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5 votes
Accepted

Problem with logarithm of tensor product of matrices

The problem is that you are using the logarithm incorrectly. If you are willing to allow for $\log(0)$, i.e., zero eigenvalues of your operators, then the correct definition must be $\log(0)=-\infty$. ...
Norbert Schuch's user avatar
4 votes

Is that right that about trace of the two density matrices multiply to each other?

Added I see this was essentially @Tobias Fünke comment which I didn't see. End Given the Hilbert-Schmidt scalar product you realize that $$\mathrm{Tr}\,(\rho \rho^{\prime}) = \langle \rho, \rho' \...
lcv's user avatar
  • 2,434
3 votes

Loss of coherence in the double-slit experiment

There is no interference because the states $|1\rangle |0\rangle$ and $|0\rangle |1\rangle$ are orthogonal. If they were not, your detector would not detect the which way information.
my2cts's user avatar
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3 votes
Accepted

Simple examples of computing the quantum discord

Perhaps first consider the two-qubit case. It is nontrivial enough that it was published in PRA, but simple enough that it can be handled analytically since the derivations just involve 4x4 matrices. ...
Anyon's user avatar
  • 2,709
3 votes

Problem with logarithm of tensor product of matrices

One problem is that the logarithm $\log(A)$ of an operator $A$ is singular if $A$ has zero eigenvalues. E.g. $$\log(|0\rangle\langle 0|)~=~\log(1)|0\rangle\langle 0|+\underbrace{\log(0)}_{=-\infty}(I-|...
Qmechanic's user avatar
  • 205k
2 votes

Is the tensor product injective on pure quantum state vectors?

As with so many things in quantum information, things get a lot easier when you consider density matrices instead of bras and kets. For density matrices, there indeed exists an operator which acts as ...
paulina's user avatar
  • 1,303
2 votes

Mach-Zehnder interferometer and superposition

Since QM is probabilistic, you can only know probability until a measurement is performed. Your doubt can be referred to a "1-input" beam-splitter since you're dealing only the branches, ...
basics's user avatar
  • 9,521
2 votes

How is information defined when considering locality in quantum mechanics?

Through this example, I understand why information in the form of bits or bit strings could not be communicated instantaneously using entanglement. However, the information about the occurance of a ...
alanf's user avatar
  • 8,290
1 vote
Accepted

Multiplication of $\mathrm{U}(3)$ matrices

$T_{3,2}$ is really a $3\times 3$ matrix as well. $\mathcal{U}(2)$ is an arbitrary $2\times 2$ unitary matrix: $$\mathcal{U}(2)=\begin{pmatrix}A&B\\C&D\end{pmatrix}.$$ The notation $$\begin{...
Riley Scott Jacob's user avatar
1 vote

Simple examples of computing the quantum discord

Define the discord as $\newcommand{\Iacc}{I_{\rm acc}}\delta(\rho) = I(\rho)-I_{\rm acc}(\rho)$, with $I(\rho)$ quantum mutual information, and $\Iacc(\rho)$ the accessible mutual information, defined ...
glS's user avatar
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1 vote
Accepted

Loss of coherence in the double-slit experiment

Just to add to @my2cts correct and concise answer (which I upvoted): This experiment has been performed, see: Young's double-slit experiment with single photons and quantum eraser Being orthogonal, ...
DrChinese's user avatar
  • 1,420
1 vote

Is the tensor product injective on pure quantum state vectors?

The tensor product $a\otimes b$ of two vectors is equivalent to a tuple $(a,b)$ up to an equivalence relation $(a,b)\sim (\lambda a, \tfrac{1}{\lambda} b)$, where $\lambda\in \mathbb C$ (or whatever ...
Norbert Schuch's user avatar
1 vote

Is two degree quantum entanglement possible?

Your implied question is effectively: can more than 2 particles be entangled? The answer to that is yes. However, 3 or more particles cannot be maximally entangled due to “monogamy of entanglement”. ...
DrChinese's user avatar
  • 1,420
1 vote

Operational meaning of mode dependence of mode-entanglement

This may not be the full answer, but after scanning the paper, here are a couple thoughts that might be of some value. If one rewrites Eq. (8) (the original Hamiltonian) the following is obtained: $$ ...
ad2004's user avatar
  • 1,132
1 vote
Accepted

Seperable Quantum States

As the comments suggest, what I didn't see is that for example $\rho^2_A$ and $\rho^2_B$ are not hermitian and therefore are no density matrices. I've also noticed, that we also have the condition $\...
Aralian's user avatar
  • 453
1 vote

Proof that $\mathrm{Tr} _ {\mathcal{Y}}[(\hat{X} \otimes \mathbf{1} _ {\mathcal{Y}}) \hat{\rho}] = \hat{X} \mathrm{Tr} _ {\mathcal{Y}} (\hat{\rho})$

You're super close! In the very last step, you note a resolution of the identity to remove the $x_i$ sum, which is correct. But you then do something strange and move the $\langle x_j|$ to the left of ...
11zaq's user avatar
  • 871
1 vote

Mach-Zehnder interferometer and superposition

Ever since Schrodinger derived the wave function for electrons academics have been trying to apply the theory to photons as well ... it leads to a lot of irrational conclusions. Dirac, Feynman and ...
PhysicsDave's user avatar
  • 2,687
1 vote
Accepted

Mach-Zehnder interferometer and superposition

The answer people give to this question depends on what interpretation of quantum theory they use. Advocates of the Copenhagen and statistical interpretations will say you can talk about what ...
alanf's user avatar
  • 8,290

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