6.15

Free Particle Solution Review

• For a free particle where l = 0, the solution looks like: constant * j0(kr) * Y{00}(\theta, \phi)

  • k is the magnitude of the wave vector: k = \sqrt{\frac{2mE}{\hbar^2}}

  • This solution is spherically symmetric because it has no \theta or \phi dependence.

  • Y{00} is a constant: Y{00} = \frac{1}{\sqrt{4\pi}}

L=1 Solutions

• For l = 1, there are three solutions corresponding to m = -1, 0, +1:

  • j1(kr) * Y{1, -1}(\theta, \phi)

  • j1(kr) * Y{1, 0}(\theta, \phi)

  • j1(kr) * Y{1, +1}(\theta, \phi)

• For these solutions, the probability density approaches zero at the origin.

  • Probability of finding the particle between r and r + dr approaches 0.

Probability at the Origin

• For l=0 solutions, the probability of finding the particle between r and r + dr is not zero.

Impact of Angular Momentum Quantum Number

• The magnitude of angular momentum quantum number l affects the radial function.

• The differential equation in the r variable contains l as a parameter, so its solution depends on l.

• Separation of variables gives a function of r times a function of (\theta, \phi), but the function of r depends on l.

Bessel Functions

• Bessel functions are sinusoidal/cosinusoidal functions that decay as r \rightarrow \infty.

• j0 starts at a constant value and oscillates, while other jl functions start at the origin.

• j_0 shows a nonzero probability of finding the particle at the origin, having a nonzero intercept.

Quantum Numbers in Spherical Coordinates

• In Cartesian coordinates (x, y, z), we specify (kx, ky, k_z).

  • Specifying the wave number vector specifies the state of the particle.

  • Energy is calculated as: E = \frac{\hbar^2 |k|^2}{2m}.

• In spherical coordinates (r, \theta, \phi), we specify energy (E), angular momentum magnitude squared (L^2), and one component of angular momentum (L_z).

Measuring Momentum Component

• If we try to measure the x component of momentum (px) when the particle is in a state with defined L^2 and Lz, we get a distribution.

  • px operator does not commute with L^2 and Lz.

  • The wave function collapses into an eigenstate of px, leading to a distribution in L^2 and Lz.

  • You lose the information about angular momentum after the measurement.

Quantum Number Considerations

• In both coordinate systems, there are three quantum numbers.

• In Cartesian coordinates, all three (kx, ky, kz) are continuous.

• In Spherical coordinates, Energy is continuous, while L^2 and L_z are discrete because angular momentum is quantized.

Fourier Transform and Wave Function

• The distribution obtained when measuring p_x can be found by doing a three-dimensional Fourier transform.

• The wave function collapses into an eigenfunction of the momentum operator.

Measuring Magnitude of Angular Momentum

• If the particle is in an eigenstate in Cartesian coordinates, then measuring the magnitude of angular momentum (L^2) leaves the energy (E) unchanged but scrambles the momentum quantum numbers (kx, ky, k_z).

• The wave function collapses into a linear combination of all YLMs with a particularlvalue. For instance, if you measure l = 3, then:

\Psi = D1 j3(kr) Y{3, -3} + D2 j3(kr) Y{3, -2} + … + D7 j3(kr) Y_{3, 3}

• This is an eigenfunction of the L^2 operator, giving an angular momentum of \sqrt{12} \hbar (since l(l+1) = 3(4) = 12).

Further Wave Function Collapse

• To further collapse the wave function into one of the seven eigenfunctions, you measure the z component of angular momentum (L_z).

• This measurement causes you to lose information you had about momentum quantum numbers (kx, ky, k_z).

• The wave function collapse puts the particle into a different state than it was before.

Commutation Relations

• The z component of momentum (pz) commutes with Lz.

  • Measuring pz doesn't affect the Lz quantum number or the energy quantum number.

  • What gets scrambled is L^2.

Linear Combination of Eigenfunctions

• A free particle state in Cartesian coordinates is a linear combination of eigenfunctions in spherical coordinates:

c e^{i(kx x + ky y + kz z)} = \sum{l=0}^{\infty} \sum{m=-l}^{l} A{lm} jl(kr) Y{lm}(\theta, \phi)

• The coefficients A_{lm} represent the probabilities of measuring certain values of l and m.

• A{lm} is computed with: A{lm} = \int jl(kr) Y{lm}^*(\theta, \phi) c e^{i(kx x + ky y + k_z z)} r^2 \sin(\theta) dr d\theta d\phi

Reverse Linear Combination

• Any solution in spherical coordinates is a linear combination of eigenstates in the Cartesian representation.

j2(kr) Y{2, -1}(\theta, \phi) = \sum{kx, ky, kz} Bk e^{i(kx x + ky y + kz z)}

• With the restriction that: |k| = \sqrt{kx^2 + ky^2 + k_z^2} and the summation includes triplets of k components that match the energy of the original state.

Fourier Transform and Coefficients

• The coefficients Bk are the Fourier transform of the function: Bk = \int e^{-i k \cdot r} j2(kr) Y{2, -1}(\theta, \phi) dxdydz

Heisenberg Uncertainty Principle

• Fourier transform of j_0 gives a distribution in k space and that distribution will have a standard deviation. The standard deviation in real space is connected to the standard deviation in k space as predicted by the Heisenberg uncertainty principle.

• Fourier transform with respect to the x variable yields a distribution in kx values, where \Delta kx \geq \frac{\hbar}{\Delta x} . This occurs because the sum includes all k values, and has a range of possible k_xvalues.

Good Quantum Numbers

• When you're in one representation, certain quantum numbers are "good" quantum numbers (needed to specify the state).

• Over-specifying leads to problems and scrambling of previous quantum numbers.

Potential Function Example

• Consider V(r) = 0 near origin, then V(r) = \infty (analogous to particle in 1D infinite well).

• The wave function has to be zero at the boundary (because it has to be zero in regions where the potential is infinite).

• The energy is discrete with k_n values (quantized magnitude of k).

• Considering only l = 0 solutions:

• Wave function has to be zero for r > b

  • The boundary condition requires the value of k which gives zero.

Eigenstates

• The first energy level is the first k1 diagram.

• The second energy level with a bigger k gives the second diagrams.

• In general, it's like \sin(\frac{n \pi x}{L}) where k_x = \frac{n \pi}{L}.

• Since j_0 = \frac{\cos(kr)}{r}, set this equal to zero when evaluated at r = b.

Finite Potential Well

• V(r) = 0 inside and has a finite Potential at r = b.

• In region 1:

\Psi = C1 j0(kr)

*In region two

\Psi = \frac{C_2 e^{-\alpha r}}{r}

*We are solving for the function u(r) = r * R(r), then you have two regions, and functions have to match with both the the slope/derivative.
*Will get a few discrete energy states before continuity.

Energy Considerations

• For E < V0, in region 1: C1 j0(kE) and in region 2: C2 j0 k(E - V0) * r. with kE = \sqrt{2mE / \hbar^2} and k{E - V0} = \sqrt{2m(E - V_0) / \hbar^2}.

Particle in Spherical Shell

• Infinite potential is A, becomes 0 at A to B, and Infinite after. Is only to be found between the ranges of a and B. And looks like a particle trapped in a box situation.

• Both j0 and n0 are included in solution because the n is not defined at origin, but our particle is not at the origin. Both functions must be considered, and The function when we graph it, would be zero at A and B, like trapped in a can with a lid at both ends. However, there is no pretty sine function.

Hydrogen Atom Potential

• Potential goes to:-\frac{kQq}{r}.

• We begin notice the Effective Potential of: Veff = \frac{l^2}{2mr^2} + \frac{-kQq}{r}.

• Thinking about Effect Potential allows for intuition from one dimensional problems

Radial Functions

• Solutions look look will look like eigenfunctions shared Hamiltonian with Ylm Functions of Thera and Phi depend on data and phi.

Central Potentials

• If the given potential, as a function of the r vector, equals the function of just the magnitude of the r vector, this is a central Potential.

Schrodinger Equation

• The Scrödinger Equation can be written:

\frac{-\hbar^2}{2m} \frac{1}{r} \frac{d}{dr} (r \Psi) + \frac{L^2}{2mr^2} \Psi + V(r) \Psi = E \Psi
*Graphing this function, Small R = This term is gonna dom and dominate Positive Infinity
So the graph has to start here for small are
*For Large R,
Graphs is gonna be a Like a negative one over R function
So it's gonna be for large R it has to be like this
*Intermediately
Right, not close to R equals zero, not close to R equals infinity, then
When you put it together
*Right so
This is the function minus KQ little q over R and this is the function l squared over two Mr squared which we can also write since we know our our Solutions are going to have l quantum number We can write this as h bar squared l times l plus one over two Mr squared but it's basically a constant over R squared
*The Energy will be Discrete because the conditions have to be met, r = 0 and when r goes to infinity

Bound Energy States

• Bound stats can be Quantized because this goes to infinity.

• Discrete energy functions, from e1 - e3, and all negative.

• We saw that en was equal to minus Z0 over n squared, a semiquantum approach with correct and incorrect.The energy levels we know are correctly because the energy values are verified by experimentation
  For an Energy lever greater than zero
  The electron has enough energy to escape and can be anything to zero.

Hamiltonian Applications

• Applying the Hamilitonian, gives us to sketch the sketch wave functions, could almost figure out what E1 has to look like
  Is has to be abandoned state to goes to infinity
  r times big are

Hydrogen Atom Schrodinger Equation

The following is the Schrodinger equation in order to find the energy of eigenvalues and energy of Eigen functions:

\frac{-\hbar^2}{2m} \frac{1}{r} \frac{d}{dr} (r \frac{dR(r)}{dr}) + \frac{L^2}{2mr^2} + \frac{-kQq}{r} = E
We expect the function to be r/r times Ylm is the central potential because solutions are shared with the squared operated if look Hamiltonian. If the potential involves Feta or phi
We will multiply everything by two Mr over -A bar squared Now this next term is gonna turn out to be positive, okay Because were multiplying were multiplying all terms on both sides by minus two Mr over H bar squared, right for the entire equation right
We will make the substitution, the same substitution we made before

Little U is equal to r times big r.