Oscillation Theorem: 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 16:17, 6 August 2013

Let 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_1\!} be a bound state of a given potential with energy 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_1\!} 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_2\!} be another bound state with energy 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_2\!} such 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 E_2>E_1\!} . Let 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\!} be a point at which both 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_1\!} 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_2\!} vanish; this is guaranteed to happen at least 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_0\to\pm\infty.\!} We will now prove that, between any two points 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\!} 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 x_1\!} at 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 \psi_1\!} vanishes, there must be at least one point at 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 \psi_2\!} vanishes. Let us begin by writing down the Schrödinger equation for each wave function:

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_1}{dx^2}+V(x)\psi_1=E_1\psi_1\!}

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_2}{dx^2}+V(x)\psi_2=E_2\psi_2\!}

Multiplying the first equation 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_2\!} and the second 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_1,\!} subtracting the second equation from the first, and simplifying, we see 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 \frac{d}{dx}\left(-\psi_2\frac{d\psi_1}{dx}+\psi_1\frac{d\psi_2}{dx}\right)=\frac{2m}{\hbar^2}\left(E_1-E_2\right)\psi_1\psi_2.\!}

If we now integrate both sides of this 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 x_0\!} to any position 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'\!} and simplify, we see 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 -\psi_2(x')\frac{d\psi_1(x')}{dx}+\psi_1(x')\frac{d\psi_2(x')}{dx}=\frac{2m}{\hbar^2}(E_1-E_2)\int_{x_0}^{x'}\psi_1\psi_2\,dx\!}

The key is to now let 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'=x_1\!} be the first position to the right 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 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 \psi_1\!} vanishes.

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_2(x_1)\frac{d\psi_1(x_1)}{dx}=\frac{2m}{\hbar^2}(E_2-E_1)\int_{x_0}^{x_1}\psi_1\psi_2\,dx\!}

Now, if we assume 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 \psi_2\!} does not vanish at or between 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\!} 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 x_1\!} , then it is easy to see that the left hand side of the previous equation has a different sign from that of the right hand side, and thus it must be true 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 \psi_2\!} must vanish at least once between 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\!} 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 x_1\!} 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 E_2>E_1.\!}

@@ Line 1: / Line 1: @@
 {{Quantum Mechanics A}}
-Let us concentrate on the bound states of a set of wavefunctions. Let <math> \frac{\hbar^2}{2m}=1</math>. Let <math>\psi_1\!</math> be an eigenstate with energy <math>E_1\!</math> and <math>\psi_2\!</math> an eigenstate with energy <math>E_2\!</math>, and <math>E_2>E_1\!</math>. We also can set boundary conditions, where both <math>\psi_1\!</math> and <math>\psi_2\!</math> vanish at <math>x_0\!</math>.This implies that
+Let <math>\psi_1\!</math> be a bound state of a given potential with energy <math>E_1\!</math> and <math>\psi_2\!</math> be another bound state with energy <math>E_2\!</math> such that <math>E_2>E_1\!</math>.  Let <math>x_0\!</math> be a point at which both <math>\psi_1\!</math> and <math>\psi_2\!</math> vanish; this is guaranteed to happen at least for <math>x_0\to\pm\infty.\!</math>  We will now prove that, between any two points <math>x_0\!</math> and <math>x_1\!</math> at which <math>\psi_1\!</math> vanishes, there must be at least one point at which <math>\psi_2\!</math> vanishes.  Let us begin by writing down the Schrödinger equation for each wave function:
-:<math>-\psi_2\frac{\partial^2 \psi_1}{\partial x^2}+V(x)\psi_2\psi_1=E_1\psi_2\psi_1\!</math>
+:<math>-\frac{\hbar^2}{2m}\frac{d^2 \psi_1}{dx^2}+V(x)\psi_1=E_1\psi_1\!</math>
-:<math>-\psi_1\frac{\partial^2 \psi_2}{\partial x^2}+V(x)\psi_1\psi_2=E_2\psi_1\psi_2\!</math>
+:<math>-\frac{\hbar^2}{2m}\frac{d^2 \psi_2}{dx^2}+V(x)\psi_2=E_2\psi_2\!</math>
-Subtracting the second of these from the first and simplifying, we see that
+Multiplying the first equation by <math>\psi_2\!</math> and the second by <math>\psi_1,\!</math> subtracting the second equation from the first, and simplifying, we see that
-:<math>\frac{\partial}{\partial x}\left(-\psi_2\frac{\partial \psi_1}{\partial x}+\psi_1\frac{\partial \psi_2}{\partial x}\right)=\left(E_1-E_2\right)\psi_1\psi_2\!</math>
+:<math>\frac{d}{dx}\left(-\psi_2\frac{d\psi_1}{dx}+\psi_1\frac{d\psi_2}{dx}\right)=\frac{2m}{\hbar^2}\left(E_1-E_2\right)\psi_1\psi_2.\!</math>
 If we now integrate both sides of this equation from <math>x_0\!</math> to any position <math>x'\!</math> and simplify, we see that
-:<math>-\psi_2(x')\frac{\partial \psi_1(x')}{\partial x}+\psi_1(x')\frac{\partial \psi_2(x')}{\partial x}=(E_1-E_2)\int_{x_0}^{x'}\psi_1\psi_2dx\!</math>
-The key is to now let <math>x'\!</math> be the first position to the right of <math>x_0\!</math> where <math>\psi_1\!</math> vanishes.
+:<math>-\psi_2(x')\frac{d\psi_1(x')}{dx}+\psi_1(x')\frac{d\psi_2(x')}{dx}=\frac{2m}{\hbar^2}(E_1-E_2)\int_{x_0}^{x'}\psi_1\psi_2\,dx\!</math>
-:<math>-\psi_2(x')\frac{\partial \psi_1(x')}{\partial x}=(E_1-E_2)\int_{x_0}^{x'}\psi_1\psi_2dx\!\!</math>
+The key is to now let <math>x'=x_1\!</math> be the first position to the right of <math>x_0\!</math> where <math>\psi_1\!</math> vanishes.
+:<math>\psi_2(x_1)\frac{d\psi_1(x_1)}{dx}=\frac{2m}{\hbar^2}(E_2-E_1)\int_{x_0}^{x_1}\psi_1\psi_2\,dx\!</math>
-Now, if we assume that <math>\psi_2\!</math> does not vanish at or between <math>x'\!</math> and <math>x_0\!</math>, then it is easy to see that the left hand side of the previous equation has a different sign from that of the right hand side, and thus it must be true that <math>\psi_2\!</math> must vanish at least once between <math>x'\!</math> and <math>x_0\!</math> if <math>E_2>E_1\!</math>.
+Now, if we assume that <math>\psi_2\!</math> does not vanish at or between <math>x_0\!</math> and <math>x_1\!</math>, then it is easy to see that the left hand side of the previous equation has a different sign from that of the right hand side, and thus it must be true that <math>\psi_2\!</math> must vanish at least once between <math>x_0\!</math> and <math>x_1\!</math> if <math>E_2>E_1.\!</math>

Oscillation Theorem: Difference between revisions

Latest revision as of 16:17, 6 August 2013

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