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14

I) The substitution $f=r\psi$ is the standard substitution to get a radial 3D problem to resemble a 1D problem, see e.g. Ref. 1. II) From the perspective of the normalization of the wavefunction $\psi(r)$, a $1/r$ singularity of $\psi(r)$ at $r=0$ is fine because $|\psi(r)|^2$ is suppressed by a Jacobian factor $r^2$ coming from the measure in 3D spherical ...


11

I) Initial value problems and boundary value problems are two different classes of questions that we can ask about Nature. Example: To be concrete: an initial value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the initial velocity $v_i$ are given, while a boundary value problem could be to ask ...


10

Not necessarily. Consider this function as an example: $$\psi(x) = \frac{C\sin x^2}{\sqrt{x^2 + 1}}$$ This function is square-integrable and asymptotes to zero as $x\to\pm\infty$, but its derivative goes to $2\cos x^2$ in the same limit. In quantum mechanics, we often assume that real systems are represented by wavefunctions which have no interesting ...


10

Indeed the problem with boundary conditions, generally speaking, is not well-posed. There are boundary conditions admitting no curves or admitting many curves, satisfying both these conditions and Euler-Lagrange equations. Examples. (1) Think of a particle constrained to stay on a smooth sphere where it can freely move. If you assign the North and the ...


9

In general, boundary conditons must be adapted to the real situation. Zero boundary conditions are just for the sake of simplicity. But they are realistic only when the field is really zero for some definite reason. If the boundary is at infinity, zero boundary conditions means that everything of interest happens in a finite domain and cannot be noticed ...


9

Loosely speaking, the gradient of a scalar field (such as the electrostatic potential) points in the direction of that field's greatest change. Since no change occurs in the field when you go along the surface, the gradient shouldn't have a component in that direction. Here is another intuitive explanation: Imagine for a moment that the electric field was ...


8

Particles do not minimize their action. Instead, they minimize their action given certain boundary conditions. We can only apply the action principle when we know the start and end points ahead of time. If we know that a particle will be at location $x_i$ at time $t_i$ and that it will be at location $x_f$ at time $t_f$, then the particle takes a path of ...


7

Actually, the extra path is not irrelevant. If you put a light bulb at A and a $4\pi$ detector (this means $4\pi$ steradian coverage, i.e. it detects incoming light in any direction) at B, the detector will see light along both paths: direct, and bounced off the mirror, which is exactly the result you got from Fermat's Principle. If you want to exclude the ...


7

In the examples you've given, the boundary conditions simply say "don't have infinite energy" and "don't be non-normalizable". These don't really have a physical interpretation. Moreover, no boundary conditions ever have an interpretation like "initial position and velocity" because the time-independent Schrodinger equation describes stationary states. ...


7

This is version two of my proof. The OP discovered a sign error in my first attempt that revealed my argument to be circular. The correct proof is below. Not surprisingly, this has to do with the signature of the spacetime metric not being positive definite. Furthermore, this issue is very subtle. It should be noted that this result is quoted partially in ...


6

The book's sentence "we can always force a system to have discrete states by enclosing it in a sufficiently large box" is perhaps a little unclearly worded. Putting the system into a box of any size will force the system to have discrete states. The author's point is that it's always possible to choose a box size that's big enough that the box's existence ...


6

Einstein's equation is $$G_{ab} = 8 \pi T_{ab}.$$ The left-hand side of the equation, $G_{ab}$, is called the Einstein tensor. It is an expression involving second derivatives of the metric, $g_{ab}.$ The right-hand side of the equation, $T_{ab}$, is called the Stress-Energy tensor. Each stress-energy tensor is associated with a unique matter ...


6

It is a very delicate matter to decide when solutions are unphysical or not. A classic example is Dirac's discovery of anti-particles: he found them as negative energy eigenstates for a relativistic Hamiltonian. A less insightful theorist might have discarded the negative energy solutions as unphysical, although we now know that those solutions mean a great ...


6

In the Lorentzian case: I am not aware of anyone studying it, and don't know explicit counterexamples off-hand. But I have doubts on the uniqueness. With the Lorentzian case, the nature candidate to draw comparisons with is the wave equation. And we see that on something as simple as the unit square $[0,1]\times [0,1]$, the wave equation with vanishing ...


6

There is a standard book which contains everything about electrostatics, the Laplace/Poisson equation and boundary conditions: Classical Electrodynamics by J. D. Jackson. Get the book from the library of your choice, read all chapters labeled "Electrostatics", and you will find the answers to all your questions (if you are simulating this, you need to know ...


6

Different boundary conditions represent different models of cooling. The first one states that you have a constant temperature at the boundary.This can be considered as a model of an ideal cooler in a good contact having infinitely large thermal conductivity The second one states that we have a constant heat flux at the boundary. If the flux is equal zero, ...


6

Topological degeneracy is only defined in the thermodynamic limit on a closed manifold. The ground state degeneracy of a finite-sized system or on an open manifold is not "topological", and can not be called topological degeneracy. Considering your examples. (1) The ground state degeneracy is ill-defined with open boundary condition. Because there might be ...


6

The two solutions are different because they have different boundary conditions. In the first case, the equation is indeed $$ \frac{\partial^2u}{\partial t^2} = c^2 \nabla^2 u = c^2 \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) u. $$ Here though we specify $u(t,x=x_0)$ to be some value ...


6

The reason is the same as why the electric field inside a conductor is zero: if it isn't zero, the free electrons undergo a force and move (rearrange) until they don't feel a force any more. If the electrons don't feel a force, the electric field must be zero. At the surface of a conductor, the free electrons feel a force perpendicular to the surface, but ...


5

One may view both (i) the infinite wall $$\tag{1} V(x)~=~\left\{\begin{array}{ccc}\infty & \text{for} & x>0, \\ 0 & \text{for} & x\leq 0, \end{array} \right. $$ and (ii) the delta function potential $$\tag{2} V(x)~=~A\delta(x),$$ as an appropriate limit of a finite barrier wall $\tag{3} V(x) ~=~ V_0 ...


5

The assumption missing in all these statements is that there are boundary conditions assumed to be given. E.g. for Poisson's equation $\Delta f = \rho$, the solution is unique for Dirichlet and/or Neumann boundary conditions, see e.g. section 1.9 in Jackson's "Classical Electrodynamics".


5

Multiple classical solutions to Euler-Lagrange equations with pertinent/well-posed boundary conditions (such solutions are sometimes called instantons) are a common phenomenon in physics, cf. e.g. this related Phys.SE post and links therein. In optics, it is well-known that already e.g. two mirrors can create multiple classical paths.


5

This is, unfortunately, not a simple task in general. My experience on non-reflecting boundary conditions is for the Navier-Stokes equations, but you should be able to do a similar approach for your system. As you noted, a fixed boundary ($u=0$) will lead to one type of wave while a free boundary ($\partial u/\partial x = 0$) leads to another wave. What you ...


4

OP's question is essentially pondering (in the context of the holomorphic/coherent state path integral) if a pair of variables is a complex conjugate pair or$^1$ truly independent variables. Notation in this answer: In this answer, let $z,z^{\ast}\in \mathbb{C}$ denote two independent complex numbers. Let $\overline{z}$ denote the complex conjugate of $z$. ...


4

Let wave function $\Psi$ be defined on domain $D \in \mathbb{R}^n$. The Neumann condition $\frac{\partial \Psi} {\partial {\bf n}} = 0$ on the boundary $\partial D$ has a simple interpretation in terms of the probability current of $\Psi$. For $\Delta \Psi = i \partial\Psi/\partial t$ (although it's usually taken as $i \partial\Psi/\partial t = - \Delta ...


4

Since it's a quiet Sunday morning let me dust off my brain cells and see if I can remember how to do this. We start by noting that if the initial and final positions for the particle as $\mathbf{r}(t_1)$ and $\mathbf{r}(t_2)$, then the action is: $$ S[\mathbf{r}(t)] = \int_{t_1}^{t_2} \tfrac{1}{2}m\dot{\mathbf{r}}^2 $$ We'll make the change $\mathbf{r} ...


4

The issue is that the underlying classical physics is determined by equations of motion (EOMs) (i.e. Newton's 2nd law), which are common for initial value problems (IVPs) and boundary value problems (BVPs). For BVPs , the EOMs can often alternatively be formulated as Euler-Lagrange (EL) equations of a stationary action principle. The latter approach does ...


4

The choice of boundary conditions fixes the domain and thus a self-adjoint extension of your Schroedinger operator $$-\frac{\hbar^2}{2m} \Delta + V\:.$$ In turn, this choice determines the spectrum of that operator, i.e., the eigenvalues $E$. (There is no analogy with initial conditions here.) No circumstances! Boundary conditions are uncorrelated with ...


4

I think that the texts you quote are referring to localized charge and current densities and the fields are defined in the whole space. The natural requirement is that far from the sources the vector fields decay as $1/r^2$ or faster, uniformly in all directions. This is the hypothesis of an induction field which is appropriate for static fields. With this ...



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