Quantum Mechanics - Eigenfunctions and Expectation Values

Eigenfunctions and Eigenvalues

Solving the eigenvalue problem yields eigenfunctions, each associated with a specific eigenvalue.
For example, \phi{a3} is the eigenfunction for the eigenvalue a_3, which is a particular value of the operator A. This operator is associated with a measurable quantity.

Eigenfunctions corresponding to different eigenvalues are orthogonal.
Analogy: Similar to unit vectors being perpendicular to each other.
Mathematically, the integral of the product of two different eigenfunctions over all space is zero:
\int \phi{ai}^* \phi{aj} dx = 0 \quad \text{if} \quad ai \neq aj
If the eigenfunctions are the same (i.e., same eigenvalue), the integral is one (normalization condition):
\int \phi{ai}^* \phi{ai} dx = 1
This is analogous to \hat{x} \cdot \hat{x} = 1 for unit vectors.
The term we use is "orthogonal".

If the wave function \Psi is one of the eigenfunctions of the operator A, then there's a 100% chance of measuring the corresponding eigenvalue.
If \Psi = \phi{a5}, then the probability of measuring a_3 is zero.
\int \phi{a3}^* \phi{a5} dx = 0
This implies that if your wave function is \phi{a5}, the probability of measuring any eigenvalue other than a_5 is zero.

Eigenfunctions are orthogonal for any operator, not just the momentum operator.
\phi{k2} and \phi{k3} represent two momentum eigenfunctions with different eigenvalues.
\int \phi{k2}^* \phi{k3} dx = 0
\int \phi{k3}^* \phi{k3} dx = 1
The eigenfunctions of the momentum operator are of the form e^{ikx}, which remains consistent across different problems.
Only the potential function changes.* The position and momentum operators remain the same across problems, but the Hamiltonian changes because it depends on the potential function V(x).
Examples of potential functions:
- Coulomb potential: V(r) = k \frac{q1 q2}{r}
- Particle in a box.

The average value of an operator A is found by "sandwiching" the operator between the complex conjugate of the wave function and the wave function:
\langle A \rangle = \int \Psi^* A \Psi dx
This gives the expectation value, which represents the average value obtained from a large number of identical experiments.
For the position operator X, the average value is:
\langle X \rangle = \int \Psi^* X \Psi dx = \int \Psi^* x \Psi dx
Where the position operator simply multiplies the function by x.
The wave function can always be expanded into a linear combination of the eigenfunctions of an operator:
\Psi = c1 \phi{a1} + c2 \phi{a2} + c3 \phi{a_3} + \dots
When the operator A acts on this wave function, it acts on each eigenfunction:
A \Psi = c1 a1 \phi{a1} + c2 a2 \phi{a2} + c3 a3 \phi{a3} + \dots

Start with the expansion of the wave function:
\Psi = c1 \phi{a1} + c2 \phi{a2} + c3 \phi{a_3}
Then, \Psi^* = c1^* \phi{a1}^* + c2^* \phi{a2}^* + c3^* \phi{a_3}^*
Calculate \int \Psi^* A \Psi dx:
\int (c1^* \phi{a1}^* + c2^* \phi{a2}^* + c3^* \phi{a3}^*) A (c1 \phi{a1} + c2 \phi{a2} + c3 \phi{a3}) dx
This results in nine terms (like multiplying two trinomials).
Three terms involve the same eigenfunction and its complex conjugate, integrating to one:
\int \phi{ai}^* \phi{ai} dx = 1
The other six terms involve different eigenfunctions and their complex conjugates, integrating to zero (orthogonality).
Result:
\langle A \rangle = |c1|^2 a1 + |c2|^2 a2 + |c3|^2 a3 + \dots
Where |ci|^2 represents the probability of measuring the eigenvalue ai. This is the probability of measuring a1 times a1, plus probability of measuring a2 times a2, etc.

When a measurement of an observable is made, the wave function "collapses" into the eigenfunction associated with the measured eigenvalue.
The act of measurement inevitably alters the state of the particle, but ideally the measurement is designed to minimize such change.

Suppose the wave function is in a state:
\Psi = \sqrt{\frac{1}{7}} \phi{a1} + \sqrt{\frac{2}{7}} \phi{a2} + \sqrt{\frac{4}{7}} \phi{a3}
The probability of measuring a_2 is \frac{2}{7} (approximately 30%).
If we measure a2, the wave function collapses to: \Psi{\text{after}} = \phi{a2}
After the measurement, a subsequent measurement of A will yield a_2 with 100% probability.
If we then decide to measure the the B operator, that can written as
\phi{a2} = d1 \phi{b1} + d2 \phi{b2} + d3 \phi{b_3}
This shows how we lock it to the state of the the A operator before switching over the the B operator.

Consider a particle trapped between x = 0 and x = L in an infinite potential well.

If a measurement of the energy yields E3, the wave function collapses to: \Psi(x) = \phi{E_3} = \sqrt{\frac{2}{L}} \sin\left(\frac{3 \pi x}{L}\right)

If we then measure the position of the particle, we will obtain a distribution of possible values.
If a particular measurement yields x = \frac{3}{4}L, the wave function collapses to a delta function at that position:
\Psi(x) = \delta(x - \frac{3}{4}L)

If instead after the energy measurement, we measure the momentum p_x of the particle, the wave function will collapse again.
However, the eigenfunction of the momentum operator, e^{ikx}, has infinite extent and does not satisfy the boundary conditions of the infinite potential well.
The collapse doesn't respect the boundaries of the of the particle.
Therefore, the wave function collapses into a wave packet, representing a distribution of momentum values, consistent with the particle remaining within the box.

The standard deviation (or spread) of an operator A is defined as:
\sigma_A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2}
Where \langle A^2 \rangle = \int \Psi^* A^2 \Psi dx is the average value of the operator squared, and \langle A \rangle^2 is the square of the average value of the operator.
The average value of A^2 is not the same as the square of the average value of A.

Suppose we make four temperature measurements: T1 = 20^\circ C, T2 = 30^\circ C, T3 = 40^\circ C, T4 = 50^\circ C.
To calculate the standard deviation:
1. Calculate the average of T^2: {\langle T^2 \rangle = \frac{20^2 + 30^2 + 40^2 + 50^2}{4} = \frac{5400}{4} = 1350.
2. Calculate the average of T: \langle T \rangle = \frac{20 + 30 + 40 + 50}{4} = \frac{140}{4} = 35.

To find the spread of position values after collapse into E_3:
1. Calculate \langle X \rangle = \int0^L \Psi^* x \Psi dx = \int0^L \sqrt{\frac{2}{L}} \sin\left(\frac{3 \pi x}{L}\right) x \sqrt{\frac{2}{L}} \sin\left(\frac{3 \pi x}{L}\right) dx = \frac{L}{2}.
2. Calculate \langle X^2 \rangle = \int0^L \Psi^* x^2 \Psi dx = \int0^L \sqrt{\frac{2}{L}} \sin\left(\frac{3 \pi x}{L}\right) x^2 \sqrt{\frac{2}{L}} \sin\left(\frac{3 \pi x}{L}\right) dx
The standard deviation is then calculated using the formula: \sigma_X = \sqrt{\langle X^2 \rangle - \langle X \rangle^2}$$