Isotropic Harmonic Oscillator: 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
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	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
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	General Formalism; Free Particle in Spherical Coordinates; Spherical Well; Isotropic Harmonic Oscillator; Hydrogen Atom; WKB in Spherical Coordinates
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Revision as of 00:14, 1 September 2013

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 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 M\!} in an isotropic harmonic oscillator potential 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(r)=\frac{1}{2}M\omega^{2}r^2} is 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 -\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}.}

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

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 0\!} , the equation 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 -\frac{\hbar^2}{2M}\frac{d^2u_{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 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 0} 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 u_{nl}(r)\simeq r^{l+1}.}

In the limit 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,} on the other hand, the equation becomes

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

whose solution is 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 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

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

Substituting this expression into the original equation, 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 \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}

Now we try the power series solution

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 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+...}

Substituting this solution into the reduced form of the equation,

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

which reduces to the equation

\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_{n}r^{n}\right]=0

For this equation to hold, the coefficients of each of the powers of r must vanish seperately.

So,when $n=0\!$ the coefficient of $r^{-2}\!$ is zero, $0.(2l+1)a_{0}=0$ implying that $a_{0}\!$ need not be zero.

Equating the coefficient of $r^{-1}\!$ to be zero, $1.(2l+2)a_{1}=0\!$ implying that $a_{1}\!$ must be zero.

Equating the coefficient of $r^{-n}\!$ to be zero, we get the recursion relation which is:

\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}

The function $f_{l}(r)\!$ contains only even powers in n and is given by:

f_{l}(r)=\sum _{n=0}^{\infty }a_{2n}r^{2n}=\sum _{n^{'}=0,2,4}^{\infty }a_{n^{'}}r^{n^{'}}

Now 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 n\rightarrow \infty\!} , 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 f_l(r)\!} diverges so that for finite solution, the series should stop after 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^{n^{'}+2}\!} leading to the quantization 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 \frac{2M}{\hbar^2}E_{n^{'}l}-\frac{Mw}{\hbar}(2n^{'}+2l+3)=0}

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 E_{n^{'}l}=\left(n^{'}+l+\frac{3}{2}\right)\hbar w, n^{'}=0,1,2,3,...}

As a result, the energy of the isotropic harmonic oscillator is 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 E_{n}=\left(n+\frac{3}{2}\right)\hbar w, n=0,1,2,3,... } with 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 n=n^{'}+l\!}

The degeneracy corresponding to the nth level 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 g=\frac{1}{2}(n+1)(n+2)}

The total wavefunction of the isotropic Harmonic Oscillator is 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 \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 )}

@@ Line 2: / Line 2: @@
 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|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:
+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:
-:<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>
+:<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>
-We look at the solutions <math>u_{nl}\!</math> in the asymptotic limits of <math> r\!</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>.
-Combining the asymptotic limit solutions we choose the general solution to the equation as
+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>
-:<math>u_l(r)=f_l(r)r^{l+1}e^-\frac{Mwr^2}{2\hbar}</math>
-Substituting this expression into the original equation,
+In the limit as <math>r\rightarrow \infty,</math> on the other hand, the equation becomes
-:<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>
+<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
+<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

Isotropic Harmonic Oscillator: Difference between revisions

Revision as of 00:14, 1 September 2013

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