6.4: Finite Square-Well Potential

• The finite square-well potential is

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• The Schrödinger equation outside the finite well in regions I and III is

or using

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yields . The solution to this differential has exponentials of the form e^αx and e^-αx. In the region x > L, we reject the positive exponential and in the region x < L, we reject the negative exponential.Then the other one decays into the classically forbidden region

Finite Square-Well Solution

• Inside the square well, where the potential V is zero, the wave equation becomes where

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• Instead of a sinusoidal solution we have

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• The boundary conditions require that

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and the wave function must be smooth where the regions meet.

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• Note that the �wave function is �nonzero outside �of the box.

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We will skip the tedious procedure of fulfilling the above boundary conditions, but discuss the results

Larger wavelength

Smaller momentum and energy

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13) Compare the results of the finite and infinite square well potential?

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1. The wavelengths are longer for the finite square well.

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• The wavelengths are shorter for the finite square well.

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• The wavelengths are the same.

Clicker - Questions

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13) Compare the finite and infinite square well potentials and chose the correct statement.

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1. There is a finite number of bound energy states for the finite potential.

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• There is an infinite number of bound energy states for the finite potential.

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• There are bound states which fulfill the condition E>V_o.

Clicker - Questions

6.5: Three-Dimensional Infinite-Potential Well

• The wave function must be a function of all three spatial coordinates. We begin with the conservation of energy
• Multiply this by the wave function to get

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• Now consider momentum as an operator acting on the wave function. In this case, the operator must act twice on each dimension. Given:

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• The three dimensional Schrödinger wave equation is

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Laplace operator

Time independent Schroedinger equation

Particle in3-D box

Use 3 quantum numbers n

Degeneracy

• Analysis of the Schrödinger wave equation in three dimensions introduces three quantum numbers that quantize the energy.

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• A quantum state is degenerate when there is more than one wave function for a given energy.

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• Degeneracy results from particular properties of the potential energy function that describes the system. A perturbation of the potential energy can remove the degeneracy (to be shown later).

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Problem6.26

Find the energies of the second, third, fourth, and fifth levels for the three dimensional cubical box. Which energy levels are degenerate?

A given state is degenerate when there is more than one wave function for a given energy

For a cubical box L1=L2=L3=L

ground state wavefunction E1 is not degenerate

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6.6: Simple Harmonic Oscillator

• Simple harmonic oscillators describe many physical situations: springs, diatomic molecules and atomic lattices.

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• Consider the Taylor expansion of a potential function:

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Redefining the minimum potential and the zero potential, we have

• =

Substituting this into the wave equation:

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Let and which yields .

© 2016 Pearson Education, Inc.

Parabolic Potential Well

• If the lowest energy level is zero, this violates the uncertainty principle.
• The wave function solutions are where H_n(x) are Hermite polynomials of order n.

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• In contrast to the particle in a box, where the oscillatory wave function is a sinusoidal curve, in this case the oscillatory behavior is due to the polynomial, which dominates at small x. The exponential tail is provided by the Gaussian function, which dominates at large x.

Analysis of the Parabolic Potential Well

• The energy levels are given by

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• The zero point energy is called the Heisenberg limit:

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• Classically, the probability of finding the mass is greatest at the ends of motion and smallest at the center (that is, proportional to the amount of time the mass spends at each position).
• Contrary to the classical one, the largest probability for this lowest energy state is for the particle to be at the center.

Hermite polynomial functions are shown above

A hydrogen molecule can be approximated as a simple harmonic oscillator with force constant k=1.1x10^3 N/m. Find (a) the energy levels and (b) the possible wavelengths of photons emitted when the H2 molecule decays from the second excited state eventually to the ground state.

Deuteron in a nucleus

h

h

3A

Rectangular box

n₁=1, n₂=2, n₃ =1

n₁=1, n₂=1, n₃ =3

Einstein: What I most admire about your art, is your universality. You don’t say a word, yet the world understands you!

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Chaplin: True. But your glory is even greater! The whole world admires you, even though they don’t understand a word of what you say.

• 7.1 Application of the Schrödinger Equation to the Hydrogen Atom
• 7.2 Solution of the Schrödinger Equation for Hydrogen
• 7.3 Quantum Numbers
• 7.4 Magnetic Effects on Atomic Spectra – The Normal Zeeman Effect

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CHAPTER 7�The Hydrogen Atom

This spherical system has very high symmetry causing very high degeneracy of the wavefunctions

Lecture a

Labelling of corresponding video

7.1: Application of the Schrödinger Equation to the Hydrogen Atom

• The approximation of the potential energy of the electron-proton system is electrostatic:

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• Rewrite the three-dimensional time-independent Schrödinger Equation.

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For Hydrogen-like atoms (He⁺ or Li⁺⁺)

• Replace e² with Ze² (Z is the atomic number)
• Use appropriate reduced mass μ

Uranium is a chemical element with the symbol U and atomic number Z=92

Application of the Schrödinger Equation

• The potential (central force) V(r) depends on the distance r between the proton and electron.

Transform to spherical polar coordinates because of the radial symmetry.

Insert the Coulomb potential into the transformed Schrödinger equation.

Application of the Schrödinger Equation

• The wave function ψ is a function of r, θ, .

Equation is separable.

Solution may be a product of three functions.

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• We can separate Equation 7.3 into three separate differential equations, each depending on one coordinate: r, θ, or .

Equation 7.3

Divide and conquer !!

7.2: Solution of the Schrödinger Equation for Hydrogen

• Substitute Eq (7.4) into Eq (7.3) and separate the resulting equation into three equations: R(r), f(θ), and g( ).

Separation of Variables

• The derivatives from Eq (7.4)

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• Substitute them into Eq (7.3)

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• Multiply both sides of Eq (7.6) by r² sin² θ / Rfg

Solution of the Schrödinger Equation

• Only r and θ appear on the left side and only appears on the right side of Eq (7.7)
• The left side of the equation cannot change as changes.
• The right side cannot change with either r or θ.

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• Each side needs to be equal to a constant for the equation to be true.

Set the constant −m_ℓ² equal to the right side of Eq (7.7)

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• It is convenient to choose a solution to be .

-------- azimuthal equation

Eq (7.8)

Properties of Valid Wave Functions

Boundary conditions

1. In order to avoid infinite probabilities, the wave function must be finite everywhere.
2. In order to avoid multiple values of the probability, the wave function must be single valued.
3. For finite potentials, the wave function and its derivative must be continuous. This is required because the second-order derivative term in the wave equation must be single valued. (There are exceptions to this rule when V is infinite.)
4. In order to normalize the wave functions, they must approach zero as x approaches infinity.

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Solutions that do not satisfy these properties do not generally correspond to physically realizable circumstances.

Not normalizable

Solution of the Schrödinger Equation

• satisfies Eq (7.8) for any value of m_ℓ.
• The solution be single valued in order to have a valid solution for any , which is

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• m_ℓ to be zero or an integer (positive or negative) for this to be true.
• If Eq (7.8) were positive, the solution would not be realized.

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• Set the left side of Eq (7.7) equal to −m_ℓ² and rearrange it.

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• Everything depends on r on the left side and θ on the right side of the equation.