Tag Info

Hot answers tagged

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 ...


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 ...


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 & := ...


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

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 ...


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 ...


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 ...


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 ...


2

The important thing is the relative sign between the potential and the Laplacian. Otherwise there are two square roots of $-1$ namely $\pm i.$ You could write $j=-i$ and then you have two perfectly good square roots of $-1$ you can use $i$ or you can use $j$. For square roots of positive numbers you get things like $\pm\sqrt 2$ and there is a way to pick ...


2

There is no physical significance because there are two equally good square roots of negative one, i.e. $\pm i$ and unlike $\pm\sqrt 2$ there is no way to tell the roots apart. So what you call $+i$ and what I call $-i$ could actually be the same. What matters is relative signs, for instance $e^{i(kx-\omega t)} is a wave traveling to the right. You ...


2

Consider a molecule of oxygen in a balloon. You know that at nonzero temperature all those molecules are bouncing around in all directions. Of course, the mass of air doesn't have any net motion in any direction. Indeed, the average velocity of each molecule is zero: $$\langle v \rangle = 0 \, .$$ Of course, the average energy of any particular molecule is ...


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 ...


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

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.


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 ...


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

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

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

$\Delta_e$ 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+ ...


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} + ...



Only top voted, non community-wiki answers of a minimum length are eligible