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Chapter 1: Problem 24

Write the time-independent Schrödinger equations for (a) the hydrogen atom, (b) the helium atom, (c) the hydrogen molecule, (d) a free particle, (e) a particle subjected to a constant, uniform force. Hint. Identify the appropriate potential energy terms and express them as operators in the position representation.

Short Answer

Expert verified
The time-independent Schrödinger equations differ by the potential energy term which depends on the system under consideration. This potential energy term then affects the Hamiltonian operator, thus changing the Schrödinger equation.

Step by step solution

01

Schrödinger Equation for The Hydrogen Atom

For a hydrogen atom, the Hamiltonian operator will consist of the kinetic energy of the electron and the potential energy due to the interaction between the proton and electron. The potential energy here is a Coulomb potential. So the time-independent Schrödinger equation is \[ \left( -\frac{{\hbar^2}}{2m} \nabla^2 - \frac{{e^2}}{{4\pi\epsilon_0 r}} \right)\psi = E\psi \]
02

Schrödinger Equation for The Helium Atom

For helium atom, the potential energy part of the Hamiltonian consists of the electron-nucleus attractions and the electron-electron repulsion. So the time-independent Schrödinger equation becomes \[ \left( -\frac{{\hbar^2}}{2m} (\nabla_1^2 + \nabla_2^2) - \frac{{2e^2}}{{4\pi\epsilon_0 r_1}} - \frac{{2e^2}}{{4\pi\epsilon_0 r_2}} + \frac{{e^2}}{{4\pi\epsilon_0 |r_1 - r_2|}} \right)\psi = E\psi \]
03

Schrödinger Equation for The Hydrogen Molecule

For a hydrogen molecule, the Hamiltonian involves the kinetic energy of the two protons and two electrons, plus the potential energy of the interactions among them. The time-independent Schrödinger equation is complex and detailed for this case, it includes terms for electron-electron, proton-proton, and electron-proton interactions.
04

Schrödinger Equation for A Free Particle

For a free particle, there are no forces acting on it, hence no potential energy. The Hamiltonian then becomes purely kinetic. The time-independent Schrödinger equation is \[ -\frac{{\hbar^2}}{2m} \nabla^2\psi = E\psi \]
05

Schrödinger Equation for A Particle Subjected to a Constant, Uniform Force

For a particle subjected to a constant, uniform force, the appropriate potential is linear with position, \( V = Fx \). The time-independent Schrödinger equation becomes \[ \left( -\frac{{\hbar^2}}{2m} \nabla^2 + Fx \right)\psi = E\psi \]

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Most popular questions from this chapter

Evaluate the commutators (a) \([x, y]\) (b) \(\left[p_{x}, p_{y}\right]\) (c) \(\left[x, p_{x}\right]\) (d) \(\left[x^{2}, p_{x}\right]\) (e) \(\left[x^{n}, p_{x}\right]\)

Show that if the Schrödinger equation had the form of a true wave equation, then the integrated probability would be time dependent. Hint. A wave equation has \(\kappa \partial^{2} / \partial t^{2}\) in place of it a chere \(\kappa\) is a constant with the appropriate dimensions (what are they?). Solve the time component of the separable equation and investigate the behaviour of \(\int \Psi^{*} \Psi d \tau\)

Confirm that the operators (a) \(T=-\left(\hbar^{2} / 2 m\right)\left(d^{2} / d x^{2}\right)\) and (b) \(l_{z}=(\hbar / \mathrm{i})(\mathrm{d} / \mathrm{d} \varphi)\) are Hermitian. Hint. Consider the integrals \(\int_{0}^{L} \psi_{a}^{*} T \psi_{b} \mathrm{d} x\) and \(\int_{0}^{2 \pi} \psi_{a}^{*} l_{z} \psi_{b} \mathrm{d} \varphi\) and integrate by parts.

Evaluate the commutators (a) \(\left[H, p_{x}\right]\) and (b) \([H, x]\) where \(H=p_{x}^{2} / 2 m+V(x) .\) Choose (i) \(V(x)=V,\) a constant, (ii) \(V(x)=\frac{1}{2} k_{t} x^{2},\) (iii) \(V(x) \rightarrow V(r)=e^{2} / 4 \pi \varepsilon_{0} r .\) Hint. For part (b), case (iii), use \(\left(\partial r^{-1} / \partial x\right)=-x / r^{3}\)

(a) Show that \(\mathrm{e}^{A} \mathrm{e}^{B}=\mathrm{e}^{A+B}\) only if \([A, B]=0 .\) (b) If \([A, B] \neq 0\) but \([A,[A, B]]=[B,[A, B]]=0,\) show that \(\mathrm{e}^{A} \mathrm{e}^{B}=\) \(\mathrm{e}^{A+B} \mathrm{e}^{f},\) where \(f\) is a simple function of \([A, B] .\) Hint. This is another example of the differences between operators \((q-\text { numbers })\) and ordinary numbers (c-numbers). The simplest approach is to expand the exponentials and to collect and compare terms on both sides of the equality. Note that \(\mathrm{e}^{A} \mathrm{e}^{B}\) will give terms like \(2 A B\) while \(\mathrm{e}^{A+B}\) will give \(A B+B A .\) Be careful with order.
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