6.13
Exam Expectations
For each of the nine experiments, expect one or two conceptual physics questions.
Mini experiments may be included.
Pre-prepared data may be provided for data analysis.
Required Skills
Skills learned in physics 4A to 4C lab classes are relevant.
Know how to do a \pm b squared split.
Proficiency in propagation of error.
Familiarity with lab skills outlined in a provided file (to be sent this weekend).
Includes standard deviation calculations.
Lab Final Details
Duration: 2.5 hours.
Involves analyzing data from different experiments.
Physical apparatus component: Each student gets 5-10 minutes on setups like ESR (Electron Spin Resonance).
Time allocation accounts for cycling through all students.
Differential Equations and Solutions
Order differential equations have two independent, orthogonal solutions.
Solutions with k \cdot l = 0 are independent.
Physicality of Solutions
Solutions that go to infinity may not be physical, even if they satisfy the differential equation.
Example: Particle in a finite box, where e^{\alpha x} was rejected for positive x because it blows up.
Accepted solutions must be normalizable, providing a definite value at specific coordinates.
Normative functions are generally rejected as non-physical because they are not defined at the origin.
Potentials and Solutions
Normative functions can be helpful when the particle is not at the origin.
For a potential where the wave function equals zero at the origin (e.g., particle trapped in a spherical shell):
The n = 0 solution's probability density does not blow up.
Effective Potential
Representation:
\frac{-\hbar^2}{2m} \frac{d^2}{dr^2} + V(r)
Can be thought of as a one-dimensional problem with an effective potential.
Free Particle Example
For a free particle, V = 0, so the effective potential is 1/r^2.
Analogy: One-dimensional problem with potential equal to some constant over x^2.
Classical Mechanics Connection
If a particle has non-zero angular momentum, it can't be found at the origin.
If the particle were at the origin, r \times p = 0, violating conservation of momentum.
For l \neq 0, the probability density must go to zero as r approaches zero.
For l = 0, the effective potential is equal to the real potential.
Bessel Functions
Goal: Show that the J_0 function gives a nonzero probability density at the origin, while other Bessel functions give zero.
Approximations
Cosine approximation for small angles:
\cos(\theta) \approx 1 - \frac{\theta^2}{2} + \text{higher order terms}
Taylor expansion used for improved approximation: 1 - \alpha^2 r^2 / r
Probability Density
Probability calculation:
|\psi|^2 r^2 dr
Objective: Show probability density at origin goes to zero, as in classical mechanics.
Small r Behavior
For small r (small \theta), sine function term:
\frac{\alpha r}{\alpha r} \approx 1
Terms:
\frac{1}{\alpha r}
\frac{-\theta^2}{2} = \frac{-\alpha^2 r^2}{\alpha r}
The \theta^2 term cannot be ignored.
Resulting term: \alpha \cdot r
Radial part of the wave function approaches zero as r goes to zero.
Justification for r²
Volume between r and r + dr is proportional to r^2.
Larger values of r have an "unfair advantage" due to volume.
In Cartesian coordinates, each interval dx occupies same amount of space.
In spherical coordinates, each interval dr occupies more volume for larger r.
Expect uniform probability density for small values of r.
J₀ Solution
Examine e^{ikx} + e^{-ikx} (proportional to cosine) and e^{ikx} - e^{-ikx}.
These solutions resemble a particle emanating from the origin.
Analogy to a point source of light, like a candle or the sun, emitting uniformly in all directions.
Free Particle Solution
Free particle solution in x, y, and z:
c1 e^{ikx x} e^{iky y} e^{ikz z}
Probability Density of J₀
Probability density is not uniform, but a cosine squared function.
Squaring J_0 and multiplying by the r^2 geometric factor yields \cos^2(\alpha r), which is not uniform, but has a uniform average value.
Average value is possibly uniform.