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I want to write a simple answer to my question based on the knowledge of one-dimensional system; which can qualitatively answer why 2D is similar to 1D while 3D is not. (1). For one dimensional system, there is at least one bound state for pure attractive potential. (Proof can be done by using a Gaussian trail wave function for variational principal) ...

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To study bound states, we have to find solutions to the Schrödinger time-independent equation $$-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi$$ Using separation of variables, in spherical coordinates $$\psi(r,\theta,\phi)=Y^m_l(\theta,\phi)\frac{u(r)}{r},$$ where $Y^m_l(\theta,\phi)$ are the spherical harmonics, the radial part can be shown after substitution ...

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Why does the statement "any negative potential supports a bound state" hold in 1D, but not in 3D? In short, this is because for a bound state to occur, any positive kinetic energy needs to be fully offset by a negative potential energy. Achieving a large negative potential energy requires the particle to be localized in the volume where the potential is ...

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The precise theorem is the following, cf. e.g. Ref. 1. Theorem 1: Given a non-positive (=attractive) potential $V\leq 0$ with negative spatial integral $$\tag{1} \int_{\mathbb{R}^n}\! d^n r~V({\bf r}) ~<~0 ,$$ then there exists a bound state$^1$ with energy $E<0$ for the Hamiltonian $$\tag{2} H~=~K+V, \qquad K~=~ -\frac{\hbar^2}{2m}{\bf ... 0 Yes, it is time independent. Every eingenstate \psi_k of the Hamiltonian H is an stationary state, since by the Schrödinger equation we have$$i\hbar\frac{\partial \psi_k(x,t)}{\partial t}=H\psi_k(x,t)=E_k\psi(x,t) \implies\\ \psi_k(x,t)=e^{-iE_kt/\hbar}\psi_k(x,0)$$Hence, the probability dristribution \rho(x,t) of position (or any other observable) ... 0 Others have provided mathematical explanations, so I'll try an intuitive explanation. You've probably seen the electric field represented like one of these: (image source) At the top of the mountain there is a positive charge, and at the bottom of the valley is a negative charge. A positive test charge in this field experiences a force analogous to a ... 0 The statement that V is at an extremum can be written \partial_i V=0. Now think about the mathematical relation between V and E in electrostatics. 1 I mean what makes the voltage remain same across two resistors connected in parallel? To be clear, in the context of ideal circuit elements, two resistors are parallel connected if all of the voltage across one resistor is across the other resistor. This defines the parallel connection which is the dual of the series connection. If the voltages ... 0 Recall the definition of voltage: ~V= \frac{Work~done}{Unit~charge} Though current gets distributed, work done per charge remains the same. This is true only if, as Ross pointed out, R_{wire}=0, If it is not, then voltage across the resistor will be less than that across the source, because wire sections will start acting like individual resistors, and ... 0 We usually assume that wires are perfect conductors. In that case, the potential is the same at both ends of a wire, so at corresponding terminals of a parallel circuit. This is an approximation, valid when the resistors in the circuit are large compared to the wire resistance. For your last question, if the source is an ideal voltage source, switching te ... 1 Radio frequency Paul traps confine charged particles without applying any magnetic fields (As in Penning traps), but since confining charged particles only using electrostatic forces is impossible according the the Earnshaw's theorem, a quasi-static approach is taken, and the particles are trapped dynamically. Radio frequency ion traps do this by forming a ... 1 Your finite square well potential looks like: where V_0 is the potential energy outside the well and V_1 is the potential inside the well. The depth of the well is \Delta V = V_0 - V_1. We normally take V_0 to be zero, in which case V_1 is negative (like your -10eV) and \Delta V = V_1. However you can add any constant value to the potential ... 0 If the electric field is conservative, you can integrate along any path between the specified points to obtain the potential. All paths will yield the same answer. This makes it easier, because you can pick any path that is convenient for you - or easiest to integrate. Recall that a conservative field is defined by:$$ \nabla \times E = 0$$So one ... 1 The main point that although a pointwise convergent Fourier series of cosine modes is an even function \psi(-x)=\psi(x), it does not have to be differentiable at x=0. A pointwise convergent infinite sum of differentiable functions is not necessarily a differentiable function. More generally, as OP already mentions, the wave function \psi(x) is not ... 1 In the 1d particle in the box the energy of the particle should be completely determined by the momentum of the particle that you observe correct? The Hamiltonian in the position basis is$$\hat H = \begin{cases}-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}, & 0\lt x \lt a\\\infty, & \text{otherwise} \end{cases}$$and the energy ... 6 The momentum operator P in the infinite well can be defined as a self-adjoint operator by infinitely many ways with respect to the boundary conditions by:$$P_\theta=-i\hbar\frac{d}{dx}\\ \mathcal{D}(P_\theta)=\left\{\psi\in \mathcal{H}^1[0,a]:\psi(a)=e^{i\theta}\psi(0)\right\},$$where \mathcal{H}^1[0,a] is the Sobolev space, on the interval [0,a]. In ... 1 There are several (equivalent) ways to look at this. One is to say that for any conservative force \mathbf{F}, one can define the potential energy Ep as an associated potential field such as \mathbf{F}=-\frac{\partial Ep}{\partial r}, or maybe more formally \mathbf{F}=-\nabla(Ep). That's no more than a definition of the potential energy. ... -2 electric field strength is$$E=\frac Fq=\frac Vdwith V=voltage, d=distance between charged plates \begin{align} \frac Fq&=\frac Vd \\ Fd&=qV \end{align} but Fd=energy\therefore {\rm energy}=qV$$2 I'll give this a shot though I'm uncertain if this will clarify or cloud the issue. electrons are people, OK current is the number of people, No, that's not the correct analogy. Current is a flow and a people (electron) current is a flow of people (electrons) and the amount of current is the number (Coulombs) of people (electrons) passing a ... 1 Since your question is kind of pictorial, i will try to stick with this and avoid any deeper mathematical / physical explanations. electrons are people ok with that current is the number of people no, that's a mistake. Actually current is is this case the number of people crossing a line on the street per time. Like Steeven already mentioned in ... 1 But what is potential in this example? In your people-in-the-street analogy, the electric potential would be a shop with sale in one end of the street, or maybe an accident or something else interesting. People tend to move towards the interesting end (less potential) and away from the other and less interesting end (higher potential). Just like a stone ... 0 The choice of the constants A and B depends on the form of the solution. You could have denote one pair of constanst by A and B, and the other by C and D. A possible solution is \psi (x)=A\sin(kx). In complex form, the sine is:$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$If you've never proved this formula, try it using e^{ix}=\cos x+i\sin x. ... 1 I also don't quite understand what you mean by "imaginary \psi", but let me give you some general, more mathematically correct view. In general, what we measure is the spectrum of a self-adjoint operator. For the energy, this operator is of course the Hamiltonian. Now, since it is self-adjoint, its spectrum will lie on the real line. To have stability, we ... 2 I ended up spending an hour or so wading through the papers, so here's my main conculsion from that: No, I don't think the critique of the papers is wrong; Nor do I think that the basic algebraic arguments in the three papers addressed by the critique are wrong either. The authors describe a particular "decomposition" of the electromagnetic fields into a ... 1 Refer to: http://www.encyclopediaofmath.org/index.php/Hulthen_potential According to this it's a model of potential which behaves similar to Coulomb potential. I don't know how it's derrived, but it was derrived by a German dude. This paper compares both the Coulomb Potential and Hulthen potential: http://adsabs.harvard.edu/full/1954AuJPh...7..365M ... 0 I think your question is really: Why is Gibbs-Duhem equation not valid for inhomogeneous and/or small systems? This is related to why Thermodynamics does not apply to these systems. Examples, discussion and a Thermodynamical theory for these systems can be found in Terrell Hill's works link. But in short: both inhomogeneous and few-body systems have more ... 2 That's probably for charged solid sphere, not a cylinder. In any case, setting the potential at infinity as zero, we have for r>R:$$V(r)-V(\infty)=-\int_\infty^rE(r')dr'\implies V(r)=\frac{Q}{4\pi \epsilon_0}\frac{1}{r}$$For r<R, we got:$$V(r)-V(\infty)=-\int_\infty^rE(r')dr'=-\int_\infty^RE(r')dr'-\int_R^rE(r')dr' \implies \\ ...

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