62

Take a gaussian (or any function that decays sufficiently quickly), chop it up every unit, and turn all the pieces sideways.


27

Let $$ \psi(x) = \begin{cases} 1 & \exists\, n \in \mathbb N: x \in [n, n+\frac 1 {n^2}]\\ 0 & \text{otherwise.} \end{cases} = \sum_{n \in \mathbb N} \mathbf 1_{[n,n+\frac 1 {n^2}]}(x) , $$ where $\mathbf 1_A$ is the characteristic function of the set $A$. Then $$ \int_{-\infty}^\infty |\psi(x)|^2 dx = \sum_{n=1}^\infty \frac 1 {n^2} < \infty, $$ ...


27

You lack complete knowledge of the system you're asking about. You only know about the jar. And if the complete system is the jar then yes, it has always been empty because there's nothing to interact with it. But if you suppose there are things which may have been in the jar, now those are added to the system. That puddle of water on the floor, was it ...


20

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 ...


20

In solving the Schroedinger radial equation there is no boundary condition applied at $r=0$. At $r=\infty$ yes, $R(r)$ must tend to zero - so we reject the positive exponential solution; any change in that would have massive consequences. But there is no constraint laid on $R(r)$ or indeed $R'(r)$ as $r \to 0$. So there's not a change in the boundary ...


19

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 ...


18

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 ...


17

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 ...


17

Emilio Pisanty and Eckhard Giere have already given discontinuous, piecewise constant counterexamples in their answers. Here we provide for-the-fun-of-it a smooth infinitely-many-times-differentiable counterexample $f\in C^{\infty}(\mathbb{R})$ of a square integrable function $f:\mathbb{R} \to [0,1]$ that does not satisfy $\lim_{|x|\to \infty}f(x)=0$. Our ...


16

Good question. Your claim that near the proton the electron's kinetic energy will be relativistic is not as straightforward as it might seem. The electron's kinetic energy $\langle \hat{T} \rangle = \langle \hat{p}^2 \rangle / (2m)$ is a nonlocal quantity that can be equivalently expressed as either of the two integrals $$\langle \hat{T} \rangle = \frac{...


16

Two ways of seeing that it’s right: Consider the case of no charge. Then nothing interesting is happening at the sheet, so the fields should be equal: both sides of the equation are zero. Gaus’s law: the sum of the fields going away, which is the outward flux, is given by the charge. Since $E_{below}$ is defined as towards the charge, it enters that ...


15

The key word here is continuity. The continuity boundary conditions for the electromagnetic field vectors sets these phenomenons. The tangential components of $\vec{E}$ and $\vec{H}$ must be continuous across an interface - the only way that they can not be is if there is a surface current flowing (which cannot happen in dielectrics). Likewise, the normal ...


15

Apart from not being sufficient to prove convergence of the integral $$\int |f(x)|^2\text dx<\infty,$$ having the vanishing limig $\lim_{x\rightarrow\infty}f(x)=0$ is only necessary for the convergence within a suitable class of "nice" functions. Consider, for example, the function $$ f(x)=\sum_{n=1}^\infty\chi_{\left[n,n+\frac1{n^2}\right]}(x)=\left\{\...


14

This is an attempt to explain, in a purely intuitive way why sound waves reflect from the end of an open pipe, and therefore can produce a standing wave. Consider a pressure wave travelling up the pipe. I've drawn just a single maximum of the pressure wave to keep the diagram uncluttered: Call the pressure maximum $P_1$ (I haven't marked $P_1$ on the ...


13

Very interesting question! I'll start by outlining some of the mathematical basics for this question: You're looking for a bounded state of some potential $V$, i.e. a non-scattering state. Mathemematically, this translates to an $L^2$-integrable eigenfunction of the Schrödinger operator $-\Delta+V$. By elliptic regularity, for these functions you instantly ...


13

First, let's start by completing the square as suggested. The equation will now be of the form $$ \frac{d^2\Psi}{dz^2}+(\nu+\frac{1}{2}-\frac{1}{4}z^2)\Psi=0 $$ with boundary conditions $\Psi(z=L)=\Psi(z=\infty)=0$ and $$ z=\sqrt{\frac{2m\omega}{\hbar}}x+L \\ \nu=\frac{E}{\hbar\omega}+\frac{L^2}{4}-\frac{1}{2} \\ L=-\sqrt{\frac{2mg^2}{\hbar\omega^3}} $$ This ...


12

Background Let $\tau$ be the tension and $\mu$ be a linear mass density (i.e., mass per unit length), then the wave equation for a string is given by: $$ \partial_{tt} \psi \left(x,t\right) - \frac{ \tau }{ \mu } \partial_{xx} \psi \left(x,t\right) = 0 \tag{0} $$ where $\partial_{jj} \equiv \partial^{2}/\partial j^{2}$ and $\psi \left(x,t\right)$ is a ...


12

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 ...


12

First of all, your field is not uniform. You have $$ \vec E(\vec x) = \begin{pmatrix} a x \\ 0 \\ 0 \end{pmatrix} \,.$$ That looks like this: This means that for larger $x$ the field is stronger and for $x \to - \infty$ the field will be very negative. The charge density is $\rho = \varepsilon \vec \nabla \cdot E$, which is indeed just $\varepsilon a$. ...


12

This usually only applies to a wall bounded flow and is normally restricted to incompressible fluids. This result usually manifests in boundary layer theory and can be obtained through order of magnitude analysis of the Navier-Stokes equations. The steady, incompressible, and constant property momentum equation in the $y$ direction takes the form, $$ u \frac{...


12

As you have noted, to solve the differential equation describing the system you need initial conditions. The condition "jar is empty" does not have a unique solution without specifying the conditions.


10

Since this has just been asked again, let me attempt an intuitive explanation. The real explanation is of course to match $\vec{E}$ and $\vec{B}$ at the interface and the direction of the reflected wave drops out, but this isn't especially intuitive. Let's calculate the ratio $E_r/E_i$ as a function of the ratio $n_t/n_i$, and let's start with the ...


10

From the Wikipedia article on sound: In physics, sound is a vibration that propagates as a typically audible mechanical wave of pressure and displacement. To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave, $$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial t^2}...


10

In fact, it is nothing but a vague and, strictly speaking, false requirement that can be found in some physically minded books (also of very good level). There are however physical situations where regularity of the used functions implies that they must vanish at infinity. If one is solving the stationary Schroedinger equation and the potential is ...


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

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 1_{[0,a]}(x)=\left\{\begin{array}...


9

Suppose now the Lagrangian $L$ is a function of $y(x), y'(x)$ and also $y''(x)$, i.e. it contains second derivatives w/r to the parameter $x$. It is straightforward to adapt the usual procedure to this case: write \begin{align} Y(x,\epsilon)=y(x)+\epsilon\,\eta(x) \end{align} for an otherwise arbitrary function $\eta$. We then have the parametrized ...


9

There are already several good answers showing the algebra. Here we will make some comments to the question (v4) concerning terminology and notation, which may clarify a thing or two. (In the following we refer to the $q$ position space as the vertical space and the $t$ time axis as the horizontal space.) Usually, the principle of least action refers to the ...


9

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. ...


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