6.11

Shared Eigenfunctions of l^2 and l_z

• l^2 and lz share eigenfunctions, termed Y{lm} functions.

• These functions are located on a provided sheet; Y_{lm} functions are on the right-hand side.

• When l = 0, the only allowed value for m is 0.

Values for m

• If l = 1, there exist three m values: -1, 0, and +1.

• If l = 2, five m values exist: -2, -1, 0, +1, +2.

• If l = 3, seven m values exist, ranging from -3 to +3.

• The number of m states is defined by 2l + 1. For instance, if l = 2, then 2(2) + 1 = 5, showing five states.

• m cannot exceed l because the vector component cannot exceed the vector's magnitude. |m| \leq l

Eigenfunctions of lx and ly

• Eigenfunctions of lx and ly exist.

• They manifest as linear combinations from the shared eigenfunctions of l^2 and lz, which are the Y{lm} functions.

• Y{xlm} eigenfunction is the linear combination of the Y{zl} eigenfunctions.

• Example: Y{x1-1} = d1 Y{z1-1} + d2 Y{z10} + d3 Y{z11}. This equation illustrates that \Delta lx \times \Delta l_y > 0.

• Measurements:

  • Measuring the magnitude of l collapses part of the wave function by fixing the quantum number l.

  • Measuring one component of l further collapses the wave function.

Sequential Collapse

• One measurement collapses the wave function to a linear combination.

• A subsequent measurement locks it into a specific state.

• Once locked into a state with definite l and one component of l, measuring another component yields a distribution.

Linear Combination

• The Y{xlmx} functions are a linear combination of the Y{z} functions, represented as: Y{xlmx} = e1 Y{zl-1} + e2 Y{zl0} + e3 Y_{zl+1}

• The value of m can be subscripted (e.g., mz, mx) to denote a measurement of a particular component of l.

• For instance:
  Y{z10} = e1 Y{x1-1} + e2 Y{x10} + e3 Y_{x11}

• You cannot specify two components of angular momentum at the same time.

• Measuring one component leads to a distribution if another component is then measured.

The YLM Function

• The variable y labels the function in terms of \theta and \phi.

• Y_{lm} = \Theta(\theta) \Phi(\phi). It is a product of functions of \theta and \phi.

• The term Y{lm} is a label or code for this function, and it's often written without parentheses: Y{lm}.

• The sheet contains functions put on the board, it includes both \theta and \phi functions, making it abstract.

• Example: For l = 0 and m_z = 0, the state is a constant, indicating spherical symmetry.

• Since it is not a function of \theta or \phi, the probability distribution is spherically symmetric.

• The solution is constant and \propto \frac{1}{\sqrt{4\pi}}. This constant normalizes the integral over all space:
  \int |\Psi|^2 dV = \int |\Psi|^2 r^2 \sin(\theta) d\theta d\phi dr = 1

• Constants ensure that integration yields a probability of 1.
  *

Eigenfunctions

• The Y{lm} functions are the shared eigenfunctions of l^2 and lz.

• Each Y_{lm} is a product of a function of \theta and a function of \phi.

• Even without a \phi dependence, the function can still be an eigenstate of the l_z operator.

• The l_z operator only affects the \phi variable.

• Every function on the chart is an eigenfunction of l^2.

• The provided l^2 operator allows direct verification.

• Each eigenfunction satisfies the eigenvalue problem for both the l^2 and l_z operators.

• Shared eigenfunctions exist because the two operators commute.

Quantum Numbers

• For l = 1, there are three eigenfunctions.

• The bottom number indicates the l quantum number, while the top number shows the ml or mz quantum number.

• For l = 2, there are five eigenfunctions.

• The bottom quantum number is l, determining the magnitude of the angular momentum vector.

• The top quantum number represents the z-component.

• Again, each function is a product of a function of \theta and a function of \phi.

• The fact that components don't commute gives rise to equations of this form.

Angular Momentum

• The component of angular momentum for y_{x1-1} is -1.

• Reference is made to a diagram showing the rotation of angular momentum.

• The representation of l_z = +1 aligns with the z-direction in the diagram.

• Cones in the diagram represent different l_z states.

• The states of l = 1 and m_z = +1 indicate the positive z-component.

• If an angular momentum vector is in the -1 state, measuring the x-component results in a distribution.

• There is no special distinction between the x and z.

• The same argument applies to x axis relative to z.

Measuring Components

• y_{x1-1}: the length is 1.

• mx = +1, mx = -1, and m_x = 0.

• The inquiry was whether certain combinations could be ruled out; the answer is that the d values aren't always zero.

• Symmetry dictates the z-component must be the same as state zero.

• There should not be a higher probability of measuring momentum with the rightward component versus the leftward component.

• These states do not inherently eliminate specific values.

• The same principles apply to x, z, and y directions.

Uncertainty

• The non-commutation of components leads to relationships such as \Delta lx \Delta ly \propto l_z.

• The x and y directions relate to the z direction.

• \Delta lx = \Delta ly.

Zero

• The relationship: \Delta lx^2 = \Delta ly^2 = \frac{1}{2} \langle l_z \rangle

• Once a particle is fixed into an lz state, uncertainties in lx and l_y can be calculated.

• The equation changes based on lz because when l is larger, the lz components increase, expanding the cone diagram.

Free Particle Problem

• The equation for the free particle problem:
  -\frac{\hbar^2}{2m} \frac{1}{r} \frac{d}{dr} \left(r \frac{d}{dr} f(r) \right) + \frac{l^2}{2mr^2} + V(r, \theta, \phi) = E f(r)

• Each operator acts on a function.

• The Hamiltonian in spherical coordinates can be expressed this way.

• The full l^2 operator is skipped because it's extensive.

• The potential must be substituted and the differential equation solved.

Potentials

Examples of potentials:

• Free particle: V = 0

• Coulomb potential: V = -k \frac{Qq}{r}

• Particle in a spherical box:

  • V = 0 for r < R

  • V = \infty for r > R

• Single step potential:V = 0 for r < R

  • jump to V = V_0 for r > R

• Spherical harmonic oscillator: V = \frac{1}{2} k r^2

• Each potential leads to different problems.

Free Particle Eigenfunctions

• Eigenfunctions for the free particle are already known.

• In Cartesian coordinates, the general solution is:
  \Psi(x, y, z) = c1 e^{ikx x} e^{iky y} e^{ikz z}

• Substituting x = r \sin(\theta) \cos(\phi), y = r \sin(\theta) \sin(\phi), and z = r \cos(\theta) would also represent an eigenfunction.

• This method isn't helpful for understanding our theta and phi.

• Choosing specific kx, ky, and k_z values breaks the symmetry of the problem.

Separation of Variables

• We solve the differential equation in spherical coordinates.
  f = R(r) Y_{lm}(\theta, \phi)

• Substituting this solution into the original equation leads to several terms.

• The potential term vanishes in the free particle case.

• l^2 operator acting on its eigenfunction yields a constant times the eigenfunction.

• The original equation simplifies to:
  -\frac{\hbar^2}{2m} \frac{1}{r} \frac{d}{dr} \left(r \frac{d}{dr} R(r) Y{lm}(\theta, \phi) \right) + \frac{l^2}{2mr^2} R(r) Y{lm}(\theta, \phi) = E R(r) Y_{lm}(\theta, \phi)

• Dividing each term by R(r) Y_{lm}(\theta, \phi) simplifies the equation.

Solutions to the Radial Equation

• The resulting differential equation depends on the value of l.

• For l = 0, the equation simplifies significantly, where u(r)=rR(r).

• The equation is:
  -\frac{\hbar^2}{2m} \frac{d^2 u(r)}{dr^2} = E u(r)

• Rearranging gives:
  \frac{d^2 u(r)}{dr^2} = -\frac{2mE}{\hbar^2} u(r)

• The constant is expected to be negative.

Trial Solutions

• Trial solution: u(r) = C \cos(\alpha r)

• Taking derivatives: \frac{d^2}{dr^2} C \cos(\alpha r) = -\alpha^2 C \cos(\alpha r)

• Equating coefficients: \alpha^2 C \cos(\alpha r) = \frac{2mE}{\hbar^2} C \cos(\alpha r)

• Therefore, R(r) = \frac{u(r)}{r} = \frac{C_1}{r} \cos(\alpha r) (for l=0).

Graphing

• One solution is proportional to \frac{\cos(\alpha r)}{r}.

• R(r) = \frac{C_1}{r} \cos(\alpha r) \frac{1}{\sqrt{4\pi}}.

• The graph of the radial function resembles a decreasing cosine wave.

• Another solution is obtained using sine: R(r) = \frac{C_2}{r} \sin(\alpha r).

• \lim_{x \to 0} \frac{\sin(x)}{x} = 1

• Though \frac{\sin(x)}{x} goes to 1, the radial solution R(r) the function goes to positive infinity.

Probability Density

• The probability of finding a particle is calculated by integrating the wave function squared.

Jacobian

• When using r, theta, and phi, the volume element is dV = r^2 \sin(\theta) dr d\theta d\phi.

• The Jacobian ensures the volume element is correctly accounted for.

• The expression is:
  P(r \text{ to } r + dr) = |\Psi|^2 r^2 \sin(\theta) dr d\theta d\phi

• The radial component of a 3D probability density has units of (length)^{-3/2}.

• * The wave function squared is a 3D probability density.

• After plotting an equation squared, the function increases with the square of the radius. Right.

• Therefore, the probability function will not blow up to infinity.

Momentum

• The momentum of the particle isn't specified.

• For the 3D problem with x, y, and z, the energy eigenfunction also determines the momentum direction.

• The original solution \Psi(x, y, z) = c1 e^{ikx x} e^{iky y} e^{ikz z} gave both energy and momentum components.

• There are two factors here, magnitude and direction of energy, and also each component of momentum.

• The Hamiltonian commutes with the px, py, and pz operators because they were shared eigenfunctions.

• Not for one particular value of solutions that are complicated when it comes to measuring momentum.

• The question is: if the momentum is measured, will a single vector be obtained, or will a distribution be observed?

Symmetry

• The function is perfectly symmetric

• The measurement of momentum would yield many values.

• This function isn't an eigenfunction of the px operator.

• All possible solutions could have momentum in any direction.