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Question

The wavefunction of a particle moving in free space is given by, $\psi = e^{ikx} + 2e^{-ikx}$

The energy of the particle is

The correct answer is
$\frac{\hbar^2k^2}{2m}$

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 $\frac{\hbar^2k^2}{2m}$.

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Important Questions from Schrödinger Equation 1D Potentials Harmonic Oscillator

  1. 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

  2. A particle is subjected to a potential 
    $V(x) = \begin{cases} \infty, & x \le 0 \\ V_0, & a \le x \le b \\ 0, & \text{elsewhere} \end{cases}$ 
    Here, $a > 0$ and $b > a$. If the energy of the particle $E < V_0$, which one of the following schematics is a valid quantum mechanical wavefunction ($\Psi$) for the system?

  3. The wavefunction for a particle is given by the form $e^{-(iax+\beta)}$, where $a$ and $\beta$ are real constants. In which one of the following potentials $V(x)$, the particle is moving?
  4. A particle of mass $m$ is moving in the potential 
    $V(x) = \begin{cases} V_0 + \frac{1}{2}m\omega_0^2x^2, & x > 0, \\ \infty, & x \le 0, \end{cases}$ 
    Figures P, Q, R and S show different combinations of the values of $\omega_0$ and $V_0$. 

    $E_j^{(P)}, E_j^{(Q)}, E_j^{(R)}$ and $E_j^{(S)}$ with $j = 0, 1, 2, ...$, are the eigen-energies of the $j$-th level for the potentials shown in figures P, Q, R and S, respectively. Which of the statement is/are true?

  5. 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) ?
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