Isotropic Harmonic Oscillator: 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 13:42, 18 January 2014

We now solve the isotropic harmonic oscillator using the formalism that we have just developed. While it is possible to solve it in Cartesian coordinates, we gain additional insight by solving it in spherical coordinates, and it is easier to determine the degeneracy of each energy level.

The radial part of the Schrödinger equation for a particle of mass $M\!$ in an isotropic harmonic oscillator potential $V(r)={\frac {1}{2}}M\omega ^{2}r^{2}$ is given by:

$-{\frac {\hbar ^{2}}{2M}}{\frac {d^{2}u_{nl}}{dr^{2}}}+\left({\frac {\hbar ^{2}}{2M}}{\frac {l(l+1)}{r^{2}}}+{\frac {1}{2}}M\omega ^{2}r^{2}\right)u_{nl}=Eu_{nl}.$

Let us begin by looking at the solutions $u_{nl}\!$ in the limits of small and large $r.\!$

As $r\rightarrow 0\!$ , the equation reduces to

$-{\frac {\hbar ^{2}}{2M}}{\frac {d^{2}u_{nl}}{dr^{2}}}+{\frac {\hbar ^{2}}{2M}}{\frac {l(l+1)}{r^{2}}}u_{nl}=Eu_{nl}.$

The only solution of this equation that does not diverge as $r\rightarrow 0$ is $u_{nl}(r)\simeq r^{l+1}.$

In the limit as $r\rightarrow \infty ,$ on the other hand, the equation becomes

$-{\frac {\hbar ^{2}}{2M}}{\frac {d^{2}u_{nl}}{dr^{2}}}+{\frac {1}{2}}M\omega ^{2}r^{2}u_{nl}=Eu_{nl}$

whose solution is given by $u_{nl}(r)\simeq e^{-M\omega r^{2}/2\hbar }.$

We may now assume that the general solution to the equation is given by

$u_{nl}(r)=r^{l+1}e^{-M\omega r^{2}/2\hbar }f_{nl}(r).$

Substituting this expression into the original equation, we obtain

${\frac {d^{2}f_{nl}}{dr^{2}}}+2\left({\frac {l+1}{r}}-{\frac {M\omega }{\hbar }}r\right){\frac {df_{nl}}{dr}}+\left[{\frac {2ME}{\hbar ^{2}}}-(2l+3){\frac {M\omega }{\hbar }}\right]f_{nl}=0.$

We now use a series solution for this equation:

$f_{nl}(r)=\sum _{n=0}^{\infty }a_{n}r^{n}=a_{0}+a_{1}r+a_{2}r^{2}+a_{3}r^{3}+\ldots +a_{n}r^{n}+\ldots$

Substituting this solution into the reduced form of the equation, we obtain

$\sum _{n=0}^{\infty }\left[n(n-1)a_{n}r^{n-2}+2\left({\frac {l+1}{r}}-{\frac {M\omega }{\hbar }}r\right)na_{n}r^{n-1}+\left[{\frac {2ME}{\hbar ^{2}}}-(2l+3){\frac {M\omega }{\hbar }}\right]a_{n}r^{n}\right]=0,$

which reduces to

${\frac {2(l+1)}{r}}a_{1}+\sum _{n=0}^{\infty }\left[(n+2)(n+2l+3)a_{n+2}+\left(-{\frac {2M\omega }{\hbar }}n+{\frac {2ME}{\hbar ^{2}}}-(2l+3){\frac {M\omega }{\hbar }}\right)a_{n}\right]r^{n}=0.$

For this equation to hold, the coefficients of each of the powers of $r\!$ must vanish seperately. Doing this for the positive powers of $r\!$ yields the following recursion relation:

$(n+2)(n+2l+3)a_{n+2}=\left[-{\frac {2ME}{\hbar ^{2}}}+(2n+2l+3){\frac {M\omega }{\hbar }}\right]a_{n}$

In addition, we have an $r^{-1}\!$ term; for it to vanish, we must set $a_{1}=0.\!$ This, combined with the above recursion relation, means that the function $f_{nl}(r)\!$ contains only even powers of $r.\!$ In other words,

$f_{nl}(r)=\sum _{n=0,2,4,\ldots }^{\infty }a_{n}r^{n}=\sum _{n'=0}^{\infty }a_{n'}r^{n'}.$

By a similar argument as the one that we employed for the one-dimensional harmonic oscillator, we find that, unless the series for $f_{nl}(r)\!$ terminates, the resulting full wave function will diverge as $r\rightarrow \infty .$ Because the series must only contain even powers of $r,\!$ the resulting quantization condition on the energy is

${\frac {2M}{\hbar ^{2}}}E_{n'l}-{\frac {M\omega }{\hbar }}(4n'+2l+3)=0,\,n'=0,1,2,3,\ldots ,$

or

$E_{nl}=\left(n+{\frac {3}{2}}\right)\hbar \omega ,\,n=0,1,2,3,\ldots ,$

where $n=2n'+l.\!$

The degeneracy corresponding to the $n^{\text{th}}\!$ level may be found to be ${\tfrac {1}{2}}(n+1)(n+2).$ We see that energy levels with even $n\!$ correspond to even values of $l,\!$ while those with odd $n\!$ have odd values of $l.\!$

The total wave function of the isotropic harmonic oscillator is thus given by

$\psi _{nlm}(r,\theta ,\phi )=r^{l+1}e^{-M\omega r^{2}/2\hbar }f_{nl}(r)Y_{lm}(\theta ,\phi )=R_{nl}(r)Y_{lm}(\theta ,\phi ).$

One may show that, in fact, $f_{nl}(r)\!$ is an associated Laguerre polynomial in ${\frac {M\omega }{\hbar }}r^{2}.$

@@ Line 1: / Line 1: @@
-The radial part of the Schrodinger's Equation for a particle of mass M in an isotropic harmonic oscillator potential  <math>V(r)=\frac{1}{2}Mw^{2}r^2</math> is given by:
+{{Quantum Mechanics A}}
+We now solve the isotropic harmonic oscillator using the formalism that we have just developed.  While it is possible to solve it in Cartesian coordinates, we gain additional insight by solving it in spherical coordinates, and it is easier to determine the degeneracy of each energy level.
-:<math>-\frac{\hbar^2}{2M}\frac{\partial^2u_{nl}(r)}{\partial r^2}+\left(\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2} + \frac{1}{2}Mw^{2}r^2\right)u_{nl}(r)=Eu_{nl}(r)</math>
+The radial part of the [[Schrödinger Equation|Schrödinger equation]] for a particle of mass <math>M\!</math> in an isotropic [[Harmonic Oscillator Spectrum and Eigenstates|harmonic oscillator]] potential <math>V(r)=\frac{1}{2}M\omega^{2}r^2</math> is given by:
-We look at the solutions <math>u_{nl}\!</math> in the asymptotic limits of <math> r\!</math>.
+<math>-\frac{\hbar^2}{2M}\frac{d^2u_{nl}}{dr^2}+\left(\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2} + \frac{1}{2}M\omega^{2}r^2\right)u_{nl}=Eu_{nl}.</math>
+Let us begin by looking at the solutions <math>u_{nl}\!</math> in the limits of small and large <math>r.\!</math>
 As <math>r\rightarrow 0\!</math>, the equation reduces to
-:<math>-\frac{\hbar^2}{2M}\frac{\partial^2u_{l}(r)}{\partial r^2}+\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2}u_{l}(r)=Eu_{l}(r)</math>
-whose nondivergent solution is given by <math>u_l(r)\simeq r^{^{l+1}}</math>.
-On the otherhand, as <math>  r\rightarrow \infty</math>, the equation becomes
+<math>-\frac{\hbar^2}{2M}\frac{d^2u_{nl}}{dr^2}+\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2}u_{nl}=Eu_{nl}.</math>
-:<math>-\frac{\hbar^2}{2M}\frac{\partial^2u(r)}{\partial r^2}+\frac{1}{2}Mw^{2}r^2u(r)=Eu(r)</math>
-whose solution is given by <math>u(r)\simeq e^-{\frac{Mwr^2}{2\hbar}}</math>.
+The only solution of this equation that does not diverge as <math>r\rightarrow 0</math> is <math>u_{nl}(r)\simeq r^{l+1}.</math>
+In the limit as <math>r\rightarrow \infty,</math> on the other hand, the equation becomes
+<math>-\frac{\hbar^2}{2M}\frac{d^2u_{nl}}{dr^2}+\frac{1}{2}M\omega^{2}r^2u_{nl}=Eu_{nl}</math>
+whose solution is given by <math>u_{nl}(r)\simeq e^{-M\omega r^2/2\hbar}.</math>
+We may now assume that the general solution to the equation is given by
+<math>u_{nl}(r)=r^{l+1}e^{-M\omega r^2/2\hbar}f_{nl}(r).</math>
+Substituting this expression into the original equation, we obtain
-Combining the asymptotic limit solutions we choose the general solution to th e equation as
+<math>\frac{d^2f_{nl}}{dr^2}+2\left(\frac{l+1}{r}-\frac{M\omega}{\hbar}r\right)\frac{df_{nl}}{dr}+\left[\frac{2ME}{\hbar^2}-(2l+3)\frac{M\omega}{\hbar}\right]f_{nl}=0.</math>
-:<math>u_l(r)=f_l(r)r^{l+1}e^-\frac{Mwr^2}{2\hbar}</math>
-Substituting this expression into the original equation,
+We now use a series solution for this equation:
-:<math>\frac{\partial^2f_l(r) }{\partial r^2}+2\left(\frac{l+1}{r}-\frac{Mw}{\hbar}r\right)\frac{\partial f_l(r) }{\partial r}+\left[\frac{2ME}{\hbar^2}-(2l+3)\frac{Mw}{\hbar}\right]f_l(r) =0</math>
-Now we try the power series solution
+<math>f_{nl}(r)=\sum_{n=0}^{\infty}a_{n}r^n= a_{0}+a_{1}r+a_{2}r^2+a_{3}r^3+\ldots +a_{n}r^n+\ldots</math>
-:<math>f_l(r)=\sum_{n=0}^{\infty}a_{n}r^n= a_{0}+a_{1}r+a_{2}r^2+a_{3}r^3+. . . +a_{n}r^n+...</math>
-Substituting this solution into the reduced form of the equation,
+Substituting this solution into the reduced form of the equation, we obtain
-:<math>
+<math>
 \sum_{n=0}^{\infty} \left[n(n-1)a_{n}r^{n-2}+2 \left( \frac{l+1}{r}-
-\frac{Mw}{\hbar} \right) na_nr^{n-1} + \left[\frac{2ME}{\hbar^2} - (2l+3)\frac{Mw}{\hbar}\right] a_n r^n\right]=0
+\frac{M\omega}{\hbar}r\right) na_nr^{n-1} + \left[\frac{2ME}{\hbar^2} - (2l+3)\frac{M\omega}{\hbar}\right] a_n r^n\right]=0,
-</math>
-which reduces to the equation
-:<math>\sum_{n=0}^{\infty}\left[n(n+2l+1)a_{n}r^{n-2}+\left(-\frac{2Mw}{\hbar}n+\frac{2ME}{\hbar^2}-(2l+3)\frac{Mw}{\hbar}\right)a_nr^n\right]=0
 </math>
-For this equation to hold, the coefficients of each of the powers of r must vanish seperately.
+which reduces to
-So,when <math> n =0 \!</math> the coefficient of <math> r^{-2} \!</math> is zero, <math>0.(2l+1)a_0=0</math> implying that <math>a_0\!</math> need not be zero.
+<math>\frac{2(l+1)}{r}a_1+\sum_{n=0}^{\infty}\left[(n+2)(n+2l+3)a_{n+2}+\left(-\frac{2M\omega}{\hbar}n+\frac{2ME}{\hbar^2}-(2l+3)\frac{M\omega}{\hbar}\right)a_n\right]r^n=0.
+</math>
-Equating the coefficient of <math> r^{-1} \!</math>to be zero, <math>1.(2l+2)a_1=0\!</math> implying that <math>a_1\!</math> must be zero.
+For this equation to hold, the coefficients of each of the powers of <math>r\!</math> must vanish seperately.  Doing this for the positive powers of <math>r\!</math> yields the following recursion relation:
-Equating the coefficient of <math> r^{-n}\!</math> to be zero, we get the recursion relation which is:
-:<math>\sum_{n=0}^{\infty}(n+2)(n+2l+3)a_{n+2}=\left[-\frac{2ME}{\hbar^2}+(2n+2l+3)\frac{Mw}{\hbar}\right]a_n</math>
+<math>(n+2)(n+2l+3)a_{n+2}=\left[-\frac{2ME}{\hbar^2}+(2n+2l+3)\frac{M\omega}{\hbar}\right]a_n</math>
+In addition, we have an <math>r^{-1}\!</math> term; for it to vanish, we must set <math>a_1=0.\!</math>  This, combined with the above recursion relation, means that the function <math>f_{nl}(r)\!</math> contains only even powers of <math>r.\!</math>  In other words,
-The function <math>f_l(r)\!</math> contains only even powers in n and is given by:
+<math>f_{nl}(r)=\sum_{n=0,2,4,\ldots}^{\infty}a_{n}r^{n}=\sum_{n'=0}^{\infty}a_{n'}r^{n'}.</math>
-:<math>f_l(r)=\sum_{n=0}^{\infty }a_{2n}r^{2n}=\sum_{n^{'}=0,2,4}^{\infty  }a_{n^{'}}r^{n^{'}}</math>
-Now as   <math> n\rightarrow \infty\!</math> , <math>f_l(r)\!</math> diverges so that for finite solution, the series should stop after   <math>r^{n^{'}+2}\!</math> leading to the quantization condition:
+By a similar argument as the one that we employed for the [[Analytical Method for Solving the Simple Harmonic Oscillator|one-dimensional harmonic oscillator]], we find that, unless the series for <math>f_{nl}(r)\!</math> terminates, the resulting full wave function will diverge as <math>r\rightarrow\infty.</math>  Because the series must only contain even powers of <math>r,\!</math> the resulting quantization condition on the energy is
-:<math>\frac{2M}{\hbar^2}E_{n^{'}l}-\frac{Mw}{\hbar}(2n^{'}+2l+3)=0</math>
+<math>\frac{2M}{\hbar^2}E_{n'l}-\frac{M\omega}{\hbar}(4n'+2l+3)=0,\,n'=0,1,2,3,\ldots,</math>
+or
-:<math>E_{n^{'}l}=\left(n^{'}+l+\frac{3}{2}\right)\hbar w,    n^{'}=0,1,2,3,...</math>
+<math>E_{nl}=\left(n+\frac{3}{2}\right)\hbar\omega,\,n=0,1,2,3,\ldots,</math>
-As a result, the energy of the isotropic harmonic oscillator is given by:
+where <math>n=2n'+l.\!</math>
-:<math>E_{n}=\left(n+\frac{3}{2}\right)\hbar w,  n=0,1,2,3,... </math> with  <math> n=n^{'}+l\!</math>
-The degeneracy corresponding to the nth level is:
+The degeneracy corresponding to the <math>n^{\text{th}}\!</math> level may be found to be <math>\tfrac{1}{2}(n+1)(n+2).</math>  We see that energy levels with even <math>n\!</math> correspond to even values of <math>l,\!</math> while those with odd <math>n\!</math> have odd values of <math>l.\!</math>
-:<math>g=\frac{1}{2}(n+1)(n+2)</math>
-The total wavefunction of the isotropic Harmonic Oscilaator is given by:
+The total wave function of the isotropic harmonic oscillator is thus given by
-:<math>\psi_{nlm}(r,\theta ,\phi )=r^{l+1}f_l(r)Y_{lm}(\theta ,\phi)e^-\frac{Mw}{2\hbar}r^2=R_{nl}(r)Y_{lm}(\theta ,\phi )</math>
+<math>\psi_{nlm}(r,\theta,\phi )=r^{l+1}e^{-M\omega r^2/2\hbar}f_{nl}(r)Y_{lm}(\theta,\phi)=R_{nl}(r)Y_{lm}(\theta ,\phi ).</math>
+One may show that, in fact, <math>f_{nl}(r)\!</math> is an [http://mathworld.wolfram.com/AssociatedLaguerrePolynomial.html associated Laguerre polynomial] in <math>\frac{M\omega}{\hbar}r^2.</math>

Isotropic Harmonic Oscillator: Difference between revisions

Latest revision as of 13:42, 18 January 2014

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