# Tag Info

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Apart from (physically motivated) asymptotic prescribed behaviour at infinity $|x|\to\infty$, we don't actually impose/demand/require any conditions on the wave function $\psi$ beyond the TDSE. Continuity and (possibly higher) smooth conditions are instead derived from a standard bootstrap argument, see e.g. my Phys.SE answer here for details.

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Because in ODE you only need $n$ conditions according the $n$-order of ODE, but these conditions could be on the same point, value of function and derivative, or two points, value of function first and last point of interval.

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Make make an analogy between gravitational potential energy, which is easy to visualize, and electrical potential energy. By doing this all the knowledge they have about a simpler subject will help them understand a more complicated one. Hold a ball and drop it. Draw what happened on the board, showing that the more potential (voltage) the fastest the ball ...

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So if I understand your question correctly, you are saying that every solution to a mathematically formulated physical problem should necessarily correspond a physical reality. Let me give you a very elementary example. Suppose I give you the following problem 'Person X and person Y are brothers. Y is younger to X by 6 years and the product of their ages is ...

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The three directions $x$, $y$ and $z$ are separable for the particle-in-a-box problem - the behaviour in each is independent of that in the others. Thus, each direction when separately considered only gives the contribution to the energy due to the limits of the box, or equivalently, the 'part' of the wavefunction, in that direction. In the case of $l_z ... 0 The charged sphere induces surface charge density on the conductor. This is necessary because field lines are always perpendicular to the surface of the conductor. This induced surface charge density modifies electric potential in the region. It will no longer be$V(d)=Q/4\pi\epsilon_0.d$. If you know the potential at the conductor(which I think is necessary ... 0 You have a slight misconception that is that higher potential i.e. Positive side to lower potential i.e. Negative side. Both sides have the same electric potential just like two masses share the same gravitational potential between each other. The direction of the current is determined by which charge is less inert (the electrons). Here the gravity analogy ... 1 What is potential energy truly? It depends on the circumstances. When you compress a spring it's stress in the bonds or electromagnetic field between the atoms. IMHO at the fundamental level it's essentially spatial stress. That might sound unfamiliar, but it shouldn't, because the stress-energy-momentum tensor "describes the density and flux of energy and ... 1 There are two important points to keep in mind when working through this problem. (1) Since the Hamiltonian for the system changes suddenly, the wavefunction just after the change is the same as the wavefunction just before the change. (2) Then energy eigenstates after the change are different from the energy eigenstates after the change. It follows that, ... 1 If some charge is given to a conductor then its potential will be remain same through out the region, because work done on every charge is same. 1 Modern electronic devices like quantum well lasers, resonant tunneling diodes, quantum cascade lasers and detectors heavily rely on the spatial and energetic position of such bound states. This defines their transport and optical properties. On a separate notice: any well, no matter how shallow or narrow, has at least one bound state. 0 In a system with non-interacting particles you may know that the right quantity to look at is$\rho \Lambda^3 \sim \left(\Lambda/l \right)^3$where$\Lambda$is the (thermal) de Broglie wavelength and$l = \rho^{-1/3}$the typical inter-particle distance in the system. Basically if$\Lambda/l \ll 1$, then your system behaves classically i.e. you do not ... 1 This is just the line integral which gives you the electric potential in two dimensions due to a charge distribution of one or more closed loops, i.e. closed charged wires. The logarithm is coming from the solution of laplace equation in two dimensions, replacing the$\frac{1}{|\vec{x} - \vec{x}'|}$of three dimensions, and$f(w)$is just the charge density ... 0 The effective potential is the potential of interaction you measure between two (or more) emergent physical objects when you forget (or "trace over" in the jargon) certain degrees of freedom of a more detailed model. If you take two pinned charges in vacuum for instance, they will interact with a "bare" Coulomb interaction in$\sim 1/r$. If you put these ... 1 As wiki says "The effective potential (also known as effective potential energy) is a mathematical expression combining multiple (perhaps opposing) effects into a single potential." Basically the concept of the effective potential simplifies the equations of motion and simplifies their analysis. 0 My textbook says: because the electric potential must be a continuous function. But why? Since, in the electrostatic case, $$\vec E(\vec r) = -\nabla V(\vec r)$$ then if the electric field is to be finite everywhere,$V(\vec r)$must be continuous. Put less rigorously, the electric field would be 'infinite' wherever$V(\vec r)$is discontinuous. ... 1 Imagine you have a point charge inside the conducting sphere. Obviously, since the electric field inside the sphere is zero (as you state), there is no force on the charge, so no work done. Therefore the potential is constant. So far so good. Now as we approach the boundary, we can imagine moving an infinitesimal amount to go from$r = R - \delta r$to$r = ...

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Firstly, when a body has no potential energy with respect to a potential present, in your case is the gravitational potential, then you are actually putting that object at the reference position. Next, if it has zero velocity then essentially it has no kinetic energy. Now you are left with another energy that can be measured by taking into account ...

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As you've worked out, since $v=0$ then the kinetic energy must also be zero. Potential energy is a little more dubious. At school you are usually taught that the gravitational potential energy is $E=mgh$ but that's not quite right; this equation is the work that you must do to lift an object of mass $m$ a height $h$ or, equivalently, the work done by ...

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The "direct" formula is $$V(r)=\frac{1}{4\pi\epsilon_0}\int\frac{dQ}{\lvert \vec{r}-\vec{R} \rvert}=\frac{1}{4\pi\epsilon_0}\iint_{sphere}\frac{\sigma(\vec{R})dS}{\lvert \vec{r}-\vec{R} \rvert}.$$ Now, think carefully about what the $\frac{1}{\lvert \vec{r}-\vec{R} \rvert}$ means---it is the reciprocal of the distance from an arbitrary point on the surface ...

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It happens because of quantum tunneling. In the first scenario, we cannot possibly find the particle beyond the barrier. in the second, however, the particle can quantum tunnel over.

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For an infinite well (actually infinitely high barriers) the probability to find an electron in the barrier vanishes. Therefore the wavefunction in the barrier has to be 0. For barriers with a finite height, the commonly used, but actually wrong boundary conditions require the wavefunction and the first derivative to be continuous. This relies on the wrong ...

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The problem can be solved by introducing the relations $$n_2-n_1=1,\:\:\:n_3-n_2=1$$ Since the resonance occurs for the energies $E_1$,$E_2$, and $E_3$ in a sequence the argument for the sine $n \pi$ also occur in a sequence in this case. This leads to one of possible solvable equations for the well potential 2 ...

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The electric potential can be negative. Both in difference and absolutely if you have chosen a gauge. To see that this must be so, just replace the charge distribution (not the test charge, all the others...) with one that has the opposite sign.

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I believe you are confusing the work done by the electric field with the work done on the particle. By the electric field: The radial force of the electric field is always pointing outwards, and the displacement of the charge in this case is going inward. Thus, the integral you've specified will be negative. That is, the work done by the electric field is ...

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The relationship between electric field and electric potential is just that the electric field is (minus) the gradient of the potential. Thus in the case of a uniform field extending from a uniformly charged plate (let's call it along the z-axis, with the late in the x,y plane) $$E_z = - \frac{dV}{dz}$$. This of course means that $$V(z) = - \int E_z\ dz = ... 1 You will have to use Kirchoff's law to get the answer. How can some of positive numbers be zero? No, the sum of all charges will be zero while that of positive plate will be finite. Now the net sum would be zero since charge is conserved on the system having all the right plates. Use Kirchoff's Loop Law and Kirchoff's current law to find charges and ... 0 Take the simplest case of a uniform magnetic field and a circular wire perpendicular to the magnetic field. You know that a changing magnetic field induces an electric field. Because the magnetic field is uniform and the wire perpendicular, the electric field has the same magnitude on all points of the wire. You can find the total emf from the change in ... 0 The key to this question is the position of the voltmeter and the flux it contains. Consider the diagram below: Let us say we are finding the voltage between A and D using the volt meter M_1. We can do this using one of two loops: One going straight from A to D. The one going from A to B to C then to D. The first one will give an answer of ... 1 Electric fields produces a difference of potential on two points with different distances of the field source. Magnetic fields induces current on a closed loop if the loop is not on parallel in relation of the lines of field and the magnitude of the field does have to change (you have to have a flux). If you have a magnetic field interfering on your ... -2 The better way of looking at it would be from the macroscopic point of view.In a equipotential surface a particle that was under acceleration due to external force now no longer needs this external force to be applied. Hence the value of force become 0, resulting in zero work done... 0 The answer is yes: Obviously it is possible to solve radially symmetric Schroedinger equations, including the specialized treatment (radial component only) that you ask about. However, you will need a well-defined problem to meaningfully begin something as rigorous as an analytical (or numerical) treatment. The differential equation you have given for the ... 0 Usually one requires, that the propability density \varrho = |\Psi(x)|^2 is integrable over the whole range (suchs that it can be normalzied). In this sense \Psi(x) = \exp(kx) is just as divergent as \Psi(x) = \exp(ikx), since the limit$$\lim_{\epsilon \rightarrow \infty } \int_{-\epsilon}^\epsilon |\Psi(x)|^2 \ dx does not exists in both cases. ...

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I am confused about why there is this difference here? What about $E$ being less than $V_o$ that we need to look if $exp(kx)$ diverges or not? Why was this not taken into account in the first case when $E>V_o$, i.e doesn't $C×exp(ikx)$ also diverge in the first case when $E>V_o$? The other answers have explored the question of why there are ...

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