Fall 2007 15 Qm

Fall 2007

Fall 2007 1 Mech
Fall 2007 2 Mech
Fall 2007 3 EM
Fall 2007 4 EM
Fall 2007 5 QM
Fall 2007 6 QM
Fall 2007 7 SM
Fall 2007 8 SM
Fall 2007 9 Misc
Fall 2007 10 Misc
Fall 2007 11 Mech
Fall 2007 12 Mech
Fall 2007 13 EM
Fall 2007 14 EM
Fall 2007 15 QM
Fall 2007 16 QM
Fall 2007 17 SM
Fall 2007 18 SM
Fall 2007 19 Misc
Fall 2007 20 Misc

A particle of mass m moves in a spherically symmetric potential well V(r) = -V₀ < 0 at r < a and V(r) = 0 at r > a. Find the smallest V₀ at which a bound state exists at zero angular momentum.

***

Answer

***

The Schrodinger equation for zero angular momentum reads

(1)

\begin{align} -\frac{\hbar^2}{2m}\left[\psi''(r)+\frac{2}{r}\psi'(r) \right] = E\psi, \;\; r>a, \end{align}

(2)

\begin{align} -\frac{\hbar^2}{2m}\left[\psi''(r)+\frac{2}{r}\psi'(r) \right] = (V_0+E)\psi, \;\; r<a, \end{align}

The sought solution for E < 0 is

(3)

\begin{align} \psi(r)=A \exp{(-\alpha r)}/r, \;\; \alpha= \sqrt{-2mE}/\hbar, \;\; r>a, \end{align}

(4)

\begin{align} \psi(r)=B \sin{(-\beta r)}/r, \;\; \beta= \sqrt{-2m(V_0+E)}/\hbar, \;\; r<a, \end{align}

The continuity of $\psi'(r)/\psi(r)$ at r = a demands

As $E \rightarrow 0^{-},\; \alpha\rightarrow 0^{+}$, and so ,$\beta a\rightarrow \pi/2$. Thus,

(5)

\begin{align} \mbox{min}V_0 = \frac{\pi^2\hbar^2}{8 m a^2}. \end{align}

Post preview:

Close preview

	or Sign in as Wikidot user
	(will not be published)

- +

Help: wiki text quick reference