# Tag Info

0

I particularly like this example of a single qubit quantum classifier. The idea is to use a variational approach to do ‘data reuploading’ and create a non-linear classifier, which has similarities to a single hidden layer neural network. There’s an example implementation from Xanadu in Python along with a basic overview.

2

It is exactly what you say: A "Choi state" which is only positive semi-definite corresponds to a completely positive map which is not necessarily trace preserving. Fixing the trace of the Choi state rescales the "success probability" when implementing the CP map via the isomorphism - if you wish, it rescales the trace of all output states $\mathcal E(\rho)$ ...

12

Some select examples of single qubit technology that has been used for decades: Atomic clocks are single qubit quantum information processors. (In fact, the natural transition frequency between the states of a specific qubit defines what humans mean by time.) NMR Spectroscopy is typically applied to single qubits. (Well, ensembles of single qubits, but ...

0

You can use $Ry$ (y-rotation gate). Its general matrix is $$Ry(\theta) = \begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\ \sin(\theta/2) & \cos(\theta/2) \end{pmatrix}$$ Applying the gate on $|0\rangle$ state, you get a state $$|\psi\rangle = \cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle$$ Hence $\cos(\theta/2) = \frac{1}{\sqrt{3}}$ ...

1

The only thing that is physical about a quantum state $|\psi\rangle$ is its square overlaps with other states. In other words, given an arbitrary complete basis $\{|u_k\rangle\}_k$, what matter are the overlaps $p_k=|\langle u_k|\psi\rangle|^2$, and these are interpreted as outcome probabilities and must therefore be normalised as $\sum_k p_k=1$. This tells ...

1

I think there's some confusion here, so I'll try to explain the principle of QKD (Quantum Key Distribution) instead. Say Alice and Bob want to communicate in a secure fashion using symmetric encryption. To do so, they require a shared secret, a key. One of them generates it and he must somehow get it to the other person without an eavesdropper, say Eve, ...

2

I would like to add general case for single qubit gate. Let us assume that our gate is described by unitary matrix $$U = \begin{pmatrix} u_{11} & u_{12} \\ u_{21} & u_{22} \end{pmatrix}$$ Eigenvalues are roots of so-called characteristic equation $$|U-\lambda I|=0$$ In particular $$\begin{vmatrix} u_{11} - \lambda & u_{12} \\ u_{21} &... 2 You can also rephase. So that gets rid of 1 more real dimension. That is quotient by \psi ~ e^{i \theta} \psi. Caution: unlike normalization where you are taking a subset, instead you are identifying many points into one. Edit: Let | \psi \rangle = \sum_{i=0}^{2^{n}-1} \alpha_i | i \rangle There are identifications | \psi \rangle \equiv z | \psi \... 0 Unitary operator is: 1. Normal, hence invertible and diagonalizable. 2. Preserve inner product and trace. 3. Take othornomal basis to orthonormal basis. 4. Eigenvalues are of form e^{i\theta}. 5. Tensor product of unitary matrices is unitary. 6. e^{-iH} is unitary if H is Hermitian. Projective measurement (observables) is: 1. Hermitian, H=H^\dagger. ... 3 First you need eigenvalues;$$X=\begin{pmatrix} 0&1 \\ 1&0\end{pmatrix}$$so you need to solve equation$$\begin{vmatrix}0-\lambda &1 \\ 1 &0-\lambda\end{vmatrix}=0$$or$$\lambda^2-1=0$$which gives eigenvalues$$\lambda_{1,2}=\pm 1Since X is Hermitian, eigenvalues are real. Now you can find eigenvectors; for example, for the first ... 0 For any unitary quantum transformation the following properties can be considered: Any quantum transformation is reversible. A unitary operator is linear by definition. A unitary operator preserves the norm or length of a state vector. It maps an orthonormal basis to another orthonormal basis If U1 and U2 are unitary operators on spaces V1 and V2 ... 4 The |+⟩ and |-⟩ are states given by the following decomposition in the Z-basis: \begin{equation} \begin{aligned} |+⟩ &= \frac{1}{\sqrt{2}} \Big(|0⟩ + |1⟩\Big)\\ |-⟩ &= \frac{1}{\sqrt{2}} \Big(|0⟩ - |1⟩\Big) \end{aligned} \end{equation} As from quantum mechanics, you can see that there's a 50% chance for both states to be found in the |0⟩ or |1⟩ ... 1 You asked "What is wrong in my calculation?" My guess is that your mistake is in forgetting that \langle \psi | = \cos{\theta}\langle 0| - i \sin{\theta}\langle 1| instead of \cos{\theta}\langle 0| + i \sin{\theta}\langle 1| (because i becomes -i when you take the complex conjugate). 2 Standard inner product on space \mathbb{C}^{n} is defined as \langle a|b \rangle = \sum_{i=1}^{n} a_{i}b_{i}^{\dagger}, $$where b^{\dagger} is complex conjugate (i.e. for x \in \mathbb{C}: if x = x_{re} + ix_{im} then x^{\dagger} = x_{re} - ix_{im}). In your case a_{1} = \cos(\theta) and a_{2} = i\sin(\theta). Since you are interested in ... 2 Yes, the state of a qubit can be described by a point inside the Bloch sphere. However, you cannot use the state vector formalism to describe it, you have to generalise to the concept of mixed states. For a qubit, these have three parameters: the two angles and the length of the vector. 3 Mixed states are represented by vectors inside Bloch sphere. Suppose you have mixed state with density matrix$$\rho=q|\psi\rangle\langle\psi| + (1-q)|\phi\rangle\langle\phi|$$where |\psi\rangle and |\phi\rangle are pure states with vectors \overset{\rightarrow}{r_{\psi}} and \overset{\rightarrow}{r_{\phi}} on Bloch sphere, and 0<q<1; then ... 0 Here is an implementation of a circuit producing state |\psi\rangle = \frac{1}{\sqrt{3}}(|00\rangle + |01\rangle + |10\rangle) on IBM Q: Note that \theta = 1.2310 for \mathrm{Ry} on q_0. \theta = \frac{\pi}{4} and \theta = -\frac{\pi}{4} for first and second \mathrm{Ry} on q_1. The \mathrm{Ry} on q_0 prepares qubit in superposition |... 3 It is known that:$$ \left| 0 \right\rangle = \frac{1}{\sqrt{2}} \left( \left| + \right\rangle + \left| - \right\rangle \right) \qquad \left| 1 \right\rangle = \frac{1}{\sqrt{2}} \left( \left| + \right\rangle - \left| - \right\rangle \right) $$By substituting in the initial state:$$ \left| \varphi \right\rangle = \frac{i}{\sqrt{3}}\left| 0 \right\rangle ...

1

If you have one parameter (one $\theta$) for both circuits then I think the first one is better... they are doing the same job, but the second one is creating extra gates. So the first one will be faster and will have fewer errors because there are fewer gates in the first circuit. But if you are obtaining the second circuit with two parameters (two ...

3

This can be done using the statevector_simulator provided with Qiskit Aer. It will return the statevector that describes the quantum state at the end of your circuit. It can be used in the same way as the qasm_simulator, only your circuit shouldn't have measurement gates at the end. There is more information about this simulator in this tutorial.

4

Just to expand on the detail of why writing out the columns works: Start by writing the action of the unitary: \begin{align*} U|0\rangle=\frac{1}{\sqrt{2}}|0\rangle+\frac{1+i}{2}|1\rangle \\ U|1\rangle=\frac{1-i}{2}|0\rangle-\frac{1}{\sqrt{2}}|1\rangle \end{align*} Before proceeding, it's always worth checking that both sides are correctly normalised. In ...

2

$$\left\langle a| b \right\rangle= \begin{pmatrix} {1} & {1} & {1} \ \end{pmatrix} \begin{pmatrix} {1} & {2} & {k} \ \end{pmatrix}$$ Since the two vectors are orthogonal,so the inner product of them is zero: $$0=1+2+k \\ k=-3$$

3

In last time there is a lot of questions how to find $\theta$ and $\phi$ for this particular state on Bloch sphere: $$\left| \varphi \right>=\frac{1+i}{\sqrt{3}} \left| 0 \right> + {\sqrt{\frac{1}{3}}} \left| 1\right>$$ I will try to demonstrate how to do so in more details in comparison with previous answer. Generally, a quantum state can be ...

0

Yes, any state can always be described as a coherent superposition of some other set of states. "Being in a superposition" is not a property of just a quantum state, but rather a property of a quantum state relative to some basis. It doesn't make sense to talk of a superposition if not relative to a basis. We often don't explicitly specify this basis simply ...

0

Generally, a quantum state can be expressed in this form: $$|\varphi\rangle = \cos\frac{\theta}{2}|0\rangle + \mathrm{e}^{i\phi}\sin\frac{\theta}{2}|1\rangle$$ Where $\theta$ and $\phi$ are coordinates on Bloch sphere. Regarding the particular state in question, we firstly have to get rid of complex amplitude before $|0\rangle$ to have only real number ...

7

Firstly simply rewrite probability amplitudes of returned states as columns of a matrix: $$U = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1-i}{2} \\ \frac{1+i}{2} & -\frac{1}{\sqrt{2}} \end{pmatrix}$$ Now do some algebra $$U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & \frac{1-i}{\sqrt{2}} \\ \frac{1+i}{\sqrt{2}} & -1 \end{pmatrix} = \frac{1}{... 0 I find the wording and notation of the question you've been given a bit odd. I think you've taken the correct strategy, and the factor of 4\pi in the question doesn't make any sense to me either. I guess it should just be a 2. 0 For a single qubit that's in a pure state (no decoherence) the probability that the state is somewhere on the surface of the sphere is 1. You don't need to measure this. If you're going to do a POVM to learn a bit of information about the state, you have to choose an axis of measurement, and you will only gain information about the qubit with respect to ... 2 If first qubit is |1\rangle there is no other possibility than second qubit to be |1\rangle as well since probability of state |10\rangle is zero. Hence probability of measuring |1\rangle in second qubit is 1. In case first qubit is |0\rangle there are two possibe results: |00\rangle or |01\rangle. Since probability of state |00\rangle is ... 2 You have normalized state$$|\psi\rangle=\alpha|0\rangle + \beta|1\rangle$$First, write the state as$$|\psi\rangle=\frac{\alpha}{|\alpha|}\left({|\alpha|}|0\rangle + \frac{\beta|\alpha|}{\alpha}|1\rangle\right)$$The factor$$\frac{\alpha}{|\alpha|}$$is a global phase and not important. Now you have$$\cos{\frac{\theta}{2}}=|\alpha|$$which gives the ... 2 See below:$$\left| \varphi \right>=\frac{1}{\sqrt{\sqrt{2}}}\left(\left(\frac{1+i}{\sqrt{2}} \right)\left| 0 \right> + \left| 1\right>\right) = \frac{1+i}{\sqrt{\sqrt{2}}}\left(\frac{1}{\sqrt{2}}\left| 0 \right> + \frac{1}{\sqrt{2}}\frac{1-i}{\sqrt{2}} \left| 1\right>\right) = \frac{1+i}{\sqrt{\sqrt{2}}}\left(\cos(\pi/4)\left| 0 \right> + \...

4

The Hadamard gate is described by this matrix \begin{equation} H=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \end{equation} Conjugate transpose of $H$ is again $H$. Hence we have to check if $HH$ is $I$. Multiplication goes as follows ($h_{ij}$ denotes elements of resulting matrix): $h_{11} = 1\cdot1 + 1\cdot1 = 2$ $h_{12} = 1\... 1$\newcommand{\ket}{\lvert #1\rangle}\newcommand{\PP}{\mathbb{P}}$Given an arbitrary state$\ket a$, let us write the corresponding density matrix/projector as$\PP_a$. Any such density matrix can be decomposed using a basis of Hermitian traceless operators (think the Bloch sphere representation for qubits) as$\PP_a\equiv 1/2(I + \sum_i a_i\sigma_i)$, ... 1 Intraphoton entanglement uses the degrees of freedom from one photon only to create entanglement. So, here either polarization and linear momentum or polarization and angular momentum can be used to create entanglement. Interphoton entanglement is the entanglement created between 2 spatially separated photons. So, naturally latter is less stable than former. ... 6 Superposition is a basis dependent concept. Namely the$|0\rangle$and$|1\rangle$states are commonly said not to be superposition states exactly because one of the two coefficients in the$\{|0\rangle,|1\rangle\}$basis expansion is zero. However, using the x-basis representation,$\{|+\rangle, |-\rangle\}$one finds$|0\rangle = (|+\rangle + |-\rangle)/\...

Top 50 recent answers are included