Spherical Well: Difference between revisions

	Quantum Mechanics A
	Schrödinger Equation; The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \|\Psi\rangle} 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(\mathbf{r}) = \begin{cases} -V_0, & 0\leq r< a \\ 0, & r>a. \end{cases} }

The Schrödinger equations for these two regions are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2l(l+1)}{2mr^2}-V_0\right)u_l=Eu_l}

for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\leq r< a \!} and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2l(l+1)}{2mr^2}\right)u_l=Eu_l}

for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r>a. \! }

The general solutions are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\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{align} }

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k' = \sqrt{\frac{2m(E+V_0)}{\hbar^2}} \!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k = \sqrt{\frac{2mE}{\hbar}}.\!}

Let us now consider bound states for the special case, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l=0.\!} In this case, the centrifugal barrier drops out and the equations become

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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^2u_0}{dr^2}=Eu_0, & r>a. \end{cases} }

The solution for this case is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\frac{\sqrt{-2mE}}{\hbar}.}

Using the boundary condition, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u(0)=0,\!} we find that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=-B.\!} The wave functions for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\leq r<a\!} thus reduces to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_0(r)=2iA\sin(kr)=\alpha\sin(kr),\!}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha=2iA.\!}

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r>a\!} , we know that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D=0\!} since, as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\rightarrow\infty,\!} the wavefunction must go to zero. Therefore, for the region in which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r>a,\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_0(r)=Ce^{-\kappa r}.\!}

Using the conditions that at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=a,\!} the wave functions and their derivatives must be continuous yields the following equations:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha\sin{ka}=Ce^{-\kappa a}\!}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\alpha\cos{ka}=-\kappa Ce^{-\kappa a}.\!}

Dividing the second equation by the first, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa^2+k^2=\frac{2mV_0}{\hbar^2},}

shows that no bound state exists if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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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