Spherical Well: Difference between revisions

	Quantum Mechanics A
	Schrödinger Equation; The most fundamental equation of quantum mechanics; given a Hamiltonian , it describes how a state evolves in time.
	Physical Basis of Quantum Mechanics
	Basic Concepts and Theory of Motion; UV Catastrophe (Black-Body Radiation); Photoelectric Effect; Stability of Matter; Double Slit Experiment; Stern-Gerlach Experiment; The Principle of Complementarity; The Correspondence Principle; The Philosophy of Quantum Theory
	Schrödinger Equation
	Brief Derivation of Schrödinger Equation; Relation Between the Wave Function and Probability Density; Stationary States; Heisenberg Uncertainty Principle; Some Consequences of the Uncertainty Principle
	Operators, Eigenfunctions, and Symmetry
	Linear Vector Spaces and Operators; Commutation Relations and Simultaneous Eigenvalues; The Schrödinger Equation in Dirac Notation; Transformations of Operators and Symmetry; Time Evolution of Expectation Values and Ehrenfest's Theorem
	Motion in One Dimension
	One-Dimensional Bound States; Oscillation Theorem; The Dirac Delta Function Potential; Scattering States, Transmission and Reflection; Motion in a Periodic Potential; Summary of One-Dimensional Systems
	Discrete Eigenvalues and Bound States
	Harmonic Oscillator Spectrum and Eigenstates; Analytical Method for Solving the Simple Harmonic Oscillator; Coherent States; Charged Particles in an Electromagnetic Field; WKB Approximation
	Time Evolution and the Pictures of Quantum Mechanics
	The Heisenberg Picture: Equations of Motion for Operators; The Interaction Picture; The Virial Theorem
	Angular Momentum
	Commutation Relations; Angular Momentum as a Generator of Rotations in 3D; Spherical Coordinates; Eigenvalue Quantization; Orbital Angular Momentum Eigenfunctions
	Central Forces
	General Formalism; Free Particle in Spherical Coordinates; Spherical Well; Isotropic Harmonic Oscillator; Hydrogen Atom; WKB in Spherical Coordinates
	The Path Integral Formulation of Quantum Mechanics
	Feynman Path Integrals; The Free-Particle Propagator; Propagator for the Harmonic Oscillator
	Continuous Eigenvalues and Collision Theory
	Differential Cross Section and the Green's Function Formulation of Scattering; Central Potential Scattering and Phase Shifts; Coulomb Potential Scattering

Latest revision as of 09:43, 26 October 2013

Let us now consider a spherical well potential, given by

$V(\mathbf {r} )={\begin{cases}-V_{0},&0\leq r<a\\0,&r>a.\end{cases}}$

The Schrödinger equations for these two regions are

$\left({\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}}{dr^{2}}}+{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}-V_{0}\right)u_{l}=Eu_{l}$

for $0\leq r<a\!$ and

$\left({\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}}{dr^{2}}}+{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}\right)u_{l}=Eu_{l}$

for $r>a.\!$

The general solutions are

${\begin{aligned}&{\frac {u_{l}(r)}{r}}=C_{l}j_{l}(k'r),&r\leq a,\\&{\frac {u_{l}(r)}{r}}=A_{l}j_{l}(kr)+Bn_{l}(kr),&r>a,\end{aligned}}$

where $k'={\sqrt {\frac {2m(E+V_{0})}{\hbar ^{2}}}}\!$ and $k={\sqrt {\frac {2mE}{\hbar }}}.\!$

Let us now consider bound states for the special case, $l=0.\!$ In this case, the centrifugal barrier drops out and the equations become

${\begin{cases}\left({\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}}{dr^{2}}}-V_{0}\right)u_{0}=Eu_{0},&0\leq r<a\\{\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}u_{0}}{dr^{2}}}=Eu_{0},&r>a.\end{cases}}$

The solution for this case is

${\begin{cases}u_{0}(r)=Ae^{ikr}+Be^{-ikr},&0\leq r<a\\u_{0}(r)=Ce^{\kappa r}+De^{-\kappa r},&r>a,\end{cases}}$

where $\kappa ={\frac {\sqrt {-2mE}}{\hbar }}.$

Using the boundary condition, $u(0)=0,\!$ we find that $A=-B.\!$ The wave functions for $0\leq r<a\!$ thus reduces to

$u_{0}(r)=2iA\sin(kr)=\alpha \sin(kr),\!$

where $\alpha =2iA.\!$

For $r>a\!$ , we know that $D=0\!$ since, as $r\rightarrow \infty ,\!$ the wavefunction must go to zero. Therefore, for the region in which $r>a,\!$

$u_{0}(r)=Ce^{-\kappa r}.\!$

Using the conditions that at $r=a,\!$ the wave functions and their derivatives must be continuous yields the following equations:

$\alpha \sin {ka}=Ce^{-\kappa a}\!$

and

$k\alpha \cos {ka}=-\kappa Ce^{-\kappa a}.\!$

Dividing the second equation by the first, we obtain

$-k\cot {ka}=\kappa ,\!$

which is just the solution for the odd states in a one-dimensional square well.

This, combined with the fact that

$\kappa ^{2}+k^{2}={\frac {2mV_{0}}{\hbar ^{2}}},$

shows that no bound state exists if $V_{0}<{\frac {\pi ^{2}\hbar ^{2}}{8ma^{2}}}.$

@@ Line 1: / Line 1: @@
 {{Quantum Mechanics A}}
-Let's consider spherical well potentials,
+Let us now consider a spherical well potential, given by
-:<math>
+<math>
 V(\mathbf{r}) =
 \begin{cases}
-V_0, & 0\leq r< a \\
+-V_0, & 0\leq r< a \\
-, & r>a
+, & r>a.
 \end{cases}
 </math>
-The [[Schrödinger equation]]s for these two regions can be written by
+The [[Schrödinger Equation|Schrödinger equations]] for these two regions are
-:<math> \left(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2l(l+1)}{2mr^2}-V_0\right)u_l(r)=Eu_l(r)
+<math> \left(\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2l(l+1)}{2mr^2}-V_0\right)u_l=Eu_l</math>
-</math>
 for <math> 0\leq r< a \!</math> and
-:<math> \left(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2l(l+1)}{2mr^2}\right)u_l(r)=Eu_l(r) </math>
+<math>\left(\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2l(l+1)}{2mr^2}\right)u_l=Eu_l</math>
-for <math> r>a \! </math>.
+for <math> r>a. \! </math>
 The general solutions are
-:<math>
+<math>
 \begin{align}
-&\frac{u_{\ell}(r)}{r} = C_{\ell} j_{\ell} (k'r), & r \leq d, \\
+&\frac{u_{l}(r)}{r} = C_{l} j_{l} (k'r), & r \leq a, \\
-&\frac{u_{\ell}(r)}{r} = A_{\ell}j_{\ell}(kr) +Bn_{\ell}(kr), & r > d,
+&\frac{u_{l}(r)}{r} = A_{l}j_{l}(kr) +Bn_{l}(kr), & r > a,
 \end{align}
 </math>
-where <math> k' = \sqrt{\frac{2m(E+V_0)}{\hbar^2}} \!</math> and <math> k = \sqrt{\frac{2mE}{\hbar}} \!</math>.
-For the <math>l=0\!</math> term, the centrifugal barrier drops out and the equations become the following
+where <math> k' = \sqrt{\frac{2m(E+V_0)}{\hbar^2}} \!</math> and <math> k = \sqrt{\frac{2mE}{\hbar}}.\!</math>
-:<math>
+Let us now consider bound states for the special case, <math>l=0.\!</math>  In this case, the centrifugal barrier drops out and the equations become
+<math>
 \begin{cases}
-\leq r< a & (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}-V_0)u_0(r)=Eu_0(r)\\
+\left (\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}-V_0\right )u_0=Eu_0, & 0\leq r< a \\
-r>a & (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2})u_0(r)=Eu_0(r)
+\frac{-\hbar^2}{2m}\frac{d^2u_0}{dr^2}=Eu_0, & r>a.
 \end{cases}
 </math>
-The generalized solutions are
+The solution for this case is
-:<math>
+<math>
 \begin{cases}
-\leq r< a & u_0(r)=Ae^{ikr}+Be^{-ikr}\\
+u_0(r)=Ae^{ikr}+Be^{-ikr}, & 0\leq r< a \\
-r>a & u_0(r)=Ce^{ik'r}+De^{-ik'r}
+u_0(r)=Ce^{\kappa r}+De^{-\kappa r}, & r>a,
 \end{cases}
 </math>
-Using the boundary condition, <math>u(r=0)=0\!</math>, we find that <math>A=-B\!</math>. The second equation can then be reduced to sinusoidal function where <math>\alpha=2iA\!</math>.
+where <math>\kappa=\frac{\sqrt{-2mE}}{\hbar}.</math>
-:<math>u_0(r)=2iA\sin(kr)=\alpha\sin(kr)=\alpha\sin\left(\frac{r}{\hbar}\sqrt{2m(E+V_0)}\right)</math>
-for <math>r>a\!</math>, we know that <math>D=0\!</math> since as <math>r\!</math> approaches infinity, the wavefunction does not go to zero.
+Using the boundary condition, <math>u(0)=0,\!</math> we find that <math>A=-B.\!</math>  The wave functions for <math>0\leq r<a\!</math> thus reduces to
-:<math>u_0(r)=Ce^{ik'r}+De^{-ik'r}=Ce^{-\frac{r}{\hbar}\sqrt{-2mE}}</math>
+<math>u_0(r)=2iA\sin(kr)=\alpha\sin(kr),\!</math>
+where <math>\alpha=2iA.\!</math>
+For <math>r>a\!</math>, we know that <math>D=0\!</math> since, as <math>r\rightarrow\infty,\!</math> the wavefunction must go to zero.  Therefore, for the region in which <math>r>a,\!</math>
+<math>u_0(r)=Ce^{-\kappa r}.\!</math>
+Using the conditions that at <math>r=a,\!</math> the wave functions and their derivatives must be continuous yields the following equations:
+<math>\alpha\sin{ka}=Ce^{-\kappa a}\!</math>
+and
+<math>k\alpha\cos{ka}=-\kappa Ce^{-\kappa a}.\!</math>
+Dividing the second equation by the first, we obtain
+<math>-k\cot{ka}=\kappa,\!</math>
+which is just the solution for the odd states in a [[One-Dimensional Bound States|one-dimensional square well]].
-Matching the conditions that at <math>r=a\!</math>, the wavefunctions and their derivatives must be continuous which results in 2 equations
+This, combined with the fact that
-:<math>\alpha\sin\left(\frac{a}{\hbar}\sqrt{2m(E+V_0)}\right)=Ce^{-\frac{a}{\hbar}\sqrt{-2mE}}</math>
-:<math>\alpha\frac{\sqrt{2m(E+V_0)}}{\hbar}\cos\left(\frac{a}{\hbar}\sqrt{2m(E+V_0)}\right)=-\frac{\sqrt{-2mE}}{\hbar}Ce^{-\frac{a}{\hbar}\sqrt{-2mE}}</math>
-Dividing the above equations, we find
+<math>\kappa^2+k^2=\frac{2mV_0}{\hbar^2},</math>
-:<math>-\cot\left(\sqrt{\frac{2m}{\hbar^2}(V_0-|E|)a^2}\right)=\frac{\sqrt{\frac{2m|E|}{\hbar^2}}}{\sqrt{\frac{2m(V_0-|E|)}{\hbar^2}}}</math>, which is the solution for the odd state in [[1D bound states|1D square well]].
-Solving for <math>V_0\!</math>, we know that there is no bound state for
+shows that no bound state exists if <math>V_0<\frac{\pi^2\hbar^2}{8ma^2}.</math>
-:<math>V_0<\frac{\pi^2\hbar^2}{8ma^2}</math>.

Spherical Well: Difference between revisions

Latest revision as of 09:43, 26 October 2013

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