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6

If $|\phi⟩$ and $|\psi⟩$ are linearly independent, then it is always possible to assign them to the column vectors $$ |\phi⟩\mapsto\begin{pmatrix}1\\0\end{pmatrix} \text{ and } |\psi⟩\mapsto\begin{pmatrix}0\\1\end{pmatrix}, $$ but if they're not orthogonal you're obviously going to need to work harder on the representation of the inner product in this basis. ...


4

In short: No, your reasoning is backwards. The reduced density matrix is mixed because of entanglement and not vise versa. If two systems A and B are in an entangled state, we can no longer separate the state description and write down individual state vectors for system A and B, i.e: $$ |\psi\rangle_A \otimes |\psi\rangle_A $$ Remember that a state is ...


3

From the way it is defined $\left| \Psi \right\rangle$ is not a vector on the sphere, but rather a vector along the z-axis between $-\hat{z}$ and $\hat{z}$, because it is a linear combination of $\left|0\right\rangle$ and $\left|1\right\rangle$ which are both vectors along the z-axis. Now we want $\left|\Psi(\theta = 0 , \phi =0)\right\rangle = ...


3

There are - even after clarifications in the comments - several ways to interpret the question. So let me start out with the way that fits all parameters here: Suppose we have a Hilbert space $\mathcal{H}$ (for qubits, $\mathcal{H}=\mathbb{C}^2$) and suppose we take a (pure) state in the Hilbert space $\mathcal{H}\otimes \mathcal{H}$. Question 1: If we ...


2

I) The main point is that the half-angle $\frac{\theta}{2}$ doubles when we go from the ket $$\tag{1} |\psi\rangle~=~\begin{bmatrix}\cos\frac{\theta}{2} \cr e^{i\phi}\sin\frac{\theta}{2}\end{bmatrix}, \qquad ||\psi||~=~1, $$ to the density matrix/operator $$\tag{2}\rho~=~| \psi\rangle \langle\psi | ~=~\frac{1}{2}\left({\bf 1}_{2\times 2}+ \vec{r}\cdot ...


2

As far as i could find with quick search EIT was first proposed by Jakob Khanin and graduate student Olga Kocharovskaya at Gorky State University, Russia, in 1986.


2

Looking at the Hilbert spaces themselves, we indeed find the puzzling equality $$\mathbb{C}^2\times\mathbb{C}^2 = \mathbb{C}^2\otimes\mathbb{C}^2 = \mathbb{C}^4$$ so the tensor product of the qubit spaces is equal to the pairs of non-entangled states. Or so it would seem. The superficial equality is wrong in the physical context, because the isomorphism ...


2

The difference The PPT criterion is an entanglement criterion whereas the Bell inequalities quantify non-locality. The PPT criterion makes no statement about hidden variables because it is a construct based on the framework of quantum mechanics! So it is really describing the entanglement (e.g. between two parties). The Bell inequalities make a statement ...


1

A qubit is simply not defined the way you define it, neither are clones. This is mostly, because you describe a state by referring to a state. It's not the problem that you are actually referring to copies of the state, you are referring to copies of the state, which is yet to be defined. Let's go to what I have learned as the Ludwig school of thought about ...


1

I'm no expert on quantum computing, but I had the same question, and found the following to help, a little bit. http://tph.tuwien.ac.at/~oemer/doc/quprog/node8.html#SECTION00323300000000000000 The gist of it is, first you 'cool' your qbit to the zero state, and then apply a unitary transformation which has the result $U|0\rangle = |S\rangle$, where ...


1

I'll answer to the second part only It is a hard problem. Read the Subhash Kak (1998-2002 12 pages ) paper on http://arxiv.org/abs/quant-ph/9805002 The problem of initializing phase in a quantum computing system is considered. The initialization of phases is a problem when the system is initially present in an entangled state and also in the application ...


1

It is a grating used in the system of X-ray differential phase contrast imaging. In front of this grating, there is a phase one, which can modulate the X-ray phase and formed a self-image at the Talbot distance. the absorption grating, generally, composed by the high aspect ratio parallel walls made of Au, used to periodically absorb X-rays. Finally,a moire ...


1

Yes. Nonweak measurements correspond to Hermitian (or self adjoint) operators. The results are 1) an eigenvalue and 2) you project the state vector onto the corresponding eigenspace. The projections onto different eigenspaces produce eigenvectors with different eigenvalues, and eigenvectors of a symmetric operator with different eigenvalues are orthogonal. ...


1

A cursory reading tells me two things: a) we consider pure states as input states and b) both states on Alice's and Bob's side are qubit states. Now, the reduced denisty matrix on either side is therefore a two-by-two matrix. If the state is pure, the entanglement is quantified by the von-Neumann entropy of the reduced density matrix, which in turn is ...


1

Let me elaborate on the answers in the comments and answer this question completely: First note that the space of Hermitian matrices is a vector space, but a real one. Real linear combinations of Hermitian matrices $A$ and $B$ are also Hermitian, i.e. $\alpha A+B$ is Hermitian for $\alpha \in \mathbb{R}$. However, it is not a complex vector space, since ...


1

The key to making a CNOT gate is being able to turn on and off the interactions between two qubits. Following the physical system you suggest, let's say our qubits are the spin up and spin down degrees of freedom of two electrons in some atoms. These spins can interact with an external potential, which is how we do the one-qubit rotations, but they will ...


1

For $z=2$ there is a study here http://arxiv.org/abs/quant-ph/0404026 in the context of ferromagnetic spin chain. The result is basically $\log L$. Swingle and Senthil argued in http://arxiv.org/abs/1112.1069 that "generally" the violation of area law for EE is at most $L^{d-1}\log L$ where $d$ is the space dimension. However, http://arxiv.org/abs/1408.1657 ...


1

The experiments of John Bush on pilot-wave hydrodynamics have evidently proven that macro-scale replication of quantum effects is entirely possible, the Copenhagen interpretation may be in trouble, and the answer to your question may end up being "yes," after all, to the surprise of most all of us. That being said, until more experiments are done, I doubt ...


1

I would re-state the problem slightly to conform to conventional notation. In any discussion of magnetic resonance -- of spin 1/2 particles, be they electrons, or say, protons-- the direction of the magnetic (polarizing) field is always chosen as +z in a laboratory coordinate frame. Then the two stationary states, which (for convenience) I write as |+> ...


1

Try S. L. McCall and E. L.Hahn -- Phys. Rev. vol. 183, p.457 (1969).



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