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0

The suggestion of Kevin Zhou, on other hand, has the advantage that we could try to control all the poles of the original well and the halved one. The polology of the square well was studied in Nussenzveig "The poles of the S-matrix of a rectangular potential well or barrier" Nuclear Physics, 11:409–521, 1959. and revisited in his book "Causality and ...


0

I'd say that the non adiabatic problem is ill-defined. Lets see. All the initial solutions $\Psi_n(x) = A_n \sin(n \pi x / 2a)$ have a probability current $\Psi'^* \Psi - \Psi^* \Psi'$ equal to zero everywhere, so in principle you can cut them at any point without having a leak of probability. The problem is that the boundary conditions for the domain of ...


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As all the answers are oriented in the theoretical side let me remind everybody of the Balmer series, an experimental observation fitted with a mathematical series, which was first modeled with the Bohr model of the hydrogen atom, and then was the cornerstone of building quantum mechanics, as it came out from the Schrodinger equation. So in this sense a ...


0

The article of D J Griffiths assumes that the delta interaction can be approximated by a sequence of even functions and then infers two boundary conditions: $$ \Psi'(0^+)-\Psi'(0^-)= (-1)^n {m c \over \hbar^2} (\Psi'^{(n)}(0^+)+\Psi'^{(n)}(0^-))$$ $$ \Psi(0^+)-\Psi(0^-)= (-1)^{n-1} {m c \over \hbar^2} n (\Psi'^{(n-1)}(0^+)+\Psi'^{(n-1)}(0^-))$$ My own ...


4

Let's simplify things, $\hbar=m=1$ and we put in the $\delta^{(n)}$ as $V(x)=V_0\delta^{(n)}/2$. Then the problem is $$-\psi'' - V_0\delta^{(n)}\psi = E \psi$$ Apart from $x=0$ the equation gives $$\psi'' = -E \psi $$ we want a bound state $E<0$ which falls off at infinity, so we get a solution $\psi_+ = A \exp(-k x)$ for $x>0$ and $\psi_- = A ...


1

How to get from $\left(\frac{-\hbar^2}{2m_e} \Delta_{r_e} + \frac{-\hbar^2}{2M_P} \Delta_{r_p} +V(r) \right)\Psi(\vec r_e,\vec r_p) = E \Psi(\vec r_e,\vec r_p)$ to: $\left(\frac{-\hbar^2}{2(m_e+M_p)} \Delta_{_{R}} + \frac{-\hbar^2}{2\mu} \Delta_{r} +V(r) \right)\Psi(\vec r,\vec R) = E \Psi(\vec r,\vec R)$ with: $\vec R=\frac{m_e \overrightarrow{r_e} + ...


5

Ok, I have a solution for $\delta'(x)$ based on a very crude limit. I'm going to neglect factors of $\hbar$, $m$, etc for the sake of eliminating clutter. Let $V_\epsilon(x)=\frac{\delta(x+\epsilon)-\delta(x)}{\epsilon}$. Then $\lim_{\epsilon\rightarrow 0}V_e(x)=\delta'(x)$. We'll solve the Schrodinger equation for finite $\epsilon$ and then take the limit ...


10

What exactly does $\Delta_{r_e}$ mean ? Your wave function isn't a field in space, it is a field on configuration space, i.e. it assigns complex numbers to a configuration. If your electron is at $(x_e,y_e,z_e)$ and your proton is at $(x_p,y_p,z_p)$ then the configuration is the point $(x_e,y_e,z_e,x_p,y_p,z_p)$ in a 6d space. A point in that 6d space ...


1

$\Delta_e$ is called the Laplacian operator and it is defined as $$\Delta_e \equiv \frac{\partial^2}{\partial e_x^2} + \frac{\partial^2}{\partial e_y^2}+ \frac{\partial^2}{\partial e_z^2}$$ The Laplacian operator is related to the Del operator $\nabla_e$ $$\nabla_e \equiv \frac{\partial}{\partial e_x}\hat e_x+ \frac{\partial}{\partial e_y}\hat e_y+ ...


5

The index of the Laplacian tells you which of the coordinates it acts on, that is, if you write $r = (r_x,r_y,r_z)^T$ and $R = (R_x,R_y,R_z)^T$ as Cartesian coordinates, then \begin{align} \Delta_r & := \frac{\partial^2}{\partial {r_x}^2} + \frac{\partial^2}{\partial {r_y}^2} + \frac{\partial^2}{\partial {r_z}^2} \\ \Delta_R & := ...


0

I wanted to bring up the same question as you raised, namely question 2). I agree with the answers to 1) and 3), but am unsatisfied with the proposed answer to 2) of buzhidao and levitopher (the argument given is also used in well-known textbooks such as Shankar, p.176). It is too sloppy in my opinion to say that "at infinity, the function is zero". This is ...


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I have the answer to this question, but it is one that is controversial in nature. I wouldn't be surprised if my answer gets removed. The Schrödinger Equation is a probabilistic Partial-Differential Equation. It is missing variables that would make it a deterministic Multivariate Equation. I used to not believe that it was a fundamental equation until I ...


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 ...


3

Let $\mathcal{H}$ be the space of states of our theory. Then, time evolution is given by a unitary operator $U(t_2,t_1) : \mathcal{H}\to\mathcal{H}$ that evolves "stuff" from time $t_1$ to time $t_2$. For time-independent Hamiltonians it is just $\mathrm{e}^{\mathrm{i}H(t_2 - t_1)}$. If we are in the Schrödinger picture, we say states "carry the time ...


3

Suppose you want to analyze the stationary behavior of a particle in a potential well that is symmetric with respect to $x=0$ (picture below). In order to simplify your calculation, you could use as a boundary condition that $\frac{\partial \psi(x)}{\partial x}=0$ and solve the Schroedinger equation for x>0 only. This boundary condition then just reflects ...


1

There is an existing subfield of physics that does exactly this, the de Broglie Bohm (dBB) version of quantum mechanics. This phase information can be expressed to look like the classical Hamilton-Jacobi equation for a parameterized family of initial value problems except the classical potential has an extra term that depends on what you call A. It is ...


1

It’s unrealistic to expect most students to able to derive the energy quantisation of a hydrogen atom on the spot. For hydrogen: $$E_n = \frac{-13.6 }{n^2}\rm eV$$ (ignore the minus sign for your problem). For a particle in a 1 D infinite potential well: $$E_n = \frac{n^2h^2}{8mL^2}$$ Set $n = 1$ to obtain the energies of both ground states. From the ...


1

Indeed the cosine function is valid for instance for other boundary conditions $$\psi'(0)=\psi'(L)=0$$ Your goal when you look for a set of solutions to the Schroedinger equation is to be able to decompose any general wavefunction as a sum over this set, and to do it consistently your boundary conditions must be the same for all solutions. So it is unlikely ...


1

The separable solutions are exactly the eigenstates of the Hamiltonian which are exactly the ones where the probability density does not change. However you can have a flux or flow of probability even if the probability doesn't change. This is like electromagnetism where you can have current floe through a wire even if the wire has no change of charge ...


0

The particle in a box is subjected to the further restriction that the potential function rapidly tends to infinity at $ x=0 $ and $ x=L $ This means that the position of the particle is restricted to being within the box. As you said, the general solution to the schrodinger equation for particle in a box is $$ \psi (x)=A\cos kx + B\sin kx $$ Because the ...


1

What you've written is only true if $\psi(x)$ is an eigenstate of $H$. For some general $\psi(x)$ that is not an eigenstate (i.e. $H \psi(x) \ne E \psi(x)$), then $\Psi(x, t)$ will be more complicated than just the time independent wavefunction multiplied by a phase factor.


0

This would not work for an arbitrary smooth central potential $V: \mathbb{R}_+ \to \mathbb{R}$ with $$V~<~0, \qquad V^{\prime}~>~0,\qquad V(0)~=~-\infty,\qquad V(\infty)~=~0,$$ because there might only be a finite number of bound states, and hence no continuous limit of bound states. For instance, one can use WKB methods to argue that if $V(r)$ ...


0

A summarised derivation for the 1 D triangular potential well can be found here. Interesting graphs of the wave functions $\psi_n(x)$ are provided. Also treated at [this *.pdf] (p.5, section 1.2.8.2)2.


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It would be better to specify that $V_0 < 0$ as otherwise your problem is that of a potential barrier (no bound states) and not that of a finite potential well. The case of a finite potential well is fully developed here (Wikipedia entry), the case of a rectangular potential barrier you can find here (Wiki).


4

You're looking for some form of differential operator that will take your plane wave $$\Psi(x,t) = \Psi(0,0)\exp\left[\frac{2\pi ix}{\lambda}-i\omega t\right]$$ and will return $p^2\Psi(x,t)$. As you've noticed, you can apply a space derivative to get $$ -ih\frac{\partial}{\partial x}\Psi(x,t) = p\Psi(x,t). $$ To get another $p$, you can apply the space ...


0

You are correct that for $n = 3$ there are $2$ non-boundary zero points. Also, the modulus of $\psi(x)$ is highest where $V(x)$ is lowest. Where you are wrong is that $\psi(A)$ is not zero and $\psi(B)$ is not zero either, as you indicate in your schematic. For $x < A$ and $x > B$, $V(x)$ is not $\infty$ (your well is a finite, not infinite well) and ...


0

This problem is somewhat similar to the ammonia inversion. In that problem the probability densities $\psi_n(x)^2$ are similar for $n =1$ and $2$, for $n = 3$ and $4$ etc. As a result the Hamiltonian levels $E_1$ and $E_2$ are close together, as are $E_3$ and $E_4$, etcetera. See for example here and here. See also Energy levels for $NH_3$.


1

The wavefunction $\psi(x)$ satisfies $$ -\frac{\hbar^2}{2m}\psi'' + V_0\left(\frac{x}{L} - 1\right) \psi = E\psi, \quad 0 \leq x \leq L\\ -\frac{\hbar^2}{2m}\psi'' = E\psi, \quad x > L $$ Since the bound states have $E < 0$ let's introduce $$ k = \frac{\sqrt{-2mE}}{\hbar}\\ \varkappa = \frac{\sqrt{2mV_0}}{\hbar} $$ Then $$ \psi'' - ...


1

Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy ...


0

I have solved my problem with some careful head scratching and reading. It turns out I misread the paragraph after Eq. (2.9). This is NOT the integration formula, just a statement of the boundary condition value for P as R -> inf and U -> constant. It provides the first two values needed for the rest of the inwards integration. The correct formula to use ...



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