CMPhysBench

1

For an oscillator with charge $q$, its energy operator without an external field is \begin{equation*} H_{0}=\frac{p^{2}}{2 m}+\frac{1}{2} m \omega^{2} x^{2} \end{equation*} If a uniform electric field $\mathscr{E}$ is applied, causing an additional force on the oscillator $f= q \mathscr{E}$, the total energy operator becomes \begin{equation*} H=\frac{p^{2}}{2 m}+\frac{1}{2} m \omega^{2} x^{2}-q \mathscr{E} x \end{equation*} Find the expression for the new energy levels $E_{n}$.

In $H_{0}$ and $H$, $p$ is the momentum operator, $p=-\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} x}$ The potential energy term in equation (2) can be rewritten as $\frac{1}{2} m \omega^{2} x^{2}-q \mathscr{E} x=\frac{1}{2} m \omega^{2}[(x-x_{0})^{2}-x_{0}^{2}]$ where \begin{equation*} x_{0}=\frac{q \mathscr{E}}{m \omega^{2}} \tag{3} \end{equation*} By performing a coordinate shift, let \begin{equation*} x^{\prime}=x-x_{0} \tag{4} \end{equation*} Because \begin{equation*} p=-\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} x}=-\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} x}=p …

E_{n} =(n+\frac{1}{2}) \hbar \omega-\frac{q^{2} \mathscr{E}^{2}}{2 m \omega^{2}}

Expression

Theoretical Foundations

$E_n$: New energy levels of the oscillator in the electric field $n$: Quantum number, $n=0,1,2, \cdots$ $\hbar$: Reduced Planck's constant $\omega$: Angular frequency of the oscillator $q$: Charge of the oscillator $\mathscr{E}$: Uniform electric field $m$: Mass of the oscillator

2

A particle of mass $m$ is in the ground state of a one-dimensional harmonic oscillator potential \begin{equation*} V_{1}(x)=\frac{1}{2} k x^{2}, \quad k>0 \end{equation*} When the spring constant $k$ suddenly changes to $2k$, the potential then becomes \begin{equation*} V_{2}(x)=k x^{2} \end{equation*} Immediately measure the energy of the particle, and find the expression for the probability of the particle being in the ground state of the new potential $V_{2}$.

(a) The wave function of the particle $\psi(x, t)$ should satisfy the time-dependent Schrödinger equation \begin{equation*} \mathrm{i} \hbar \frac{\partial}{\partial t} \psi=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} \psi+V \psi \tag{3} \end{equation*} When $V$ undergoes a sudden change (from $V_{1} \rightarrow V_{2}$) but with a finite change quantity, $\psi$ remains a continuous function of $t$, implying that $\psi$ does not change when $V$ changes abruptly. Denote $\psi_{0}(x)$ and $\phi_{0}(x)$ as the ground state wave functions of the potential $V_{1}$ and $V_{2}$, res …

\frac{2^{5 / 4}}{1+\sqrt{2}}

Expression

Theoretical Foundations