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4

I think there is some trick? The electric field is horizontal thus the electric potential varies in the horizontal direction only, not the vertical direction. I wouldn't call this a trick but it does appear that the question tests your conceptual grasp of the relationship between the electric field and electric potential. In particular, you should be ...


3

The nucleon-nucleon interaction has a short range, roughly 1 fm. Therefore if there were to be a bound dineutron, the neutrons would have to be confined within a space roughly this big. The Heisenberg uncertainty principle then dictates a minimum uncertainty in their momentum. This amount of momentum is at the edge of what theoretical calculations suggest ...


3

Why does the statement "any negative potential supports a bound state" hold in 1D, but not in 3D? In short, this is because for a bound state to occur, any positive kinetic energy needs to be fully offset by a negative potential energy. Achieving a large negative potential energy requires the particle to be localized in the volume where the potential is ...


2

To study bound states, we have to find solutions to the Schrödinger time-independent equation $$-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi$$ Using separation of variables, in spherical coordinates $$\psi(r,\theta,\phi)=Y^m_l(\theta,\phi)\frac{u(r)}{r},$$ where $Y^m_l(\theta,\phi)$ are the spherical harmonics, the radial part can be shown after substitution ...


2

The precise theorem is the following, cf. e.g. Ref. 1. Theorem 1: Given a non-positive (=attractive) potential $V\leq 0$ with negative spatial integral $$\tag{1} \int_{\mathbb{R}^n}\! d^n r~V({\bf r}) ~<~0 ,$$ then there exists a bound state$^1$ with energy $E<0$ for the Hamiltonian $$\tag{2} H~=~K+V, \qquad K~=~ -\frac{\hbar^2}{2m}{\bf ...


2

Consider the following proof by contradiction: Suppose you find a state with $E < -V_0$, then its kinetic energy becomes negative at every point $x$ (classically the velocity is imaginary at every point), which means the whole wavefunction (at all $x$) is an evanescent (exponentially decaying) wave, but such a solution is not a physically stable solution ...


1

Your boundaries are at $r=a$ and $r=b$. Notice that the potentials at these two surfaces are independent of $\theta$ (they are spherically symmetric). Look at a list of the first few Legendre Polynomials $P_{l}(\cos{\theta})$. For what value of $l$ does $P_{l}(\cos{\theta})$ not depend on $\theta$? Further, notice that $V(a) = V(r=a,\theta) = V_{0}$, and ...


1

OP is considering the Bohr-Sommerfeld quantization rule $$\tag{1} \oint k(x) \mathrm{d}x ~=~2\pi (n + \frac{1}{4}\sum_i\mu_i) , \qquad n\in\mathbb{N}_0, $$ where the sum $\sum_i$ is over turning points $i$ with positions $x_i$ and where $\mu_i\in\mathbb{Z}$ is the metaplectic correction/Maslov index of the $i$th turning point. See also this Phys.SE post ...


1

Hint to the question (v2): For a velocity-dependent force ${\bf F}$ (such as e.g. the Lorentz force), the relationship between force ${\bf F}$ and potential $U$ is $$ {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}. $$ See e.g. Goldstein, Classical Mechanics, Chapter 1. See also e.g. this and this Phys.SE ...


1

Given the potential $U(r)=A^3/r^2 + 2B^3 r$, the effective spring constant can be defined as the second derivative of $U(r)$ evaluated at the equilibrium point. Hence, $$U''(r)= \frac{6A^3}{r^4}$$ If $r_0$ is our equilibrium point, then $k_{\mathrm{eff}} = 6A^3/r^4_0$. Another way to perform the calculation is to compute the Taylor Series about the ...


1

No, $x$ can be measured from any arbitrary origin. Nature doesn't care where you set your coordinate system. At times there are choices of origin that make the math easier, but as far as the physics goes, it makes no difference at all.


1

Unfortunately I cannot tell you what went wrong on your first try, since I don't exactly know what you did. However, I sat down and tried to solve the system you describe: We are looking for Eigenstates of the Hamiltonian $$ H = \frac{p^2}{2} + V(x),\qquad \text{where }V(x)=\left\{\begin{array}{ll}\infty & \text{if } x<0 \\ x & ...



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