The Dirac Delta Function Potential: Difference between revisions

From PhyWiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 1: Line 1:
{{Quantum Mechanics A}}
{{Quantum Mechanics A}}
A delta potential, eg. <math>V_0\delta(x-a)\!</math>, is a special case of the finite square well, where the width of the well goes to zero and the depth of the well goes to infinity, while the produce of the height and depth remains constant. For a delta potential, the wavefunction is still continuous across the potential, ie. <math>x=a\!</math>. However, the first derivative of the wavefunction is discontinuous across the potential.
A delta function potential, <math>V(x)=V_0\delta(x-a)\!</math>, is a special case of the symmetric finite square well; it is the limit in which the width of the well goes to zero and the depth of the well goes to infinity, while the product of the height and depth remains constant. For such a potential, the wave function is still continuous across the potential, ie. <math>x=a\!</math>. However, the first derivative of the wave function is not.


For a particle subject to an attractive delta potential <math> V(x) = -V_0\delta(x)\!</math> the Schrödinger equation is
For a particle subject to an attractive delta function potential, <math>V(x)=-V_0\delta(x),\!</math> the Schrödinger equation is


:<math>\frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}-V_0\delta(x)\psi(x)=E\psi(x)</math>
:<math>-\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}-V_0\delta(x)\psi(x)=E\psi(x).</math>


For <math> x \neq \!0 </math> the potential term vanishes, and all that is left is
For <math> x \neq \!0 </math> the potential term vanishes, and all that is left is


:<math>\frac{d^2 \psi(x)}{dx^2} + \frac{2mE}{\hbar^2}\psi(x) = 0</math>
:<math>\frac{d^2\psi(x)}{dx^2}+\frac{2mE}{\hbar^2}\psi(x)=0.</math>


A bound state(s) may exist when <math> E < 0 \! </math>, and <math> \psi(x) \! </math> vanishes at <math> x = \pm \infty \!</math>.  The bound state solutions are therefore given by:
A bound state(s) may exist when <math>E<0\!</math>, and <math> \psi(x) \! </math> vanishes at <math> x = \pm \infty \!</math>.  The bound state solutions are therefore given by:


:<math> \psi_{1}(x) = Ae^{kx} x < 0  \!</math>  
:<math> \psi_{I}(x) = Ae^{kx},\, x < 0  \!</math>  


:<math> \psi_{2}(x) = Be^{-kx} x > 0 \!</math>
:<math> \psi_{II}(x) = Be^{-kx},\, x > 0 \!</math>


where  
where <math>k=\frac{\sqrt{-2mE}}{\hbar}.</math>


The first boundary condition, the continuity of <math> \psi(x) \!</math> at <math> x = 0 \!</math>, yields <math> A = B \! </math>.  
The first boundary condition, the continuity of <math> \psi(x) \!</math> at <math> x = 0 \!</math>, yields <math> A = B \! </math>.  
Line 24: Line 24:
Integrating the whole equation across the potential gives
Integrating the whole equation across the potential gives


:<math>\int_{-\epsilon}^{\epsilon} \frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}dx+\int_{-\epsilon}^{+\epsilon} (-V_0\delta(x))\psi(x)dx=\int_{-\epsilon}^{\epsilon} E \psi(x)dx</math>
:<math>-\frac{\hbar^2}{2m}\int_{-\epsilon}^{\epsilon} \frac{d^2 \psi(x)}{dx^2}dx+\int_{-\epsilon}^{+\epsilon} (-V_0\delta(x))\psi(x)dx=\int_{-\epsilon}^{\epsilon} E \psi(x)dx</math>


In the limit <math>\epsilon \rightarrow 0 \!</math>, we have
In the limit <math>\epsilon \rightarrow 0 \!</math>, we have


:<math>\frac{-\hbar^2}{2m}\left[\frac{d \psi_2(0)}{dx}-\frac{d \psi_1(0)}{dx}\right]+V_0\psi(0)=0</math>
:<math>-\frac{\hbar^2}{2m}\left[\frac{d \psi_{II}(0)}{dx}-\frac{d \psi_{I}(0)}{dx}\right]-V_0\psi(0)=0,</math>


which yields the relation: <math> 2kA = \frac{2mV_0}{\hbar^2}A \!</math>.
which yields the relation, <math> k = \frac{mV_0}{\hbar^2}. \!</math>.


Since we defined  <math> k = \sqrt{\frac{2m|E|}{\hbar^2}} \!</math>, we have <math> \sqrt{\frac{2m|E|}{\hbar^2}} = \frac{mV_0}{\hbar^2} \!</math>. Then, the energy is
Since we defined  <math> k = \frac{\sqrt{-2mE}}{\hbar}, \!</math>, we have <math> \frac{\sqrt{-2mE}}{\hbar} = \frac{mV_0}{\hbar^2} \!</math>. Then, the energy is <math> E = -\frac{mV_0^2}{2\hbar^2} \!</math>
<math> E = -\frac{mV_0^2}{2\hbar^2} \!</math>


Finally, we normalize <math> \psi(x) \!</math>:
Finally, we normalize <math> \psi(x) \!</math>:
Line 51: Line 50:
The Schrödinger equation is
The Schrödinger equation is


:<math>\frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}+V_0\delta(x-a)\psi(x)=E\psi(x)</math>
:<math>-\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}+V_0\delta(x-a)\psi(x)=E\psi(x)</math>


Integrating the whole equation across the potential gives
Integrating the whole equation across the potential gives


:<math>\int_{a-\epsilon}^{a+\epsilon} \frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}dx+\int_{a-\epsilon}^{a+\epsilon} V_0\delta(x-a)\psi(x)dx=\int_{a-\epsilon}^{a+\epsilon} E \psi(x)dx</math>
:<math>-\frac{\hbar^2}{2m}\int_{a-\epsilon}^{a+\epsilon} \frac{d^2 \psi(x)}{dx^2}dx+\int_{a-\epsilon}^{a+\epsilon} V_0\delta(x-a)\psi(x)dx=\int_{a-\epsilon}^{a+\epsilon} E \psi(x)dx</math>


In the limit <math>\epsilon \rightarrow 0 \!</math>, we have
In the limit <math>\epsilon \rightarrow 0 \!</math>, we have


:<math>\frac{-\hbar^2}{2m}\left[\frac{d \psi(a^+)}{dx}-\frac{d \psi(a^-)}{dx}\right]+V_0\psi(a)=0</math>
:<math>-\frac{\hbar^2}{2m}\left[\frac{d \psi(a^+)}{dx}-\frac{d \psi(a^-)}{dx}\right]+V_0\psi(a)=0</math>


Hence the first derivative of the wave function across a delta potential is discontinuous by an amount:
Hence the first derivative of the wave function across a delta potential is discontinuous by an amount:

Revision as of 16:22, 24 April 2013

Quantum Mechanics A
SchrodEq.png
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.
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
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
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
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
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

A delta function 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(x)=V_0\delta(x-a)\!} , is a special case of the symmetric finite square well; it is the limit in which the width of the well goes to zero and the depth of the well goes to infinity, while the product of the height and depth remains constant. For such a potential, the wave function is still continuous across the potential, ie. 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 x=a\!} . However, the first derivative of the wave function is not.

For a particle subject to an attractive delta function 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(x)=-V_0\delta(x),\!} the Schrödinger equation 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 -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}-V_0\delta(x)\psi(x)=E\psi(x).}

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 x \neq \!0 } the potential term vanishes, and all that is left 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 \frac{d^2\psi(x)}{dx^2}+\frac{2mE}{\hbar^2}\psi(x)=0.}

A bound state(s) may exist when 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<0\!} , 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 \psi(x) \! } vanishes 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 x = \pm \infty \!} . The bound state solutions are therefore 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_{I}(x) = Ae^{kx},\, x < 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 \psi_{II}(x) = Be^{-kx},\, x > 0 \!}

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=\frac{\sqrt{-2mE}}{\hbar}.}

The first boundary condition, the continuity of 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(x) \!} 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 x = 0 \!} , yields 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 second boundary condition, the discontinuity of 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{d\psi(x)}{dx} \! } 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 x = 0 \!} , can be obtained by integrating the Schrödinger equation from 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 -\epsilon \!} 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 \epsilon \!} and then letting 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 \epsilon \rightarrow 0 \!}

Integrating the whole equation across the potential gives

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}\int_{-\epsilon}^{\epsilon} \frac{d^2 \psi(x)}{dx^2}dx+\int_{-\epsilon}^{+\epsilon} (-V_0\delta(x))\psi(x)dx=\int_{-\epsilon}^{\epsilon} E \psi(x)dx}

In the limit 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 \epsilon \rightarrow 0 \!} , we have

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}\left[\frac{d \psi_{II}(0)}{dx}-\frac{d \psi_{I}(0)}{dx}\right]-V_0\psi(0)=0,}

which yields the relation, 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 = \frac{mV_0}{\hbar^2}. \!} .

Since we defined 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 = \frac{\sqrt{-2mE}}{\hbar}, \!} , we have 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{\sqrt{-2mE}}{\hbar} = \frac{mV_0}{\hbar^2} \!} . Then, the energy 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 E = -\frac{mV_0^2}{2\hbar^2} \!}

Finally, we normalize 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(x) \!} :

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 \int_{-\infty}^{\infty} |\psi(x)|dx=2|B|^2\int_{0}^{\infty} e^{-2kx}dx=\frac{|B|^2}{k}=1 }

so,

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 B=\sqrt{k}=\frac{\sqrt{mV_{0}}}{\hbar} }

Evidently, the delta function well, regardless of its "strength" 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} \!} , has one bound 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(x)=\frac{\sqrt{mV_{0}}}{\hbar}e^{\frac{-mV_{0}|x|}{\hbar^{2}}} }

Similarly, for a delta potential of the form 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\delta(x-a)\!} , the discontinuity of the first derivative can be shown as follows:

The Schrödinger equation 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 -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}+V_0\delta(x-a)\psi(x)=E\psi(x)}

Integrating the whole equation across the potential gives

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}\int_{a-\epsilon}^{a+\epsilon} \frac{d^2 \psi(x)}{dx^2}dx+\int_{a-\epsilon}^{a+\epsilon} V_0\delta(x-a)\psi(x)dx=\int_{a-\epsilon}^{a+\epsilon} E \psi(x)dx}

In the limit 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 \epsilon \rightarrow 0 \!} , we have

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}\left[\frac{d \psi(a^+)}{dx}-\frac{d \psi(a^-)}{dx}\right]+V_0\psi(a)=0}

Hence the first derivative of the wave function across a delta potential is discontinuous by an amount:

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{d \psi(a^+)}{dx}-\frac{d \psi(a^-)}{dx}=\frac{2m V_0}{\hbar^2}\psi(a)}

Problem

Problem 1

External Link

Additional information on the dirac delta function

Retrieved from "http://wiki.physics.fsu.edu/index.php?title=The_Dirac_Delta_Function_Potential&oldid=14882"