Question

Consider a potential well defined as U(x) =  for x < 0, U(x) = 0 for 0 < x < L, and U(x) = U0 > 0 for x > L (see the following figure).

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Consider a potential well defined as \(U(x)=\infty\) for \(x<0, U(x)=0\)for \(0<x<L,\)and \(U(x)=U_{0}>0\) for \(x>L\) (see the following figure). Consider a particle with mass \(m\) and kinetic energy \(E<U_{0}\)that is trapped in the well. (a) The boundary condition at the infinite wall ( \(x=\)

0) is \(\psi(x)=0\). What must the form of the function \(\psi(x)\) for \(0<x<L\)be in order to satisfy both the Schrödinger equation and this boundary condition? (b) The wave function must remain finite as \(x \rightarrow \infty\). What must the form of the function \(\psi(x)\) for \(x>L\) be in order to satisfy both the Schrödinger equation and this boundary condition at infinity? (c) Impose the boundary conditions that \(\psi\) and \(d \psi / d x\) are continuous at \(x=L\). Show that the energies of the allowed levels are obtained from the solutions of the equation \(k \cot (k L)=-\alpha,\) where \(k=\sqrt{2 m E} / \hbar\) and \(\alpha=\sqrt{2 m\left(U_{0}-E\right)} / \hbar\)

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Answer #1

Given that

The potential well is defined as

\(U(x)=\infty\)

for \(x<0\)

\(U(x)=0\)

for \(0

\(U(x)=U_{o}>0\)

for \(x>L\)

(a)

We have a boundary condition at the infinite wall \((x=0)\) is \(\Psi(x)=0\)

The form of wave function satisfying boundary conditions and Schrodinger equation is

\(\Psi(x)=A \sin k x\)

Here \(A\) is constant and \(k^{2}=\frac{2 m E}{\hbar^{2}}\)

(b)

The wave function is required to have the form of equation as

\(\frac{d^{2} \Psi(x)}{d x^{2}}=\frac{2 m\left(U_{o}-E\right)}{\hbar^{2}} \Psi(x) \quad\) for \(x>L\)

The condition also exists that \(C=0\) in order for the wave function to remain finite as \(x\) approaches \(\infty\) the constant is

\(k^{2}=\frac{2 m\left(U_{o}-E\right)}{\hbar}\)

(c)

We have that at \(x=L, \quad A \sin k L=D e^{-k L}\)

The first derivative of the this equation is

\(k A \cos k L=-k D e^{-k L}\)

From the above two expressions we have

\(\frac{k A \cos k L}{A \sin k L}=\frac{-k D e^{-k L}}{D e^{-k L}}\)

\(k \cot k L=-k\)

This happens to be a transcendental equation which has to be solved numerically for various values of the length \(L\) as well as the ratio \(\frac{E}{U_{o}}\)

answered by: Accesem
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Consider a potential well defined as U(x) =  for x < 0, U(x) = 0 for 0 < x < L, and U(x) = U0 > 0 for x > L (see the following figure).
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