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1

From the equations, $\phi$ is the operator acting on the variable/state $\xi$. It is important to notice also that in the factorization the $a_i$ numbers are real and the $c_i$ are complex. This factorization comes just from the mathematical fact that for a polynomial equation of degree $n$, such as the first equation, there are $n$ complex roots.


1

This makes sense to me. $l$ and $m$ are two different summation indices. If you used the same index, say $l$, you would have only the $c^*_{lu}$ $c_{lu}$ terms, without the cross terms when $l\neq m$.


1

No $\hat\phi|0\rangle$ is not an eigenvector of $\hat\phi$. You can see this, for example, by writing out $\hat\phi$ in terms of creation and annihilation operators, then compare $\hat\phi|0\rangle$ against $\hat\phi^2|0\rangle$, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct To obtain some analogy of ...


2

You might be interested in this question or this question (and some others I cannot track down now). The basic problem is this: It is not clear what we exactly mean by "deterministic". If you mean that we can in principle determine the future state of a system solely from initial conditions, then the time evolution given by the Schrödinger equation is ...


0

To say the least, they are inseparable. The "indeterminacy" is meant to be a synonym of the "uncertainty" (original in German: Unschärfe oder Unbestimmtheit), e.g. the nonzero values of $\Delta x$ (uncertainty of position) and $\Delta p$ (uncertainty of momentum) that obey $$ \Delta x \cdot \Delta p \geq \frac\hbar 2$$ This is a consequence of the nonzero ...


0

So there is no wave particle duality. By particle we tend to use the classical macroscopic definition of a particle, which means a shape( a volume) a center of mass that describes absolutely the particle's position in (x,y,z,t). A billiard ball, for example. Electron is always a particle. But the location of electron is represented as wave, because ...


2

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230. Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $\hat{p}$, but $\hat{\pi}$ (the canonical conjugate momentum of the field $\hat{\phi}$). With slightly more precision: the Fock space ...


0

See this answer and the comments. There is an explanation how occures fringes due to the EM field from the surface of the slit or wire: https://commons.m.wikimedia.org/wiki/File:Moellenstedt_biprisma_voltage_shadow.JPG It shows the influence of an electrical field to fringes. ...


3

Isn't it just generally true that in the absence of any potential, the momentum eigenfunctions are also energy eigenfunctions? In other words, when there is no potential, (in the right units) $$ H = p^2/2m + V(x) = p^2/2m $$ Since the Hamiltonian is proportional to the momentum operator squared, it's easy to see that any eigenket of the momentum operator ...


1

These relations are based on the fact that both the position and the momentum distributions are centred around zero, which is in turn due to the symmetry of the atom. Given that, the width of the position and momentum distributions ($\Delta x$ and $\Delta p$) is of the same order as a typical position or momentum within those distributions ($r$ and $p$).


1

First of all, it is indeed correct to model decoherence the system has to interact with what is called the "environment". Basically you have a joint CLOSED (unitary) evolution of system+environment, after which you discard the environment (technically called a partial trace), and you are left with the state of the system. Your "observer" can be taken as part ...


5

Saying that things "are both waves and particles" is a vestige of the 18th century way of thinking, and really ought to be done away with. Everything is described by a wavefunction. Period. What is a wavefunction? It is a complex-valued function. If you are interested in an electron's position, it is easy to think of it as a complex-valued function of ...


3

An electron can interfere with itself. there have been experiment of interference with single electron. Saying that an electron is alway a particule is then wrong. The wave function is the "best" way we have find to describe electrons and other quantons. object at the quantum level are not wave or particles. They just follow quantums rules and waves or ...


2

In the special case where $\ell^2$ or $s^2$ has eigenvalue zero, then $j^2$ is fixed. Otherwise you must know the projections $m_\ell,m_s$ to find $j$.


1

The way this is justified is as follows: We start with the uncertainty principle, which can be roughly stated as $$\Delta x \Delta p \geq \hbar$$ For this rough estimate, we will ignore some factors of perhaps $2$ or $\pi$, but we're interested in some order of magnitude, not the exact result. Now, our second assumption will be that the ground state of the ...


1

Perhaps you are starting by the wrong end. Your concern seems to be related in the first term with the totally misleading notation of integrals in quantum mechanics, and this is more related with the spectral theorem than with distributions itself. Distributions only appear in Quantum mechanics when certain operators has empty spectrum in the usual Hilber ...


2

There seems to be no way to proceed unless more information is given. In fact, from what you have above (correcting the typo pointed out by Bernhard and ticster): $$ \vec{j}=\vec{l}+\vec{s} \quad \Rightarrow \quad \vec{j}^2=\big(\vec{l}+\vec{s}\big)^2 = \vec{l}^2+\vec{s}^2 + 2\, \vec{l}\cdot\vec{s}\,, $$ meaning that knowledge of the eigenvalues of ...


3

$$D(\hat{X}_i):= \left\{ \psi \in L^2(K, d^nx)\:\left|\: \int_K x_i^2|\psi(x)|^2 d^nx< \right.+\infty \right\} = L^2(K, d^nx)$$ where the last identity holds true if $K$ is bounded (in particular compact) because $x_i^2$ is bounded as well thereon (in this case the operator is bounded, too). Moreover $$D(\hat{J}^2):= \left\{\psi \in L^2(\mathbb S^2, ...


1

In perturbation theory perturbed eigenstates expanded by unperturbed eigenstates, Yes, they are! but we know when the system perturbed its Hilbert space altered and hence its basis changed, Having a basis means that any vector is expanded by those. Thus, in general, the new eigenvectors are a linear combination of the former. then we can't ...


0

I think in quantum mechanics we assign to each system a specific Hilbert space i.e. if systems are different then their Hilbert spaces are different. Is this true? If not why? For differernt system I mean their hamiltonians are different. I found really interesting the question and @joshphysics, however I desagree with his point of view. It is true ...


0

To answer the question I think you meant to ask, $V(x,t)$ is a real-valued function. But you can also think of it as an operator. An operator is anything that maps functions to other functions, and multiplication by a fixed function is one way to do that. To be a little more pedantic with the notation, if you use $\hat{V}$ to represent the operator, then ...


0

As usually, physicists are being a bit sloopy with notation here. Mathematically, there's a big difference between a function (such as $V$) and a function's value for some arguments (such as $V(x,t)$). The latter is indeed simply a number, scalar if you like. However, in physics, functions are often used just as "families of values" and values with some free ...


3

In the representation given by you, the $V(x,t)$ is a scalar function depending on $x$ and $t$. However, you already have choses a basis (the $x$-basis). In general, the $\hat{V}$ or rather $\hat{H}$ in the Schödinger Equation is an operator. This operator can then be evaluated in a basis of your choice. If you are familliar to the dirac notation one can ...


3

The potential is, without any ambiguity, an operator. There is no other way, because it changes the spatial dependence of the wavefunction: $$ \Psi(x,t)\text{ changes to }V(x,t)\Psi(x,t). $$ It's unclear to me what you mean by "scalar", though, because this word can have multiple meanings. For example, it is indeed a scalar operator - as opposed to vector ...


2

Indeed, the Slater-type orbitals (radial wave functions) are not orthonormal – they are not even orthogonal to each other. The $\delta_{n,n'}$ Kronecker delta symbol doesn't appear in the inner product and it can't because the $r$-dependent integrand is positively definite and there is no room for cancellation. Their not being orthogonal physically means ...


0

So one detail I omitted from the question was that: $$ \psi_{sc}(k)=\frac{g+g I}{2\pi(k^2-p^2)} $$ Where: $$ I=\int^{\infty}_{-\infty}\psi(q)dq \space\space\space\space (1) $$ (I had used in arbitrary prescription in the original description of the problem, this is what I obtain before solving for $I$)$$\\$$ Using equation (1) we can solve for I, obtaining: ...


2

The planck length is not necessarily an absolute limit to how small thing can be sub divided. The planck length is theoretical and it is empirically defined by dimensional analysis. At this length scale our knowledge of physics makes no sense. The planck length $\ell_P$ is defined as: $\ell_\text{P} =\sqrt\frac{\hbar G}{c^3} \approx 1.616\;199 (97) \times ...


1

No that has nothing at all to do with dark matter. The nature of dark matter is currently unknown, that does not imply that it is opposite of normal matter in any sense. Dark matter is simply a hypothetical particle(s) that don't participate in any interactions except for gravitational interactions, or potentially, the nuclear forces.


2

Not every Hermitian operator is an observable. The identity on a space is certainly Hermitian, but it is no observable in any physical sense - its eigenvalue is $1$, and that's it. The eigenvalue of scalar operators gives you precisely zero information about the state considered, and usually, we would like an observable to at least give us some information ...


4

In the simple case of a pure state, the density matrix is the projection $\rho_\psi=\lvert\psi\rangle\langle\psi\rvert$, for some $\psi\in \mathscr{H}$ (the Hilbert space). Measuring it on another general state $\rho$ (i.e. calculating $\mathrm{Tr}[\rho_\psi\rho]$) gives you the probability of $\psi$ being the "configuration" (or Hilbert space vector) ...


3

All the definitions you've posted are correct, and they aren't in conflict with each other, although they are a bit imprecise. I'll try to explain in more detail what these concepts are. Probability I'll assume we don't need to attempt to define what probability is. :) I'll just note that it's formalized in mathematics under the aegis of measure theory. ...


3

To understand the difference between probability and probability density consider the difference between mass and density. Density is the mass per unit volume, so to find the mass you multiply the density by the volume: $$ mass = density \times volume $$ In some cases the density will be a function of position and we have to write it as a function of the ...


-1

It has been many years since I have used this information (so please forgive my inaccuracies, but I wanted to get you PART of the way there with an answer (since no one has responded yet). The probability density is the integral (area under the curve) of the probability. I think the reason for the funky discrepancy between the two definitions is because.. ...


1

The Stanford Encyclopedia of Philosophy's article on Everett gives a much better description than Wikipedia. "Everett's solution to the [measurement] problem was to drop the collapse postulate from the standard formulation of quantum mechanics then deduce the empirical predictions of the standard collapse theory as the subjective experiences of observers who ...


0

Answer to question one : The Principle of Quantum Mechanics by R. Shankar page 149 reads "Barring a few exceptions, the schrodinger equation is always solved in a particular basis. Although all basis are equal mathematically, some are more equal that others. First of all, since H = H(X,P) the X and P basis recommend themselves....The choice between the two ...


3

$$ \newcommand{\ket}[1]{|{#1}\rangle} \newcommand{\bra}[1]{\langle{#1}|} \newcommand{\braket}[2]{\langle{#1}\,|\,{#2}\rangle} \newcommand{\bracket}[3]{\langle{#1}\,|\,{#2}\,|\,{#3}\rangle} $$ Well, I think by specifying mass $m$ and charge $q$ you simply define your system, a single electron, and then its complete non-relativistic description is indeed ...


2

First of all, light waves and matter waves may be treated together, using the same maths, because the waves associated with light and the waves associated with matter are fundamentally the same thing. Second, all the waves before they interfere and after they interfere may be written in terms of the probability current $j^\mu (x,y,z,t)$, and its ...


1

"Now, is there an increase of gravitational mass corresponding to this increase of inertial mass?" Yes, both increase by the same amount. For any object inertial = gravitational. Even antimatter will fall down at the same speed. "If the original frequency and energy of a photon “launched” from a body M were equal in magnitude to the potential energy, then ...


1

When solving for the spectrum, you assume the solid has infinite extent, and so the electron is not really bound. Thus you can get a continuum of momentum. Consider for example the trivial case where there are no atoms, and you just have a free electron in space. Then the energy is proportional to momentum squared, and momentum can be arbitrarily big, so ...


0

I think this problem is similar to the problem of finding modes of rectangular dielectric waveguide. In this case, you can use the effective-index method for finding the approximated solution (For your problem, we can call it effective-potential method). For more information about effective-index method see the following articles: Effective-index analysis ...


1

Ok, I am by no means and expert on solid state but I might be a little helpful. Band structure in solids arises due only to periodicity of the lattice. It all comes down to this periodicity. Periodicity of the lattice makes the potential also periodic. This periodicity has many (interesting) consequences (Bloch states and bla bla bla) but the one that ...


0

You could think of making a solid by slowly bringing all the atoms closer together. As they get closer the discrete electron orbits start to bump into each other. But the electrons can't be in the same state, and you start to get new states that are a combination of the discrete states.. And these form bands.(Yeah that last is a bit of a cop out, see a ...


0

Because in solid states, the electrons are not bound. A simple derivation of the continuous bands can be made if you look at particle in a box (Wikipedia). If we approach the number of an infinite box with infinite particles (and constant density), there are infinite states, and the states get closer, as the energy gap goes as $L^{-2}$. Therefore a ...


1

If your electron is in a pure state then it's an eigenfunction, $\psi_e$, of the Hamiltonian describing it, $H_e$. The measuring system will also, in principle at least, be described by some wavefunction, $\psi_m$. If the two don't interact then the total wavefunction will just be a product: $$ \Psi = \psi_e\psi_m $$ and the system won't change with time. ...


0

You have to clarify the term "basis" in infinite dimensions. By definition, finite-dimensional spaces have a finite basis (so that, for instance, $\mathbb{R}^n\approx\mathbb{R}^m$ iff $n=m$). But for infinite-dimensional spaces, there are various types of bases (assuming you can take infinite sums). A natural one (using finite sums) is the Hamel basis, but ...


5

First, the term "basis spaces" isn't standard in quantum physics but let us assume that we understand what the sentences approximately mean. Second, the momentum is continuous (not quantized) if the position space is noncompact (infinite). The momentum only becomes quantized if the position space is compact (or periodic), and indeed, it's been ...


1

As the previous post mentioned, forget about the concept that the electron is actually spinning. Spin, like rest mass and electric charge, is an intrinsic property of subatomic particles. Yes, it's angular momentum. No, nothing is spinning. Although many physicists today do not like this explanation, special relativity introduces a useful analogy with mass. ...


2

In a comment you write space time symmetries don't fit into the framework of the action since the action is a functional on the fields only not also on space time (space time here appears merely as a dummy variable This isn't quite right. A given spacetime transformation often induces a transformation on fields themselves, and in this way, spacetime ...


3

First, the electron isn't actually spinning. Physical objects made up of collections of electrons and protons (and neutrons) can have angular momentum because they rotate; the electron does not get its angular momentum for the same reason. Second, the magnetic moment of an object with angular momentum L is proportional to $$ \mu \propto \frac{qL}{M} $$ ...


2

"Multiplying the wavefunctions" is a pretty nebulous term. Let's work with some definite vocabulary here, shall we? $(1)$ The states of one QM particle are elements of some Hilbert space $\mathcal{H}$. If we care only about position on a line as completely defining the state (which we can for a scalar boson), i.e. demand that the space be spanned by the ...



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