chapter 41 1d wavefunctions. topics: schrödinger’s equation: the law of psi solving the...
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Chapter 41
1D Wavefunctions
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Topics:
• Schrödinger’s Equation: The Law of Psi
• Solving the Schrödinger Equation
• A Particle in a Rigid Box: Energies and Wave Functions
• A Particle in a Rigid Box: Interpreting the Solution
• The Correspondence Principle
• Finite Potential Wells
• Wave-Function Shapes
• The Quantum Harmonic Oscillator
• More Quantum Models
• Quantum-Mechanical Tunneling
Chapter 41. One-Dimensional Quantum Mechanics
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The wave function is complex.
ih
∂∂t
ψ (x,t) =−h2
2m∂2
∂x2 ψ (x,t) +U(x)ψ (x,t)
What is the PDF for finding a particle at x ?
P(x, t) =ψ (x,t) 2
Step 1: solve Schrodinger equation for wave function
Step 2: probability of finding particle at x is P(x, t) =ψ (x,t) 2
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Stationary States - Bohr Hypothesis
ih
∂∂t
ψ (x,t) =−h2
2m∂2
∂x2 ψ (x,t) +U(x)ψ (x,t)
ψ (x,t) = ψ̂ (x)e−1Et /h
Eψ̂ (x) =−
h2
2m∂2
∂x2 ψ̂ (x) +U(x)ψ̂ (x)
ω =
E
h
Stationary State satisfies
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Eψ̂ (x) =−
h2
2m∂2
∂x2 ψ̂ (x) +U(x)ψ̂ (x)
Stationary states
∂2
∂x2ψ̂ (x) = −β 2 (x)ψ̂ (x)
Rewriting:
β 2 (x) =
2m
h2E − U(x)( )
Dependence on x comes from dependence on potential
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∂2
∂x2ψ̂ (x) = −β 2 (x)ψ̂ (x)
β 2 (x) =
2m
h2E − U(x)( )
Requirements on wave function
1. Wave function is continuous2. Wave function is normalizable
E −U(x)( ) =K
Classically
K= kinetic energy
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∂2
∂x2ψ̂ (x) = −β 2 (x)ψ̂ (x)
β 2 (x) =
2m
h2E − U(x)( )
β 2 (x) =
2m
h2K( )
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∂2
∂x2ψ̂ (x) = −β 2 (x)ψ̂ (x)
β 2 (x) =
2m
h2E − U(x)( )
β 2 (x) =
2m
h2K( )
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For 0<x<L ∂2
∂x2ψ̂ (x) = −β 2ψ̂ (x)
Solution: ψ̂ (x) =Asinβx+ Bcosβx
Boundary Conditions: ψ̂ (0) =0 ψ̂ (L) =0
ψ̂ (0) =0 ψ̂ (0) =Asin0 + Bcos0 =B→ B=0
ψ̂ (L) =0 ψ̂ (L) =AsinβL =0
Must have
βL = nπ
βn
2 =2m
h2En( )
En =
h2
2mβn
2 =h2
2mnπL
⎛⎝⎜
⎞⎠⎟
2
Energy is Quantized
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A Particle in a Rigid BoxA Particle in a Rigid BoxThe solutions to the Schrödinger equation for a particle in a rigid box are
For a particle in a box, these energies are the only values of E for which there are physically meaningful solutionsto the Schrödinger equation. The particle’s energy is quantized.
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A Particle in a Rigid BoxA Particle in a Rigid BoxThe normalization condition, which we found in Chapter 40, is
This condition determines the constants A:
The normalized wave function for the particle in quantum state n is
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EXAMPLE 41.2 Energy Levels EXAMPLE 41.2 Energy Levels and Quantum jumpsand Quantum jumps
QUESTIONS:
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EXAMPLE 41.2 Energy Levels EXAMPLE 41.2 Energy Levels and Quantum jumpsand Quantum jumps
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EXAMPLE 41.2 Energy Levels EXAMPLE 41.2 Energy Levels and Quantum jumpsand Quantum jumps
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EXAMPLE 41.2 Energy Levels EXAMPLE 41.2 Energy Levels and Quantum jumpsand Quantum jumps
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EXAMPLE 41.2 Energy Levels EXAMPLE 41.2 Energy Levels and Quantum jumpsand Quantum jumps
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EXAMPLE 41.2 Energy Levels EXAMPLE 41.2 Energy Levels and Quantum jumpsand Quantum jumps
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The Correspondence PrincipleThe Correspondence Principle• Niels Bohr put forward the idea that the average
behavior of a quantum system should begin to look like the classical solution in the limit that the quantum number becomes very large—that is, as n ∞.
• Because the radius of the Bohr hydrogen atom is r = n2aB, the atom becomes a macroscopic object as n becomes very large.
• Bohr’s idea, that the quantum world should blend smoothly into the classical world for high quantum numbers, is today known as the correspondence principle.
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As n gets even bigger and the number of oscillations increases, the probability of finding the particle in an interval x will be the same for both the quantum and the classical particles as long as x is large enough to include several oscillations of the wave function. This is in agreement with Bohr’s correspondence principle.
The Correspondence PrincipleThe Correspondence Principle
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∂2
∂x2ψ̂ (x) = −β 2 (x)ψ̂ (x)
β 2 (x) =
2m
h2E − U(x)( )
Between x=0 and x=L
β 2 (x) > 0
Outside
β 2 (x) < 0
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Finite Potential WellsFinite Potential WellsThe quantum-mechanical solution for a particle in a finite potential well has some important properties:
• The particle’s energy is quantized.
• There are only a finite number of bound states. There are no stationary states with E > U0 because such a particle would not remain in the well.
• The wave functions are qualitatively similar to those of a particle in a rigid box, but the energies are somewhat lower.
• The wave functions extend into the classically forbidden regions. (tunneling)
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The wave function in the classically forbidden region of a finite potential well is
The wave function oscillates until it reaches the classical turning point at x = L, then it decays exponentially within the classically forbidden region. A similar analysis can be done for x ≤ 0.We can define a parameter η defined as the distance into the classically forbidden region at which the wave function has decreased to e–1 or 0.37 times its value at the edge:
Finite Potential WellsFinite Potential Wells
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∂2
∂x2ψ̂ (x) = −β 2 (x)ψ̂ (x)
β 2 (x) =
2m
h2E − U(x)( )
Outside β 2 (x) < 0
1
η2 =2mh2 U0 −E( )
∂2
∂x2ψ̂ (x) =
1
η 2ψ̂ (x)
Solution, x>L
ψ̂ (x) =ψ̂ edgeexp[−(x−L) / η]
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The Quantum Harmonic The Quantum Harmonic OscillatorOscillatorThe potential-energy function of a harmonic oscillator, as
you learned in Chapter 10, is
where we’ll assume the equilibrium position is xe = 0.The Schrödinger equation for a quantum harmonic oscillator is then
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The Quantum Harmonic OscillatorThe Quantum Harmonic OscillatorThe wave functions of the first three states are
Where ω = (k/m)1/2 is the classical angular frequency, and n is the quantum number
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Quantum-Mechanical Quantum-Mechanical TunnelingTunneling
Once the penetration distance η is calculated using
probability that a particle striking the barrier from the left will emerge on the right is found to be
1
η2 =2mh2 U0 −E( )
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THE END !
Good Luck on the remaining exams