6.16

Approximations for Energy Level Solutions

• When considering energy levels, focus on the limits as r approaches zero.

Large r Approximation

• For large r, ignore terms that become insignificant compared to others.

• Terms with a smaller power dependence are negligible compared to dominant terms.

Small r Approximation

• For small r, disregard terms that are less significant.

• This simplifies the differential equation, making it easier to solve.

• Solutions can be found by considering a function u proportional to r^{l+1}.

Proposed Solution

• Try a solution of the form u = c
  ^{l+1}, where c is a constant.

• First derivative: \frac{du}{dr} = c (l+1) r^l

• Second derivative: \frac{d^2u}{dr^2} = c (l+1) l r^{l-1}

Verification of the Solution

• Substitute the second derivative into the simplified differential equation.

• The differential equation is:\frac{d^2u}{dr^2} - \frac{l(l+1)}{r^2} u = 0

• After substitution: c (l+1) l r^{l-1} - \frac{l(l+1)}{r^2} c r^{l+1} = 0

• Simplify the equation to verify the solution.

• For small r, expect solutions to behave like c r^{l+1}.

• Since the radial functions are u/r, expect solutions like c r^l for small r.

Combining Solutions

• Combine solutions for small and large r using a polynomial times an exponential function.

• For small r, the exponential function e^{-\beta r} approaches one.

• For large r, the exponential function dominates over any polynomial function.

Trial and Error Function

• Try a function u(r) = c r e^{-\beta r}.

• It satisfies the small r condition, requiring a polynomial type function.

• Take two derivatives of the trial function to verify it satisfies the Schrodinger equation.

Second Derivative of the Trial Function

• \frac{d^2u}{dr^2} = c e^{-\beta r} (\beta^2 r - 2\beta)

• Consider the case when l = 0.

Simplified Equation for l=0

• \frac{d^2u}{dr^2} + \frac{2}{a_0 r} u = \alpha^2 u where \alpha^2 = -\frac{2mE}{\hbar^2}

• Substitute the second derivative and the trial function into the equation.

• c e^{-\beta r} (\beta^2 r - 2\beta) + \frac{2}{a_0 r} c r e^{-\beta r} = \alpha^2 c r e^{-\beta r}

• The equation is dimensionally correct, with terms having units of inverse meters squared.

Parameters and Length Scales

• The equation involves two different lengths: \frac{1}{\alpha} and a_0.

• The solution may depend on the dimensionless number \frac{r}{a_0}.

• Simplify the equation by canceling out common terms.

• \beta^2 r - 2\beta + \frac{2}{a_0} = \alpha^2 r

• Since the equation must hold for all r, equate the coefficients of corresponding terms.

Solving for Parameters

• Equate coefficients to find relationships between parameters.

• \alpha = \beta

• -2\beta + \frac{2}{a_0} = 0

• Solve for \beta: \beta = \frac{1}{a_0}

Solution for l=0

• The solution for l = 0 is R(r) = c e^{-\frac{r}{a_0}}.

Solutions for l ≠ 0

• For l \neq 0, a polynomial times an exponential function is still a valid solution.

• The polynomial function becomes more complicated.

• Try a solution of the form (a + b r) e^{-\beta r}.

Hydrogen Atom Solutions

• The solutions are shown on the handout.

• The first solution is for n = 1, l = 0, which is R{10}(r) Y{00} = c e^{-\frac{r}{a_0}}.

• Y_{00} is a constant function.

Quantum Numbers

• The quantum number n is 1, l is 0, and m is 0.

• The m quantum number ranges from -l to +l.

n=2 Solutions

• For n = 2, l = 0, the solution involves a polynomial 1 - \frac{r}{2a0} times an exponential function e^{-\frac{r}{2a0}}.

• The exponential function decays slower than the n = 1 solution.

Bohr's Analysis

• Bohr's analysis indicates that the radius of the circle increases with higher n values.

• rn = a0 n^2

• The quantum analysis shows that the exponential function extends further out for higher n values.

• The denominator of the exponential function is two times the Bohr radius.

• The second solution is further extended, consistent with Bohr's analysis.

Spherical Symmetry and Orbitals

• s orbitals (l=0) are spherically symmetric.

• They do not depend on \theta and \phi.

• p orbitals (l=1) are represented by the l = 1 solutions.

• The probability density is quantized.

• There are regions where the particle is unlikely to be found.

• Quantum solutions are probability densities, not fixed radius values.

Radial Wave Functions

• The right-hand column in the handout shows the radial part of the wave function.

• Isotropic distributions only occur for l=0.

• Solutions with l=1 or l=2 have angular dependence given by spherical harmonics. These functions give a higher probability to certain angles.

• The number of nodes increases by one for each bound state.

• The first state has no nodes except at infinity.

• The n=2, l=0 solution has one node between zero and infinity.

• The n=3, l=0 solution has two nodes.

• There is a specific value of r where the probability density goes to zero.

• The electron in the 2s orbital has a probability of being near the center and at a larger radius value.

Probability Density

• The left column gives the probability density P(r) between r and r + dr.

• P(r) = |R(r)|^2 r^2

• Multiply the function by r^2 to get the probability density.

• The probability density goes to zero at the origin.

• This is due to the volume element being zero at the origin.

• The volume element is 4 \pi r^2 dr.

• The geometric factor is r^2.

• For the 1s solution, the most likely value of r is the Bohr radius.

Proving Most Probable Radius

• Take r^2 times the square of the function, take the derivative, and set it equal to zero to find the most probable value of r: \frac{d}{dr} (r^2 |R(r)|^2) = 0 .

Coordinate Systems

• In Cartesian coordinates, each box has the same size: dx dy dz.

• In spherical coordinates, each box does not have the same size.

• The volume element in spherical coordinates is dV = r^2 \sin \theta dr d\theta d\phi.

• The probability includes the coordinate factor because the "boxes" in the spherical coordinate system are of unequal size. If you used Cartesian coordinates the boxes have equal size and that factor would not be there.

Derivative for Maximum Probability

• The function to maximize is [c r^2 e^{-\frac{2r}{a_0}}]

• Set the derivative equal to zero to find the maximum.

• The most probable value of r is a_0.

Average Radius

• Calculate the average value of r to compare with Bohr's analysis.

• \langle r \rangle = \int0^{\infty} r P(r) dr / \int0^{\infty} P(r) dr

• Integrate the probability times r to get the average value of r.

• The integration involves squaring the wave function and multiplying by r.

Bonus Problem

• Bonus: Calculate the average value of r for the n = 2, l = 0 solution.

• Show if the average value is equal to 4 a_0, as predicted by Bohr's model.

• Use an integration table to find the integral of r^n e^{-\beta r}.

• Determine if Bohr's analysis is consistent with the full quantum result.