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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 ...


7

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.$^1$ ...


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, ...


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 ...


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

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 ...


0

If you want an example of a quantity of entanglement, it's always a good idea to check the qubit maximally entangled Bell states first. Here, this immediately gives you the answer. A proof that this state does what you want is that in the qubit case (i.e. four dimensional Hilbert space of the joint system qubit x qubit) the Peres criterion states that ...


0

The first lines of the Wikipedia entry Quantum logic give an impression. To cut it short, the essence of the example is this: You have a particle smeared out in some box of length $d$. If you split $d$ into two parts, $d_\text{Left}$ and $d_\text{Right}$, then "the particle is in the union of $d_\text{Left}$ and $d_\text{Right}$" is true by definition, while ...


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 ...


0

This page has a good simple summary of the experiment, they present the following simple schematic of a delayed choice quantum eraser (the same as Figure 1 in the original paper by Kim et al.): In this case, if the entangled "idler photon" is detected by Alice at D3 or D4 then she will know whether the "signal photon" went through slit A or slit B, but if ...


0

Since there still isn't an answer but the question has attracted a few upvotes, let me elaborate on my comment. This is more maths, than physics, but anyway. Writing $\sum_{jk} \mathrm{tr}(E_j^{\dagger}E_i)|j\rangle\langle k|$ doesn't give you anything. This is indeed a rank-one decomposition, but the theorem does not tell you that ANY rank-1 decomposition ...


0

The measurement procedure described by the measurement is as follows: You have a quantum state and do some measurement on it. Doing a measurement means you get one of a set of results. For example, if you measure the component of the spin of an electron in a certain direction, you have two possible results: Spin up, or spin down. The state of the electron ...


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 ...


0

I don't think there is a good answer to your question. For me, the fidelity is a distance measure (not a metric though). This means, it just tells you how "far" two states are from each other. The further they are away (e.g. the smaller the fidelity) the less they behave like each other, the closer the fidelity is to one, the better the states match. In ...


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 ...


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

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 ...


0

$\sigma_y$ isn't an observable, but $S_y$ is, so let's focus on that. If you know that $S_y | \psi \rangle$ can be written $a |\psi\rangle$, where $a$ is a number, then it only tells you that $| \psi \rangle$ a measurement of the $y$-component of spin will definitely yield $a$. That's all. I don't think you can consider $S_y|\psi\rangle$ a measurement of ...


0

Any operator, $A$, acting one of its eigenvector, $| \psi_i \rangle$, will give, $$ A | \psi_i \rangle = \lambda_i | \psi_i \rangle, $$ where $\lambda_i$ is the corresponding eigenvalue. The eigenvalues of the Pauli matrices are $\pm 1$ corresponding to either spin up or down in the corresponding direction. The eigenvalues are possible results of measuring ...


0

Quantum Computing since Democritus was useful to me, coming from a similar background. It first observes that one only needs a few principles of QM to think about quantum computation, and then launches into the complexity theory underlying the abilities of theoretical quantum computers.


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.


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

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 ...


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 ...


2

I think what you say is correct. By saying it should be unitary, he means it should be time-reversible. It is indeed time-reversible if before reaching the detector 2 it is in the same state as the initial state. It starts off with the superposition, and then only vertical polarization in the top part and horizontal in the lower part of the arms, which ...



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