see why, let's consider the first expression
$$
\|U-V\|_1=\text{Tr}(\sqrt{2I-VU^\dagger-UV^\dagger}).
$$
Now, $VU^\dagger$ is a unitary, and hence as a spectral decomposition. Let the eigenvectors be $|\lambda_i\rangle$ with eigenvalues $e^{i\lambda_i}$. $UV^\...
1
The fundamental problem is that you've transposed the matrix of eigenvectors compared to what you need. Yes, as you've written it, the eigenvectors are the columns of the matrix. But I think you must be using the rows in your actual code.
One of the basic checks that I did (and this is crucial whenever you code something) is to be able to do simple test ...
2
Trace out everything except the qubit you are interested in. Do this by computing the outer product of the state of the target qubit, for each possible value of the other qubits, and summing up all those outer products. This will produce the 2x2 density matrix of the target qubit.
Get the x, y, z coordinates of the Bloch vector from the 2x2 density matrix $D$...
0
You can use a method described in Transformation of quantum states using uniformly controlled rotations.
Authors of the article introduced method how to change an arbitrary state $|a\rangle$ to state $|b\rangle$ with so-called uniformly controlled rotation, i.e. a rotation which rotational angle depends on combination of zeros and ones in controlling ...
0
What you seem to be asking for is the general construction of an arbitrary permutation. There’s nothing quantum about this; the equivalent question in classical is how to construct an arbitrary reversible circuit of n bits. There are $2^n!$ such circuits, from which it is hopefully obvious that there is not a compact description for the vast majority.
I ...
4
Your states differ in global phase only, hence they are indistinguishable (or in other words they are equivalent). Therefore you do not need to apply gate $-I$.
Note that the global phase is $\pi$ as $-1 = \mathrm{e}^{i\pi}$
However, in case the state is produced by controlled gate, global phase cannot be neglected. In that case you can implement ...
3
First it is instructive to ask oneself: "how does classical data get into my computer?" In a classical computer, your data is always stored in bits. Because calculations in base 2 are not very straightforward for most people there are abstractions like int types for integers and float types for rational numbers with the associated math operations readily ...
2
The unitary matrix of the Fourier transform is
$$
U_{QFT}=\frac{1}{\sqrt{2^n}}\sum_{x,y=0}^{2^n-1}e^{2\pi i\frac{xy}{2^n}}|y\rangle\langle x|.
$$
So, if I apply this to the input state $|0\rangle$, I get the answer
$$
U_{QFT}|0\rangle=\frac{1}{\sqrt{2^n}}\sum_{y=0}^{2^n-1}|y\rangle.
$$
The amplitude for finding the system is state $y=0$ is $1/\sqrt{2^n}$. ...
2
The whole point of the system/environment description is that the "system" is the bit that you can characterise and control. The environment is the bit that you cannot characterise or control. It could be anything. Ultimately, it's the whole of the rest of the Universe. For your particular physical realisation, you might have some approximation of what the ...
2
An arbitrary 1 qubit pure state can be represented as:
$$|\psi \rangle = \cos(\theta /2) |0\rangle + e^{i\varphi} \sin(\theta/2) |1\rangle$$
where $0 \leq \theta \leq \pi$, $0 \leq \phi < 2 \pi$. The amplitude of the $|0\rangle$ state is $\cos(\theta /2)$ and the amplitude of the $|1\rangle$ state is $e^{i\varphi} \sin(\theta/2)$. The phase of the qubit ...
1
There is no measurement in the picture; the picture shows how to construct 4 Bell states $|\Phi^+\rangle$, $|\Phi^-\rangle$, $|\Psi^+\rangle$ and $|\Psi^-\rangle$ using Hadamard-CNOT circuit.
3
When input state to QFT is $|00\dots0\rangle$ then no controlled gates in QFT work as each control qubit is set to $|0\rangle$. However, there are also Hadamard gates on each qubit. As a result, QFT behaves as $H^{\otimes n}$ and produces superposition consisting of all basis state with same probability $\frac{1}{2^n}$, where $n$ is number of input qubits.
...
4
By looking to the circuit for the QFT presented in the M. Nielsen and I. Chuang textbook (Figure 5.1.) we can notice that all controlled rotations can be neglected because for each control rotation gate the control qubits are in $| 0\rangle$ state (for the case described in the question). Here is the Figure from the book:
So effectively, in this case, we ...
2
Your state is a generalization of so-called W-state.
Here is a implementation of the state for three qubits.
You can also use method described in paper Transformation of quantum states using uniformly controlled rotations for preparing W-state with any qubits you want.
When you use this method, set probability of states $|10\dots0 \rangle$, $|01\dots 0\...
0
In classical computing, each additional bit adds one dimension to the total state-space. The state-space defined by $n$ bits is a direct sum of $n$ one-dimensional binary spaces ($X_i$). So the total space ($X$) is just $$X = X_1 \oplus X_2 \oplus ... \oplus X_n,$$ and every point in this space can be specified by an $n$-tuple: $$x=(x_1, x_2, ... , x_n), \;...
2
There is no sense in which the classical and quantum state spaces are both exponential in the number of (qu)bits. That misconception comes from conflating two different notions of the "size" of the state space. Depending on how you think about it, either the classical space is linear and the quantum space is exponential, or the classical space is exponential ...
1
When you have $n$ qubits (or classical bits as well) you can represent $2^n$ basis states (or numbers in classical sense). Hence you need $2^{n+1}$ real numbers to desribe a quantum state of $n$ qubits (real and imaginary part of amplitude for each member of the state).
Consider for example $n=3$. In this case, 3 bits can represent numbers 000, 001, 010, ...
2
Qubit is described by a vector $|\psi\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}$ where $\alpha,\beta \in \mathbb{C}$. So, you would need four real numbers for representing it. Because of condition $|\alpha|^2 +|\beta|^2 = 1$, number of degrees of freedom is reduced, so you need only three parameters, hence you can describe the qubit in spacce $\...
1
There's one more way:
Create two identical qubits on the Bloch sphere (this is start with 0 and apply the same random rotational gate). Then, apply a CNOT gate followed by a Hadamard gate to the first qubit to perform a measurement in the Bell basis. You will measure 00, 01, 10 (which are equivalent to the 3 symmetric Bell states) with equal probability ...
2
Your question has most of the components of the answer: indeed you need to take the square of the absolute value for the probability amplitude.
You have to remember that $\cos \theta$ is a real number, and its absolute value is just $|\cos \theta|$, but $e^{i\phi}$ is a complex number (Euler's formula: $e^{i\phi}=\cos \phi + i \sin \phi$), so its absolute ...
3
1) In this step, you connect a teleported qubit with entangled qubits between Alice and Bob. This means, Bob now has an "access" to the teleported qubit.
2) Here you get some information about the teleported qubit and "partially colapse" Bob's qubit according to a state of the teleported qubit.
3) In this last step you bring information about the ...
1
There is more information towards the end of the tutorial here but essentially how you do this is you run both circuits on the state_vector simulator and then you can use the function state_fidelity to work out the fidelity between the two states.
The code to do this should look something like this
from qiskit.quantum_info import state_fidelity
# set up ...
1
What you're talking about is called a "projection". The projection onto basis state $|\psi\rangle$ is given by:
$P_{\psi} = |\psi\rangle\langle\psi|$.
Any measurement operator $M$, such as $Z$ can be written as the sum over projectors, weighted by their respective eigenvalues, such that
$M = \sum_i \lambda_i P_i$
where $\lambda_i$ is the real-valued ...
2
A qubit is a two-level quantum system and these two-levels in some hardware can be implemented by spins of the electron, but it's not the only option (can be just two "stabile" quantum states not essentially associated with spins). In this sense it is not only about the spin (directional) correlation between qubits, it's about state correlation between them. ...
1
By taking into account this representation of the CNOT gate:
$$CNOT = | 0 \rangle \langle 0 | \otimes I + | 1 \rangle \langle 1 | \otimes X$$
We can write:
$$CNOT(1, 3) = | 0 \rangle \langle 0 | \otimes I \otimes I + | 1 \rangle \langle 1 | \otimes I \otimes X$$
$$CNOT(1, 2) = | 0 \rangle \langle 0 | \otimes I \otimes I + | 1 \rangle \langle 1 | \otimes ...
2
Your mental model of how cirq works is slightly off. You don't invoke operations on a qubit in one line and then check how that qubit changed in another line. You create a circuit in one line and then ask about its properties, such as its effect on a qubit, in another line.
In this case, you want to ask what the final_wavefunction of your circuit is and ...
4
Hadamard gate can be interpreted as a rotation in 3D Euclidean space (on Bloch sphere) by angle $\pi$ around X+Z axis.
The qubit rotation by angle $\theta$ around axis pointed by unit vector $\textbf{n}=\{n_x,n_y,n_z\}$ is described by rotation operator ($X$, $Y$ and $Z$ are Pauli matrices)
\begin{align}
R_{\textbf{n}}(\theta)=&n_xe^{-i\frac{\theta}{2}X}...
1
The first postulate of quantum mechanics that can be found in the M. Nielsen and I. Chuang textbook:
Postulate 1: Associated to any isolated physical system is a complex vector space with the inner product (that is, a Hilbert space) known as the state space of the system. The system is completely described by its state vector, which is a unit vector in the ...
2
An example of constructing (with help of Qiskit) a controlled version of some simple 4x4 unitary matrix:
$$
U = \begin{pmatrix}
\mathrm{e}^{i g_1} & 0 & 0 & 0 \\
0 & \mathrm{e}^{i g_2} & 0 & 0 \\
0 & 0 & \mathrm{e}^{i g_3} & 0 \\
0 & 0 & 0 & \mathrm{e}^{i g_4} \\
\end{pmatrix}
$$
where $g_s$ are some given ...
2
$\mathcal{H}_A \neq \mathcal{H}_B$, they are two distinct physical systems (even if they have the same dimension).
That formula for a reduced density matrix is a shortcut for
$$
\rho_A = \sum_j \big(I_A \otimes \langle j |_B\big) \cdot | \Psi\rangle \langle\Psi| \cdot \big(I_A \otimes |j\rangle_B \big)
$$
You can check the dimensions. If $d_A = \text{dim}...
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