# Tag Info

## New answers tagged quantum-mechanics

0

Complex conjugation has nothing to do with charge conjugation. Charge conjugation flips quantum numbers, which don't appear at all in the standard Schrodinger equation. The actual symmetry related to complex conjugation is time reversal. However, to actual perform time reversal, you must also replace $i$ with $-i$, so the time reversed Schrodinger equation ...

1

There is absolutely no "freedom for interpretations" here. Both books should – and all other books that are not completely wrong do – agree with these formulae and agree that $a_n$ never indicates an eigenvalue. In both books and all of science, $a_n$ is the complex probability amplitude such that $|a_n|^2$ represents the probability that the system has ...

1

The eigenstates of a time-independent Hamiltonian $$H |\psi_j \rangle = E_j|\psi_j\rangle$$ have the usual rotating-phase time dependence in the Schrödinger picture: $$|\psi_j(t)\rangle = |\psi_j(t_0)\rangle \cdot \exp(E_j(t-t_0)/i\hbar)$$ However, your formula indicates that $H_0$ and $V$ are time-dependent. So if the state vector is an eigenvector of ...

0

I cannot be sure, but I suspect that you can get analytical solutions of the Pauli equation by taking a non-relativistic limit of analytical solutions of the Dirac equation. The latter can be found in many books, say Bagrov, Vladislav G. / Gitman, Dmitry, The Dirac Equation and its Solutions (http://www.degruyter.com/view/product/177851) (you can find a ...

2

The spacetime geometry may be indirectly read from the differential operators in the equations controlling other objects – particles and (matter) fields. Schrödinger's equation contains the differential operator $$\Delta = \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$ which is the operator ...

6

First, there's no physical $c\to 0$ limit of relativity because that limit basically says $v\gg c$, velocities are much larger than $c$. But relativity prohibits $v\gt c$, let alone $v\gg c$, so this limit $c\to 0$ i.e. $v\gg c$ cannot be taken. Relativity generalizes non-relativistic physics so that it allows large $v$, $v\sim c$, but at the same moment, it ...

2

The first part of the sentence is right, given some assumptions, the rest is not. The most precise description of every physical system is in terms of a very general superposition of a priori possible states. There is never any collapse. But the rest isn't true. An off-diagonal element of the density matrix doesn't correspond to any single state. It ...

1

I am adding a few points that Lawrence's answer didn't yet address, but should help clarify the nature of the quantum tunneling process: According to the QM postulates, the unitary evolution of such a system should by definition keep it reversible, so it is only when measured that a decay can be observed or not. But this would make the decay rate ...

0

We want two things from quantised physical conserved charges. The quantisation construes the charge as a linear operator on the Hilbert space of physical states. We require this operator to be a Hermitian operator that commutes with the Hamiltonian $H$. Exercise: prove that, if $A,\,B$ are Hermitian operators that commute with $H$, then $i\left[A,\,B\right]$ ...

0

The argument is wrong, because the set of eigenstates of H and A may be complete. For H and B too. As a trivial (even if artificial) example use H=I and some arbitrary pair of operators which define a complete base even if taken alone, say, A=Q, B=P. Of course, AB-BA will commute with H anyway, because H commutes with A and B. But this may not give ...

1

You're looking for a set of wavefunctions $\psi_n(x)$ that are solutions to $H\psi_n=E\psi_n$. The index $n$ does not refer to a discretized position. For numerical solutions, you will have to discretize $x$, for example with constant steps $h$ (you call them $\Delta x$); then the second derivative will be approximated as {\partial^2\psi\over\partial x^2} ... 1 What you quote there is not the equation for weak hypercharge, but for strong hypercharge, a concept from the early days when we didn't understand how all those particles were formed from quarks. Yes, the quarks do not possess the same strong hypercharge, strong hypercharge is a rather arbitrary (but conserved) combination of the quark numbers that was ... 2 I agree with CuriousOne that you would be better off ditching both of these viewpoints and looking for something more modern. However, this is instructive because it does illustrate a common problem in QM education: many authors are invested in a particular interpretation, and present that interpretation (disingenuously, to my thinking) as the only correct ... 1 The Fox-Goodwin reference is to a method for numerical integration of ordinary differential equations, which must refer to the algorithm you were to use. Possibly the problem was taken from a paper which referenced this method; that means you can use a scientific citation index to find papers which cite Proc. of Cambridge Phil. Soc. 45 (1949) 373. You can ... 2 Let the magnetic field be in the \hat z direction. If you calculate the expectation values of S_x and S_y, you find that they have time dependence like \cos(\omega t),\sin(\omega t) while the expectation value of S_z is constant. Explicitly, the Hamiltonian is H = -\omega \sigma_z. Using the Heisenberg equation of motion, \begin{align} \dot ... 0 As far as I know, the external magnetic field that produces a torque on the magnetic moment is not necessarily the reason for the precession of it around the direction of the magnetic field. The magnetic moment of an electron is proportional to it's spin and its revolving motion around the nucleus. so when an external field acts on it, it tends to align in ... 0 It's spin-orbit coupling in QCD--ala hyperfine splitting; however, in QCD it's not so fine. 2 \def\kk{\mathbf{k}} \def\ii{\mathrm{i}} \def\qq{\mathbf{q}} You can prove this using translation invariance, without any other assumptions on the nature of the interacting ground state. I am going to give the argument in D dimensions, which obviously holds in D=1 as a special case. Spatial translations of the system as a whole are generated by the ... 0 While reading Dicke&Wittke, pp. 274-275, it occured to me that a (maybe more) convincing derivation of the "transition rate formula", as an experimentalist would interpret such a thing, can be given as follows. As usual, we assume that the system is in a pure state \vert i \rangle at t=0, when the perturbation (which for simplicity I assume to be ... 2 I thought I would indicate how radiactive decay can be argued to occur from quantum tunneling. This derives in a very “back of the envelope” way the phenomenological equation for radioactive decay. From there I can argue some about the role of observing radioactive decay. Quantum tunneling can be see with the square potential barrier. A square potential ... 1 You are misunderstanding (and thus misquoting) the book of Nielsen and Chuang. What they do is to first define the reduced density matrix \rho_A of a bipartite state \rho_{AB} as\rho^{A} =\mathrm{tr}_B(\rho^{AB})$$Then, they explain what a partial trace is: It is the linear map defined by its action$$ \mathrm{tr}_B \left( |a_1 \rangle \langle a_2| ...

0

A mixed state just describes our ignorance about a particular system I don't think you can call our inability to access a pure state of any system an ignorance about a particular system. Because I think of pure state as a mathematical abstruction that can only be related to reality by application of Born rule - wich either reflects our fundamental ...

2

Engtangled photons: first you must understand that the photon is the particle obtained when the modes of the electromagnetic field are quantized, and that they are created and destroyed as discrete quanta of energy, in agreement with Planck's relation, $E=hf$, where $f$ is the frequency of the electromagnetic field corresponding to the quantized mode; that ...

0

PART A : Plane Phase Waves and Lorentz Transformations In the following it's necessary to use a more general form of Lorentz Transformation between two systems $\:S\:$ and $\:S^{\prime}\:$. So, let the system $\:S\:$ translating with constant velocity $\:\mathbf{v}\:$ with respect to $\:S^{\prime}\:$ (see Figure in the bottom). With finite variables ...

0

"What entangled photons really are?" Photons are measurements of the state (changes) of a quantum field. Let's take your knowledge about the hydrogen atom as a starting point. Let's say your atom is in a p-state. The atom then changes into an s-state. Its angular momentum and energy change. Angular momentum and energy conservation demand that both ...

1

What you have here could be described as a subset sum problem. Given $n$ can take any integer value (not including zero), you have the set of squares up to $36$, $S = \{1,4,9,16,25,36\}$ and you wish to find subsets of three which sum to $41$. Looking at the subset sum problem this can not be solved analytically but algorithms can be employed. To do this ...

2

If you are happy with Bob being right only in some percentage of the cases, there is a much easier protocol which does not even require entanglement: Just let Bob guess Alice's bits. He will be right in 50% of the cases. Note that once the probability to guess the right result is above 50%, communication is possible, e.g. by doing majority voting. In ...

4

In the quantum mechanical description of any physical system, including a quantum field or a collection of interacting quantum fields, there is always one state vector – one collection of numbers (probability amplitudes) that generalizes what is referred to as the "wave function" in quantum mechanics of particles. In quantum field theory, a better name is a ...

2

It's neither a classical wave nor a classical particle. I think any attempts to describe it as either of those need to be qualified like this. It might look like one or the other, but both are only approximations. The best theories we have describe quantum fields, and a particle is a field quantum. I don't really know how to describe a field quantum in ...

0

I think I understand what you are getting at. I will take my best shot. Particles are subatomic and waves of particles are classic physics. Quantum physics and classic physics do not agree. Einstein saw the power in matter while Tesla saw the power in empty space that was just proven with the Higgs Boson and Quantum Entanglement. Both are right or wrong, ...

2

The short answer is that only if the Berry curvature is defined by: (in matrix notation): $$F_{\mu \nu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} + [A_{\mu}, A_{\nu}]$$ it becomes gauge covariant, i.e., for a gauge transformation: $$A_{\mu} \rightarrow g^{-1}A_{\mu} g+g^{-1}\partial_{\mu}g$$ $g \in U(N)$ ($N$ is the degeneracy of the level), the ...

5

The answer to your question is NO. The simplest counter-example is the identity matrix. There is no solution to $|a\rangle\langle b| = I$. For example, try to solve $\begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} z & t \end{bmatrix} = \begin{bmatrix} 1&0 \\0&1 \end{bmatrix}$. Since $xt=0$ we know that $x=0$ or $t=0$. But clearly $x$ ...

2

Spontaneous emission and stimulated emission are cases related to measurement, but they are not identical to measurement. We might ask how the measurement of states and the preparation of states are related to measurement, or the decoherence of states. Both are necessary aspects of quantum physics. In order to do experiments we need to prepare identical ...

6

Group 2 has some non-objectable contents (“for these other particles, we do not even have a seemingly concrete macroscopic property to associate with the wave”), but is otherwise inconsistent (“We arrive at the conclusion...”: how does the “conclusion” relate to the previous statement in any way?) and wrong in the main aspect with which you're concerned ...

7

Not every tensor is a simple tensor. The simple tensors of the form $a\otimes b\in A\otimes B$ span the tensor product, which means that a general element $t$ of the tensor product is a linear combination of these simple tensors, i.e. $$t = \sum c_{ij} a_i\otimes b_j$$ for some basis $a_i,b_j$ of $A$ and $B$ respectively. If $A$ itself is a space of ...

6

No. Counterexample: choose a basis in 2D quantum space, so the basis states are written $\left(\begin{array}{c}1\\0\end{array}\right)$ and $\left(\begin{array}{c}0\\1\end{array}\right)$. Now think of a mixture, with a state's having probability $p$ to be in state 1. The density matrix is then $\mathrm{diag}(p,\,1-p)$, which is not singular for ...

9

There is no contradiction. This: "The de-Broglie wave and the particle are the same thing; there is nothing else. The real particle found in nature, has wave properties and that is a fact." is a more general statement. Note that it does not define what is "waving". It just states that the particle is characterized by a wave. This goes into the ...

1

Photo-emission is not a simple one step process. The incident photon excites an electron, but the momentum of the initial photoelectron is in the same direction as the incident photon i.e. down into the bulk of the metal. For the electron to be emitted as a photo-electron it has either to backscatter off another electron in the metal or it has to transfer ...

5

Phenomena in quantum mechanics may be expressed using any basis. It doesn't mean that all bases are equally useful for a given situation. In particular, a fundamental postulate of quantum mechanics says that right after every measurement, the system is found in one of the eigenstates of the observable that was just measured. That's why the basis of the ...

1

There are several factors which determine the energy acquired by a photo-electron: the structure of the crystal surface determines the work function, which means that (111) surface is different from (100); only a perfect crystal has a single surface structure. There is always some bandwidth to the optical source, though it may be quite small. One should ...

0

There is a similarity. In quantum mechanics the uncertainty relation is determined as an anticommutator. There is also non commutativity in Lorentz transformation boosts so there is an order dependency of (some) repeated Lorentz transformations. Understood in a similar way as quantum mechanics this could be seen as an uncertainty relation with associated ...

1

Here is probably the simplest argument that I have heard of. Detection of a photon from a thermal source gives a rise to a probability to detect several more in a short interval of time due to stimulated emission. Assume that you have some atoms in a medium that emits light, and they are in an excited state. If you know that one atom emitted a photon, this ...

3

For your initial check, you used the formula $c = \nu \lambda$, giving the incorrect formula $E = pc$. This is wrong, matter waves don't travel at the speed of light; instead, $\nu \lambda$ is equal to the phase velocity $v_p$. You then use $E = pc$ in all the other derivations, making them all wrong. Hypothesis #1 is correct for relativistic quantum ...

2

Every complex number can be written in the form $re^{i\theta}$ for a real number $r$. We call $e^{i\theta}$ the phase. For example, if $$|\psi \rangle = \frac{1}{\sqrt{2}} ( |0 \rangle + i |1 \rangle)$$ then the phases of the $|0 \rangle$ and $|1 \rangle$ components are $1$ and $i$, and their relative phase is $i$. Now consider $$|\psi' \rangle = ... 0 Anna, you are more or less right. It is pressure from matter at higher densities that can stop the gravitational collapse. It depends on the state of the matter, and in a simplistic description the equation of state. As it collapses as a hot gas after it exhausts it nuclear fuel, and maybe after a supernova explosion, it'll collapse. The first point at ... 0 Your Hamiltonian is much the same actually. If I write e^{i\omega t}~=~cos(\omega t) + i sin(\omega t) the quoted driving term is$$ H_d = \epsilon \left(a(cos(\omega t) + i sin(\omega t)) + a^\dagger(cos(\omega t) - i sin(\omega t))\right)  = \epsilon \left((a + a^\dagger)cos(\omega t) + i(a - a^\dagger)sin(\omega t))\right) $$You can check that ... 2 I'm just guessing here, but one way to interpret it goes like this. If you have the amplitude \Phi for an outcome that can be achieved in multiple indistinguishable ways each described by an amplitude \phi_i with i \in 1, 2, \dots , n we write$$\Phi = \sum_{i=1}^n \phi_i \,.That makes the probability for the outcome \begin{align*} P &= ... 0 Take for example the double-slit experiment interpreted in the Copenhagen sense. The particle leaves as an object with mass, yet passes through the slits as a massless wave, only to collapse again as a particle. We can consider this example as a generalisation of the principle of anti-realism posited by Bohr. Where does the energy "go" when the ... 4 The basis is still \{|\boldsymbol r\rangle\}. The abstract Schrödinger equation is i\frac{\mathrm d}{\mathrm dt}|\psi\rangle=H|\psi\rangle $$where |\psi\rangle is a set of four kets, (with a slight abuse of notation)$$ |\psi\rangle=\begin{pmatrix}|\psi_1\rangle\\|\psi_2\rangle\\|\psi_3\rangle\\|\psi_4\rangle\end{pmatrix}  Time is still a ...

2

Short answer: no. I'll give some context with the details of the simplest examples. In the context of conservation laws, "energy" refers to the Hamiltonian. In classical mechanics, a quantity without explicit time dependence is conserved iff its Poisson bracket with the Hamiltonian is 0. In quantum mechanics, quantities are promoted to operators on a ...

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