Potential Barrier Wavefunction Analysis
We are considering an electron with energy $E$ approaching a potential barrier. The potential is defined as $V(x) = 0$ for $x < 0$ and $V(x) = V_0$ for $x > 0$. The condition given is $E < V_0$. We need to determine the form of the wavefunction $\psi(x)$ for $x > 0$.
Schrödinger Equation in the Barrier Region ($x > 0$)
For the region where $x > 0$, the potential is constant $V(x) = V_0$. The time-independent Schrödinger equation takes the form:
$ -\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} + V_0\psi(x) = E\psi(x) $Rearranging this equation, we get:
$ \frac{d^2\psi(x)}{dx^2} = \frac{2m(V_0 - E)}{\hbar^2}\psi(x) $Wavefunction Form for $E < V_0$
Since $E < V_0$, the quantity $(V_0 - E)$ is positive. Let's define a positive real constant $\alpha$:
$ \alpha = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}} $Note that $\alpha$ is a real positive quantity as stated in the problem.
Substituting $\alpha^2 = \frac{2m(V_0 - E)}{\hbar^2}$ into the Schrödinger equation gives:
$ \frac{d^2\psi(x)}{dx^2} = \alpha^2\psi(x) $The general solution to this second-order differential equation is:
$ \psi(x) = A e^{\alpha x} + B e^{-\alpha x} $where $A$ and $B$ are constants.
Physically, for $x \to \infty$ (deep inside the barrier region), the wavefunction must be bounded. The term $A e^{\alpha x}$ increases exponentially as $x$ increases, which is not physically plausible as it implies infinite probability. Thus, we must set $A = 0$.
The physically valid wavefunction in the region $x > 0$ is therefore of the form $B e^{-\alpha x}$.
Final Solution
The space part of the wavefunction for $x > 0$, given $E < V_0$, must be of the form $e^{-\alpha x}$ (or proportional to it).
Important Questions from Schrödinger Equation 1D Potentials Harmonic Oscillator
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- 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
