Particle Energy Calculation in Free Space
The problem asks for the energy of a particle described by the wavefunction $\psi = e^{ikx} + 2e^{-ikx}$ in free space. In free space, the potential energy $V(x)$ is zero. The time-independent Schrödinger equation for a free particle is:
$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi$
Here, $\psi$ is the wavefunction, $\hbar$ is the reduced Planck's constant, $m$ is the mass of the particle, and $E$ is the energy.
Hamiltonian Operator Analysis
The Hamiltonian operator for a free particle is $\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$. Let's apply this operator to the given wavefunction $\psi$. The wavefunction is a sum of two terms:
- Term 1: $\psi_1 = e^{ikx}$
- Term 2: $\psi_2 = 2e^{-ikx}$
We calculate the second derivative for each term:
- For $\psi_1$: $\frac{d^2}{dx^2}(e^{ikx}) = (ik)^2 e^{ikx} = -k^2 e^{ikx}$
- For $\psi_2$: $\frac{d^2}{dx^2}(2e^{-ikx}) = 2(-ik)^2 e^{-ikx} = 2(-k^2) e^{-ikx} = -2k^2 e^{-ikx}$
Now, apply the Hamiltonian operator $\hat{H}$ to the total wavefunction $\psi = \psi_1 + \psi_2$:
$\hat{H}\psi = \hat{H}(e^{ikx} + 2e^{-ikx})$
$= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}(e^{ikx} + 2e^{-ikx})$
$= -\frac{\hbar^2}{2m}\left(\frac{d^2}{dx^2}(e^{ikx}) + \frac{d^2}{dx^2}(2e^{-ikx})\right)$
$= -\frac{\hbar^2}{2m}\left(-k^2 e^{ikx} - 2k^2 e^{-ikx}\right)$
$= -\frac{\hbar^2}{2m}(-k^2)(e^{ikx} + 2e^{-ikx})$
$= \frac{\hbar^2k^2}{2m}(e^{ikx} + 2e^{-ikx})$
$= \frac{\hbar^2k^2}{2m}\psi$
Energy Eigenvalue Determination
Comparing the result $\hat{H}\psi = \frac{\hbar^2k^2}{2m}\psi$ with the Schrödinger equation $\hat{H}\psi = E\psi$, we see that the wavefunction $\psi$ is an eigenstate of the Hamiltonian with the energy eigenvalue:
$E = \frac{\hbar^2k^2}{2m}$
Both components ($e^{ikx}$ and $2e^{-ikx}$) represent states with the same kinetic energy, $\frac{\hbar^2k^2}{2m}$. Therefore, the superposition state also has this energy.
Final Answer
The energy of the particle is $\frac{\hbar^2k^2}{2m}$.
Important Questions from Schrödinger Equation 1D Potentials Harmonic Oscillator
- The energy $E$ and degeneracy $d$ of the second excited state of a three-dimensional, isotropic quantum harmonic oscillator with angular frequency $\omega$ are
- A particle of mass $ m $ is in a potential $ V(x) = \frac{1}{2}m\omega^2x^2 $ for $ x > 0 $ and $ V(x) = \infty $ for $ x \leq 0 $, where $ \omega $ is the angular frequency. The ratio of the energies corresponding to the lowest energy level to the next higher level is
- Young's double slit experiment is performed using a beam of $C_{60}$ (fullerene) molecules, each molecule being made up of 60 carbon atoms. When the slit separation is 50 nm, fringes are formed on a screen kept at a distance of 1 m from the slits. Now, the experiment is repeated with $C_{70}$ molecules with a slit separation of 92.5 nm. The kinetic energies of both the beams are the same. The position of the 4th bright fringe for $C_{60}$ will correspond to the $n^{th}$ bright fringe for $C_{70}$. What is the value of $n$ (rounded off to the nearest integer) ?
- Consider a particle in a two dimensional infinite square well potential of side $L$, with $0 \le x \le L$ and $0 \le y \le L$. The wavefunction of the particle is zero only along the line $y = \frac{L}{2}$, apart from the boundaries of the well. If the energy of the particle in this state is $E$, what is the energy of the ground state?
The wavefunction of a particle in an infinite one-dimensional potential well at time $t$ is
$\Psi(x, t) = \sqrt{\frac{2}{3}} e^{-iE_1t/\hbar}\psi_1(x) + \frac{1}{\sqrt{6}} e^{i\pi/6}e^{-iE_2t/\hbar}\psi_2(x) + \frac{1}{\sqrt{6}} e^{i\pi/4}e^{-iE_3t/\hbar}\psi_3(x)$
where $\psi_1, \psi_2$ and $\psi_3$ are the normalized ground state, the normalized first excited state and the normalized second excited state, respectively. $E_1, E_2$ and $E_3$ are the eigen-energies corresponding to $\psi_1, \psi_2$ and $\psi_3$, respectively. The expectation value of energy of the particle in state $\Psi(x, t)$ is
