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10

Call $u_1, u_2, u_3, u_4$ the eigenvectors described by you, respectively. Your claims are all right, but realize that both $u_1$ and $u_2$ share the same eigenvalue, that is $1$, i.e., $Pu_1=u_1$ and $Pu_2=u_2$. Hence, any linear combination of $u_1$ and $u_2$ will also be eigenvectors with the same eigenvalue $1$. Try to find eigenvectors of $H$ of the ...


9

Hint: When an eigenvalue for an operator $P$ is degenerate, there are more than one way to chose a set of eigenvectors. If the other commuting operator $H$ lifts that degeneracy, there will be a preferred choice of common eigenvectors. More generally, a set of diagonalizable operators commutes if and only if the set is simultaneously diagonalizable.$^1$ ...


4

There are a lot of misconceptions here so let's take it one step at a time. The entropy in classical mechanics is called the Gibbs entropy, $$S = - k_B \sum_i p_i \ln p_i,$$ where $p_i$ is the probability of some microstate $i$. This is essentially the same thing as Shannon entropy for physical systems. With this concept one can view knowing ...


3

To expand the comment, spectral measures, or projection valued measures are introduced to characterize self-adjoint operators. They are families of orthogonal projections on the Hilbert space that, when acting on vectors suitably, define a measure. If you denote by $\{P_\lambda\}_{\lambda\in\mathbb{R}}$ this family, a self adjoint operator $A$ corresponding ...


3

Let me expand a bit on the intuition part and write down an example. This is all essentially already covered by yuggib's answer. Your confusion about positive operator valued measures, as also pointed out, is that they are not to be confused with measurement outcomes. The problem with measurement outcomes is that they are rather arbitrary. Often, they rely ...


3

If you have another model, then use it to generate a set of predictions, and let's compare those to the predictions we get in the world we see. If you think there's some ether filling the universe, then write down some properties it has, and make some predictions with those properties. Talking in vague generalities will only lead you down a rabbit hole of ...


2

The most elementary formulation of Quantum Mechanics (the one usually formulated in Hilbert spaces) can be constructed starting form a lattice of all the elementary propositions which can be tested on a given quantum system obtaining, as the outcome, YES or NOT. This can be similarly done for classical mechanics and, in that case the elementary ...


1

Your Hilbert space is finite-dimensional (specifically, 4-dimensional). It means that you need no fancy way of deriving the spectrum, just proceed with the standard approach: Write your Hamiltonian in the matrix form and solve the characteristic equation $ \det ( H - \lambda \cdot 1_{4 \times 4} ) = 0 $ with respect to $\lambda$, which would give you ...


1

As Lelesquiz points out, that looks like a somewhat standard matrix differential equation to me. The Wikipedia link does give a solution method for the matrix, but I think that it might be easier to do some remapping: \begin{align} \rho_{00}&\to v_1\\ \rho_{01}&\to v_2\\ \rho_{10}&\to v_3\\ \rho_{11}&\to v_4 \end{align} and write it as ...


1

John Baez, a well-known mathematical physicist presently (2014) in the Department of Mathemetics at the University of California, Riverside is trying to find more and more ways that category theory connects with physics, chemistry, electrical engineering, network theory, etc. You can read more about his efforts: At his UCR homepage here; and In particular, ...


1

Why can't we simply assign to every point of the Bloch sphere a phase $e^{i\phi}$? This is the idea of a section of a fibre bundle. You are considering in this case a base space $S^{2}$ with fibre $S^{1}$. Locally the fibre bundle looks like $S^{2}×S^{1}$. However you want to consider a fibration such that the global space is not the trivial product ...


1

Realize that an arbitrary rotation around the axis $\mathbf{n}$ is given by $R_\mathbf{n}=\cos(\alpha/2)I-i\sin(\alpha/2)\hat{\mathbf{n}}\cdot\mathbf{\sigma}$ and an arbitrary unitary operator can be written as $U=\exp{(i\gamma)}R_\mathbf{n}$ with $\gamma$ some phase factor. Thus, in general, any operations on the qubit can be seen as a rotation with some ...


1

Two states $\rho_0$ and $\rho_1$ are perfectly distinguishable if there exists a POVM measurement $\{P_0,P_1\}$ such that $$Tr(\rho_0P_0)=1$$ $$Tr(\rho_0P_1)=0$$ $$Tr(\rho_1P_0)=1$$ $$Tr(\rho_1P_1)=0$$ and $P_0+P_1=\mathbf{I}$, where $\mathbf{I}$ is the identity matrix. You can interpret the above equations as follows: If we obtain outcome $P_0$, we know ...


1

No. Local unitary operations do not change the spectrum of the reduced density matrix. Thus, all eigenvalues of the reduced density operator must be the same in both states. On the other hand, if all eigenvalues of the RDM are equal, the states can be transformed into each other by local unitaries, since they have the same Schmidt coefficients. Note that ...


1

I do know something about the history of the problem, but I don't know the answer. The first reference is S. Lloyd, Almost any quantum logic gate is universal, Phys. Rev. Lett. 10, 346–349 (1995). which gives a "proof" of this... but it isn't strictly correct. I don't know what exactly was wrong with it, because I never looked at it in detail. It doesn't ...


1

I have recently found a reference proving the statement from my question. It is: Universal quantum gates Jean-Luc Brylinski and Ranee Brylinski In Mathematics of Quantum Computation, Chapman & Hall (2002) arXiv:quant-ph/0108062 Theorem 1.4, proven in section 8.



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