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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 ...

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 ...

0

I) Ignoring the metaplectic correction/Maslov index, the Bohr-Sommerfeld quantization rule reads $$\tag{1} N ~\approx~\int_a^b \!\frac{\mathrm{d}x}{\pi\hbar} p(x) ~=~ \int_a^b \! \frac{\mathrm{d}x}{\pi\hbar} \sqrt{2m(E-V(x))},$$ so that $$\tag{2} \frac{dN}{dE} ~\approx~\int_a^b \! \frac{\mathrm{d}x}{\pi\hbar} {\sqrt{\frac{m}{2(E-V(x))}}}~=~\int_a^b ... 1 OP is considering the Bohr-Sommerfeld quantization rule$$\tag{1} \oint k(x) \mathrm{d}x ~=~2\pi (n + \frac{1}{4}\sum_i\mu_i) , \qquad n\in\mathbb{N}_0, $$where the sum \sum_i is over turning points i with positions x_i and where \mu_i\in\mathbb{Z} is the metaplectic correction/Maslov index of the ith turning point. See also this Phys.SE post ... 4 I think there is some trick? The electric field is horizontal thus the electric potential varies in the horizontal direction only, not the vertical direction. I wouldn't call this a trick but it does appear that the question tests your conceptual grasp of the relationship between the electric field and electric potential. In particular, you should be ... 3 The nucleon-nucleon interaction has a short range, roughly 1 fm. Therefore if there were to be a bound dineutron, the neutrons would have to be confined within a space roughly this big. The Heisenberg uncertainty principle then dictates a minimum uncertainty in their momentum. This amount of momentum is at the edge of what theoretical calculations suggest ... 0 Situation 1., the ideal wire with an ideal voltage source, is an idealization: it is not a situation that can occur in nature. One should not be surprised that physics, which aims to describe nature, cannot describe a situation that cannot occur in nature. I will note that you specified a battery, not an ideal voltage source. In this case one can develop ... 1 Unfortunately I cannot tell you what went wrong on your first try, since I don't exactly know what you did. However, I sat down and tried to solve the system you describe: We are looking for Eigenstates of the Hamiltonian$$ H = \frac{p^2}{2} + V(x),\qquad \text{where }V(x)=\left\{\begin{array}{ll}\infty & \text{if } x<0 \\ x & ...

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

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 ...

0

In gravitation and in electromagnetism, the introduction of potentials greatly enhances our problem solving power because it is usually much easier to sum the potentials due to sources and apply relationships between the relevant fields and potentials, than it is to try and find the vector sums of the fields they produce. In gravitation this is quite obvious ...

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 ...

0

I want to write a simple answer to my question based on the knowledge of one-dimensional system; which can qualitatively answer why 2D is similar to 1D while 3D is not. (1). For one dimensional system, there is at least one bound state for pure attractive potential. (Proof can be done by using a Gaussian trail wave function for variational principal) ...

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 ...

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

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 ...

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