Quantum Mechanics and Angular Momentum

Kinetic Energy and Momentum

• Kinetic energy can be expressed as \frac{p^2}{2m}, where p is momentum and m is mass.
• This can be broken into radial and tangential parts:
  • KE = \frac{p{radial}^2}{2m} + \frac{p{tangential}^2}{2m}
  • The radial part relates to the particle's movement toward or away from the origin.
  • The tangential part relates to the particle's motion around the origin.

Angular Momentum

• Defined as the cross product of the radius vector (r) and the momentum vector (p).
  • \L = r \times p
• The magnitude of angular momentum is given by:
  • \L = |r| |p| sin(\theta), where \theta is the angle between (r) and (p).
• The kinetic energy can be expressed in terms of angular momentum (L) as:
  • \KE = \frac{L^2}{2mr^2}

Quantum Mechanical Operators

Radial Momentum Operator

• In classical mechanics, we transition to quantum mechanics by replacing classical quantities with operators.
• A naive guess for the radial momentum operator might be:
  • \p_r = -i\hbar \frac{\partial}{\partial r}
  • However, this operator does not have real eigenvalues, which are required for physical quantities.
• The correct radial momentum operator is:
  • \p_r = -i\hbar \frac{1}{r} \frac{\partial}{\partial r} r
  • This operator is dimensionally correct and has real eigenvalues.
• When squared, this operator appears in the Hamiltonian. Squaring it involves an exercise to prove:
  • \p_r^2 = -\hbar^2 \frac{1}{2m} \frac{1}{r} \frac{\partial}{\partial r} r

Angular Momentum Operator

• Classically, the x-component of angular momentum is:
  • \Lx = ry pz - rz p_y
• This comes from the cross product of (r) and (p).
• In Cartesian coordinates:
  • \A \times \B = (Ay Bz - Az By)\hat{x} + (Az Bx - Ax Bz)\hat{y} + (Ax By - Ay Bx)\hat{z}
• In quantum mechanics, we replace the components with their corresponding operators:
  • \L_x = y(-i\hbar \frac{\partial}{\partial z}) - z(-i\hbar \frac{\partial}{\partial y})
  • \L_x = -i\hbar (y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y})
• This operator is experimentally verified.
  • Measurements of (L_x) yield eigenvalues of this operator.
• The eigenvalue problem to solve is:
  • \Lx \psi = \lambda \psi, where \{lambda} corresponds to allowed values after measurement of (Lx).
• The square of the angular momentum operator is:
  \L^2 = Lx^2 + Ly^2 + L_z^2
• This is needed to determine the energy.

Coordinate Transformations

• We want the Hamiltonian in terms of spherical coordinates (r, \theta, \phi), but the derived operators are in Cartesian coordinates (x, y, z).
• This requires converting derivatives from (x, y, z) to (r, \theta, \phi).
• Using the chain rule:
  • \frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} + \frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x}
• We hope for wave functions of the form:
  • \psi(r, \theta, \phi) = R(r) \Theta(\theta) \Phi(\phi)
• Cartesian coordinates can be converted to spherical coordinates by:
  • x = r \sin(\theta) \cos(\phi)
  • y = r \sin(\theta) \sin(\phi)
  • z = r \cos(\theta)

Conversion of Derivatives

• Each derivative with respect to x, y, or z transforms into three terms involving derivatives with respect to r, \theta, and \phi.
• The transformation equations give:
  • r = f(x, y, z)
  • \theta = f(x, y, z)
  • \phi = f(x, y, z)
• The Laplacian operator in Cartesian coordinates is:
  • \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}

Polar Coordinates

• In polar coordinates:
  • x = r \cos(\theta)
  • y = r \sin(\theta)
• After simplification, the Laplacian only has three terms.

Simplified Form

• The full operator, after miraculous simplifications, reduces to the following form, with only three terms in the very end.

Commutation Rules

• The operator (L^2) commutes with (Lx), (Ly), and (L_z).

  • This means they share eigenfunctions.

• However, (Lx) and (Ly) do not commute.

  • This leads to a Heisenberg uncertainty principle relationship between them.

• \sigma{Lx} \sigma{Ly} \geq \frac{1}{2} \hbar |

  • It is impossible to simultaneously know the magnitude of angular momentum and all three components.
  • If a system is in an eigenfunction of (L^2) and (Lx), measurements of (Ly) and (L_z) will yield a distribution of values.

• A state that is an eigenfunction of (L^2) and (Ly) is a linear combination of eigenstates in (Lx) and (L_z).

• An eigenstate for (Lz) is a linear combination of eigenstates for (Lx).
  \psi{Lz} = c1 \psi{L{x1}} + c2 \psi{L{x2}} + c3 \psi{L_{x3}} + …

• The coefficients represent the probability of measuring certain values of (L_x).

• The commutator of (Lx) and (Ly) is:

  • [Lx, Ly] = Lx Ly - Ly Lx = i\hbar L_z

• This is similar to the commutator of (x) and (px), which is [x, px] = -i\hbar

  • This is relevant to the Heisenberg uncertainty principle.

• So that the size of the product of the spreads depends on which state you're in.

Eigenfunctions of Angular Momentum

• We seek an eigenfunction of just the \theta and \phi part of the Hamiltonian.

• This involves solving the eigenvalue problem:
  \hat{L}^2 \, f(\theta, \phi) = constant \cdot f(\theta, \phi)

• If we solve the eigenvalue problem, then the values that can be found when you make a measurement are these eigenvalues.

• Trying (f(\theta) = \cos(\theta))

Pugging into the original equation:

\frac{-1}{r^2 sin(\theta)} \frac{\partial}{\partial \theta} sin(\theta) \frac{\partial}{\partial \theta} f + \frac{1}{r^2 sin^2(\theta)} \frac{\partial ^2}{\partial \phi^2} f = constant \cdot f

We try plugging in cosine theta for the eigenfunction:

\frac{-1}{r^2 sin(\theta)} \frac{\partial}{\partial \theta} sin(\theta) \frac{\partial}{\partial \theta} cos(\theta) + \frac{1}{r^2 sin^2(\theta)} \frac{\partial ^2}{\partial \phi^2} cos(\theta) = constant \cdot cos(\theta)

• This gives:
  • \L^2 = 2\hbar^2
• In general, the magnitude of (L^2) is quantized:
  • \L^2 = \hbar^2 l(l+1), where (l = 0, 1, 2, 3…)
• The allowed values of (L^2) are:
  • \l = 0: L^2 = 0
  • \l = 1: L^2 = 2\hbar^2
  • \l = 2: L^2 = 6\hbar^2
  • \l = 3: L^2 = 12\hbar^2
• The angular momentum is quantized for all problems and the eigenvalue of the problem.

Handout Summary

• The handout summarizes what we have learned today.
• There are different solutions which are combinations of theta and phi.
• It included the operators for angular momentum.
• It gives the forms of (Lx), (Ly), and (L_z).
• Plus the one for the (L^2), the square of the angular momentum operator to the line on the page.
• It has h bar squared in it, since h bar is a unit of angular momentum.
• It gives all the different solutions for the different l values. So it has h bar squared.
• Values of angular momentum are also quantized.