# Tag Info

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Let us start from the beginning. Elementary particles are quantum mechanical entities. They can be described with the quantum mechanical solutions of the appropriate equations for the set up under consideration with the constants taken from the boundary conditions of the problem. In this it is not different than the situation with classical mechanics ...

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Nielsen and Chuang are referring to a scheme known as measurement-based quantum computation, for which many good learning resources are a short google search away. The idea is that you prepare a large, highly entangled state involving a large number of qubits at the start of the experiment. You then proceed to make measurements on the state, potentially ...

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In today's understanding of Nature, there is nothing completely isolated. So technically there will always be interaction with the surrounding, at least from a quantum physical perspective. Here vacuum is not empty i.e. it does allow for electromagnetic interaction and there will be heat loss due to these vacuum effects. Furthermore also the other concepts ...

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Although it has been said in other comments and answers, it bears repeating succinctly: photons (as far as any experiment can tell) are massless and therefore always move at the universal, invariant speed of light. There is NO non-relativistic description of the photon. Even the "classical" description of light - Maxwell's equations - can be interpreted as ...

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People are addressing the speed question, but just to be clear: a photon can be very low energy. For instance, radio waves are much lower energy than gamma rays, even though both are made of photons (and, in vacuum, both travel at the speed $c$). What determines the energy of a photon is the frequency of the excitation (frequency of the corresponding light ...

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Refractive Index is when light travels more slowly in a medium. Here is an example of light being slowed down to 38 miles per hour. The speed of a photon does not affect its energy. It has zero mass, therefore zero kinetic energy. The energy it has is due to its frequency (color), and nothing else. (However, it does have momentum!)

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Alas, it was very simple! $$\delta U_j =-i\Delta t \delta u_k(j)\mathcal{H}_kU_j \, . \tag{1}$$ $$\frac{\delta U_j}{\delta u_k(j)} =-i\Delta t \mathcal{H}_kU_j \, . \tag{2}$$ Now, looking at $\Phi_o$, we see the only terms that actually have a $u_k(j)$ dependence are $U_j$ and $U_j^{\dagger}$, hence we have the following via the product rule: ...

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Why is not really important, how is. If you ask yourself why then the answers can be many, for example Why does gravity make two masses attract each other? The answer is because it does, what is really important is how and for that you have a first theory, Newton's Law of Gravitation, this theory is only true for relatively small masses (or masses with ...

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Well, we could say, yes, that is simply how quantum mechanics works. But these are not the axioms of quantum mechanics, and the exclusion principle in particular is really only understood in the context of quantum field theory. The electron does not "spiral in" because it doesn't move in the classical sense at all. At the scale of the size of an atom, ...

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Let's start with questions 1, for which the answer is "no". An isometric quantum channel $T_{A\to A^{\prime}}$ is a channel $T_{A\to A^{\prime}}:\mathcal{B}(\mathcal{H}_A)\to\mathcal{B}(\mathcal{H}^{\prime}_A)$ such that $T_A(\rho)=U\rho U^{\dagger}$ with $U^{\dagger}U=1$ and $UU^{\dagger}=P$ with some projection onto a subspace. The conditions on $U$ mean ...

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With this transformation $\rho = kr$ you can't possibly change the form of the equation, because it's a scale transformation that does nothing special to the derivatives, so I think that you have computed the derivatives wrong.

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If there is evidence of some balance problem that involves energy the missing energy must be somewhere and the extra energy must come from somwhere. A good way to see this was detailed in the Fyenman Lectures, where Richard Feyman explained conservation of energy in a very simple way. So the answer is: No, Energy cannot be Created or Destroyed.

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It is not the out of phase states that get dissipated to wider environment. Rather, it is only the interference components with those out of phase states that get dissipated into the wider environment. I know you got to this question from a question about the Everett ("many worlds") interpretation. If you find Timaeus' more technical description difficult ...

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@Couchyam gave a very nice, detailed explanation. I'll just add that the density of states is just the continuum limit of the degeneracy of energy states. For a discrete system $\Omega\left(E \right)$ gives the number of microstates with energy $E$. When calculating average values of bulk thermodynamic properties, this acts as a weighting function. To find ...

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Expanding on physicsphile's answer, there is an alternative way of computing the expectation value of $f(q)$, and that is to sum the possible values of $f(q)$ weighted by their respective probabilities as follows. If $f(q)$ represents a physical dynamical variable, then it is a Hermitian (self-adjoint) operator and can therefore be diagonalized by some set ...

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Since state $\sigma$ is not in thermal equilibrium I don't think one can use your definition of "thermodynamical" entropy. In fact, one should instead use Von Neumann entropy, which is a correct measure of statistical (so not quantum!) uncertainty. There is no other "classical" or "thermodynamical" entropy in quantum systems. As you mentioned, for a thermal ...

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They give you more detail. Using the the eigenfunctions you can work out how probable different measurement results are. The probability of eigenvalue $i$ being measured is: $$\left|\int \psi \phi_i^*dq\right|^2$$ where $\phi_i$ is the corresponding eigenvalue.

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In general, the density of states can be computed as follows: Find eigenstates of the Hamiltonian, $\psi_{s}$, so that $H\psi_s=E_s\psi_s$ (except in special cases, this is usually the hardest part). Compute $N(\epsilon)$, defined as the number of states $\psi_s$ with energy $E_s<\epsilon$. $D(\epsilon)=\frac{dN}{d\epsilon}$ is the density of states. ...

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The question is really one of definition. In the math literature on self adjoint opertors the "discrete spectrum" is by definition that part of the spectrum which consists of normalizable states, while the "continuous spectrum" is that part where they are non-normalizable. It is possible to have a physical system (a random potential on the entrire real ...

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Looking at simple cases might help. You can have a vector $\vert\Psi\rangle$ in a Hilbert Space and represent it as a pure state such as $\vert\Psi\rangle\langle\Psi\vert$ and you could do the same for the vector $\vert\Phi\rangle$ in a Hilbert Space and represent it as a pure state such as $\vert\Phi\rangle\langle\Phi\vert.$ You could also imagine that ...

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The standard procedure is the following: starting from $\langle nlm|\,\partial^2_z\,| n'l'm'\rangle$ insert the identity operator with respect to the position basis $$1 = \int d\textbf{r} |\textbf{r}\rangle\otimes\langle \textbf{r}|$$ to have $$\int d\textbf{r}\, \langle nlm\, |\,\partial^2_z\,|\textbf{r}\rangle\cdot\langle \textbf{r}|n'l'm'\rangle. ... 0 The article isn't bad, but the author rather presents hypothesis as fact. For example look at this bit: If a particle interacts with just a single photon, for example, then the two particles will enter an entangled state and that will be enough to trigger the onset of decoherence (for example a single photon entering the double-slit experiment will be ... 5 The inner product is always between two (ket) vectors. However, let \mathcal H be the space in which they live. This is a complex vector space with a Hermitian inner product. The inner product defines a map \mathcal H\to\mathcal H^\vee, where \mathcal H^\vee is the dual of \mathcal H, and consists of linear functionals on \mathcal H. The map is ... 0 A density matrix quantum mechanically is a way to easily model the many body wave function of a many particle system. It is a matrix whose off diagonal elements keep the phases between the disparate wavefunctions and the amplitude of projecting particle A to particle N. This is the "information" that particle A has for the existence of particle N. Thus ... -3 You can take inner product of any 2 vectors if they have same dimension. However, we consider the physics problem in quantum mechanics, where inner product of bra and ket vector (of course is complex conjugate of bra vector) of wave function means probability density of finding particle. Inner product of 2 arbitrary vectors (with same dimension) is no ... 1 Is it possible for two events happen at the exact same time? No. Even at any one event itself there can be several (or in though-experimental principle even arbitrarily many) distinct participants (encountering and passing each other, momentarily). All their individual distinct times (indications) are attributable to this one event of their meeting. ... 1 Right before the part you boldfaced and right after the part you didn't boldface there was a parenthetical part. And the parenthetical included a link. In the link the author is pretty clear that the phases are in the reduced density matrix. A reduced density matrix is about reproducing the statistics and usually about adding some classical statistics on ... 2 Is it possible for any two thing to occur at the exact same time This is a physics question and answer site. In physics our examination of nature has shown that there exist many frameworks for defining "events" , as in your title, or "things" as in your question. The main frameworks where "simultaneity" and "event" have to be defined so as to make the ... 0 Events are points (x,t)_S onto a chart S on some space-time manifold and in this respect whenever two such points P_1 = (x_1,t), P_2=(x_2,t) have the same t-coordinate in that reference frame then yes, they do occur at the same time for the observer described by the chart S. For another observer, represented by a different chart S', the ... -1 In a non-relativistic point of view, yes. Events occurring in less than 5.39106 × 10^{-44} s (the Planck time) are considered to be simultaneous. The Plank time is the time it takes light to travel 1.1616 × 10^{-35}m, about 1/(2.67×10^{24}) the size of the hydrogen atom. The Plank time is also, considered to be the smallest time lapse that could ever ... 0 It's important to understand that the size of the state space in the many worlds or Everett theory is constant: it doesn't increase when the individual "worlds" or eigenstates "split up" or become macroscopically distinguishable. A consequence is that just as the "worlds" can "split up", they can also "merge", such as when the two spin eigenstates of an ... 3 A simpler example than putting humans inside your equations might be more clear. Imagine a single electron with some spin. It is going to enter a Stern-Gerlach device as a beam going in the positive y direction and the Stern-Gerlach device will deflect a spin up beam entirely left. And the Stern-Gerlach device will deflect a spin down beam entirely right. ... 1 Solving time-dependent SE as danielsmw mentioned above good starting point.$$ i\hbar\frac{\partial\psi}{\partial t}=H\psi  \frac{d\psi}{\psi}=\frac{H}{i\hbar}dt  log(\psi)\mid^{\psi}_{\psi_{0}}=-\frac{iHt}{\hbar}\mid^{t}_{t_0} $$suppose \psi=|\alpha,t> and \psi_{0}=|\alpha, t_{0}>$$ \psi=e^{-\frac{iH(t-t_{0})}{\hbar}}\psi_0 $$Under ... 2 The assumption behind the statement you ask about is that one's consciousness of the world corresponds to a single classical state and not a quantum superposition: that is, if you looked at schrodinger's cat, you would "see" it in either a "dead" state or an "alive" state and not a mixture of both. (You may want to look up schrodinger's cat if you don't ... 0 Why doesn't Many-Worlds interpretation talk about many worlds? Because it's popscience pseudoscience that even after sixty years has no evidential support whatsoever. I'm afraid it's one of those things that has gained some acceptance because people have grown up with it, rather than for any scientific reason. I'm not really understanding the reason "It ... 5 Yes excited states have a non-zero lifetime. Electronically excited states of atoms have lifetimes of a few nanoseconds, though the lifetime of other excited states can be as long as 10 million years. The decay probability can be calculated using Fermi's golden rule. The lifetime is then an average lifetime derived from the decay probability. The lifetime ... 0 The particle and antiparticle pair don't emerge from nothing, but rather the field (e.g. lepton field for electrons and positrons) that permeates the vacuum over all space. So pair creation an annihilation isn't tied to the vacuum but to the quantum field, and it happens everywhere - not just in an experimental vacuum. I.e it happens in the nucleus of an ... 1 The characteristic time of interaction - energy of interaction relation between two systems is usually written as \delta E\cdot\delta t\sim\hbar/2 (do NOT mix with the uncertainty principle). So the characteristic time would be about \delta t\sim\hbar/(2\delta E), where for \delta E we can take the difference of energies between two states. 0 This resource ended up basically explaining the answer to my question: https://wiki.physics.udel.edu/phys824/Discretization_of_1D_Hamiltonian 1 The quantity minimized during the geometry optimization is the total energy U of the molecular system with nuclei being fixed. It consists of the kinetic energy of electrons, the potential energy of the Coulomb repulsions between the electrons, the potential energy of the Coulomb attractions between electrons and nuclei, and the potential energy of the ... 3 Your matrix is wrong. It should be det {\begin{pmatrix} -\lambda & \hbar/2 (1-i) \\ \hbar/2 (1+i) & -\lambda \end{pmatrix}} = 0 Now when you compute the determinant, it should be okay. -3 If we calculate the energy required to assemble an electron from a thinly spread out cloud of charge as e, then e is almost equal to the rest mass of the electron. (This is taking into account the experimental size of the electron from collision experiments and uniform charge density in the electron). Though this information does not explain the stability ... 4 Everything follows is valid at least when the various operators appearing below are defined on the dense subspace finitely spanned by the standard vectors |n\rangle, n=0,1,2,\ldots. From [a,a^\dagger]=I and [A,BC]= B[A,C]+[A,B]C, by induction, one easily finds that$$[a,(a^\dagger)^n] = n(a^\dagger)^{n-1}\:. If $\alpha_k \in \mathbb C$ and ...

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There is no such thing as "proving a postulate" theoretically, because your proof will always need to start from somewhere. What you can hope for is to show that a set A of postulates, which seems clunky and ad hoc, actually follows from a simpler set B of postulates. The clearest example of this is when Einstein showed that the Lorentz transformations, ...

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The very useful solutions of the Shrodinger equation that are usually taught in beginning quantum mechanics are not Lorenz invariant and therefore paradoxes with respect to special relativity may be constructed. The relativistic equations of Dirac: the Dirac equation is a relativistic wave equation derived by British physicist Paul Dirac in 1928. In its ...

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There have been conducted many experiments in which light impulses travelled faster than light Are you aware of any scientific publication on major journals officially acknowledging such event? As far as the scientific community knows, there is no such evidence. Even though there were any, this would have nothing to do with quantum entanglement anyway, ...

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As another side note, if you know the ground state, it IS possible to use the variational method to get the next-lowest energy. You simply restrict your trial wavefunctions to be orthogonal to the ground state, and then minimize as usual. If you know the two lowest ground states, you can use the variational method to find the third-lowest energy, etc. Also, ...

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Comment on the answer of @Michael: The short answer is that you need antiparticles is false. In Quantum Field Theory you have perfectly working solutions also without antiparticles, i. e. for real fields. Even if you do want to consider antiparticles, always have in mind that despite the misleading name they are in fact different particles from the ...

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This is just an additional remark – the answer by NowIGetToLearnWhatAHeadIs is fine. But you can get all states from the variational method, just not with the trial function method (unless your trial function space contains the exact excited states). While the excited states are usually not extrema of the variation functional, they are stationary points! ...

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