Quantum Mechanics - Commutation Relationship and Angular Momentum

Kinetic Energy Operator

The kinetic energy operator in Cartesian coordinates (x, y, z) is given by $-\frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right)$ .
This operator is then converted into spherical coordinates (r, θ, φ).
The conversion results in terms involving r and terms involving θ and φ.

Spherical Coordinates

The kinetic energy operator in spherical coordinates can be separated into radial and tangential components.
The tangential kinetic energy is identified as $\frac{L^2}{2mr^2}$ , where $L^2$ is the angular momentum operator.
The radial kinetic energy is related to the radial momentum squared over $2m$ .

Angular Momentum Operator

The angular momentum operator $L^2$ can be derived by comparing the kinetic energy operator in Cartesian and spherical coordinates.

Commutation Relationship

The LX, LY, and LZ operators represent the components of angular momentum along the x, y, and z axes, respectively.
The LX operator does not commute with the LY operator, meaning you cannot simultaneously specify two components of angular momentum.
Measuring one component (e.g., LX) scrambles the others (LY and LZ), resulting in a distribution of possible values.
Only one component of angular momentum can be precisely specified at a time in quantum mechanics.

Magnitude and Z-Component

It is possible to know the total magnitude of the angular momentum vector (L) and one component, typically the z-component (LZ), simultaneously.
The x and y components (LX and LY) will have a distribution of values.

Visual Representation

Angular momentum vector can be visualized as a cone where the magnitude (L) and the z-component (LZ) are known.
The x and y components are distributed around the base of the cone.
The relationship between the components is given by $Lx^2 + Ly^2 = L^2 - L_z^2$ , derived from the Pythagorean theorem.

Importance of Commutation Relations

Commutation relations impose restrictions on what can be measured simultaneously in nature.
This is analogous to the Heisenberg Uncertainty Principle, which restricts simultaneous measurements of position and momentum.

Proof of Commutation Relation

LX is defined as $Lx = ry pz - rz py$ , and LY is defined as $Ly = rz px - rx pz$ .
The commutator $[Lx, Ly]$ is calculated as $Lx Ly - Ly Lx$ .
Expanding the commutator results in eight terms, which are then simplified.

Simplification of Terms

Terms are simplified using the commutation relations between position and momentum operators.
For example, $[x, py] = 0$ and $[x, px] = i\hbar$ .
After simplification, the commutator $[Lx, Ly]$ is shown to be equal to $i\hbar L_z$ .

Uncertainty Principle for Angular Momentum

The uncertainty in measuring LX and LY is related to the value of LZ by $\Delta Lx \Delta Ly \geq \frac{1}{2} |\langle i\hbar L_z \rangle|$ .
This means that the product of the standard deviations of LX and LY must be greater than or equal to half the magnitude of the average value of LZ.
As LZ increases, the uncertainties in LX and LY also increase, making the circle bigger.

Observations

None of the components of angular momentum can be greater than the magnitude of L.

Eigenfunctions and Eigenvalues

The $L^2$ operator has certain eigenfunctions, each with its own eigenvalue.
The case where l = 0 is allowed, representing a system with no angular momentum.

Particle in a Box Analogy

In the particle in a box problem, the n = 0 solution was rejected because it implied the particle had zero energy, which didn't make sense.
However, the l = 0 solution is allowed because it corresponds to a particle moving directly towards or away from the origin, which is physically plausible.

Magnitude of L Squared

The magnitude of $L^2$ is given by $L^2 = \hbar^2 l(l+1)$ , where l is an integer (0, 1, 2, …).

Eigenfunctions of LZ

The eigenvalue problem for the LZ operator is given by $Lz \Psi{lz} = constant \times \Psi_{lz}$ .
Solving this problem yields the possible values that can be measured.
The LZ operator is related to the partial derivative with respect to the azimuthal angle φ.

Boundary Conditions

The solution to the eigenvalue problem is $\Psi = e^{i \frac{L_z}{\hbar} \phi}$ .
The values that are allowed need to satisfy that $\Psi(\phi) = \Psi(\phi + 2\pi)$ . This implies that $e^{i \frac{Lz}{\hbar} \phi} = e^{i \frac{Lz}{\hbar} (\phi + 2\pi)}$ .
Which then can be reduced to, $1 = e^{i \frac{L_z}{\hbar} (2\pi)}$ .

Quantization

Because of the boundary condition requiring the wave function to return to the same value after a full rotation ( $\phi + 2\pi$ ), the values of LZ are quantized.
The allowed values of LZ are given by $L_z = m \hbar$ , where m is an integer.

Restrictions

The conditions lead to $L_z = m \hbar$ , where m can be any integer but must be less than l

Special Cases

If l = 0, then the magnitude of angular momentum is zero, and LZ must also be zero.
If l = 1, then the magnitude of angular momentum is $\sqrt{2} \hbar$ , and LZ can be 0, +1, or -1.

Angle Quantization

The angle between the angular momentum vector L and the z-axis is quantized.
For example, the cosine of the angle is equal to 1/1.4 \approx 45 degrees.

Eigenfunctions of L Squared and LZ

The discussion focuses on finding the eigenfunctions of the $L^2$ operator and then finding the eigenfunctions of $L_z$ .

Solving Problems

The next step is to solve a problem involving a certain potential is solving the eigenvalue problem for the full Hamiltonian.

Hamiltonian

To deal with the Hamiltonian, all three terms (r, theta, phi) must be accounted for to create one differential equation.

Commuting Operators

If two operators commute, they share eigenfunctions.
Since the Hamiltonian includes the $L^2$ operator, it means the eigenfunctions of the $L^2$ operator should be considered when solving the full equation.

Practice Material

Practice problems for quiz four have been provided, along with solutions.