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:
  • $KE = \frac{p<em>{radial}^2}{2m} + \frac{p</em>{tangential}^2}{2m}$
  • 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 $(L</em>x)$.
• 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 $(L<em>x)$, $(L</em>y)$, and $(L_z)$.

  • This means they share eigenfunctions.

• However, $(L<em>x)$ and $(L</em>y)$ do not commute.

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

• $\sigma<em>{Lx} \sigma</em>{Ly} \geq \frac{1}{2} \hbar |<[L<em>x, L</em>y]>|$

  • 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 $(L<em>x)$, measurements of $(L</em>y)$ and $(L_z)$ will yield a distribution of values.

• A state that is an eigenfunction of $(L^2)$ and $(L<em>y)$ is a linear combination of eigenstates in $(L</em>x)$ and $(L_z)$.

• An eigenstate for $(L<em>z)$ is a linear combination of eigenstates for $(L</em>x)$.
  $\psi<em>{L</em>z} = c<em>1 \psi</em>{L<em>{x1}} + c</em>2 \psi<em>{L</em>{x2}} + c<em>3 \psi</em>{L_{x3}} + …$

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

• The commutator of $(L<em>x)$ and $(L</em>y)$ is:

  • $[L<em>x, L</em>y] = L<em>x L</em>y - L<em>y L</em>x = i\hbar L_z$

• This is similar to the commutator of $(x)$ and $(p<em>x)$, which is $[x, p</em>x] = -i\hbar$

  • 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 $(L<em>x)$, $(L</em>y)$, 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.