Eigenvalues & Spectral Theory
The spectrum of observables and measurement outcomes
3.1 Eigenvalue Problem
The eigenvalue equation is central to quantum mechanics:
where $a$ is the eigenvalue and $|a\rangle$ is the corresponding eigenvector (or eigenstate).
Physical Interpretation:
When we measure an observable $\hat{A}$, we get one of its eigenvalues $a$. If the system is in eigenstate $|a\rangle$, the measurement always yields $a$ with certainty!
Discrete Spectrum
For bound systems (e.g., electron in atom), eigenvalues are discrete:
Continuous Spectrum
For unbound systems (e.g., free particle), eigenvalues form a continuum:
3.2 Spectral Theorem
For a Hermitian operator $\hat{A}$, eigenvectors corresponding to different eigenvalues are orthogonal:
Completeness Relation
For discrete spectrum, eigenstates form a complete orthonormal basis:
Any state can be expanded:
where $c_n = \langle n|\psi\rangle$ are probability amplitudes and $|c_n|^2$ is the probability of finding the system in state $|n\rangle$.
3.3 Example: Harmonic Oscillator
The quantum harmonic oscillator has Hamiltonian:
Eigenvalues (energy levels) are:
Note the zero-point energy $E_0 = \frac{1}{2}\hbar\omega$ even for the ground state! This is a purely quantum effect with no classical analog.
Key Concepts
- • Eigenvalues of Hermitian operators are measurable values
- • Eigenstates of different eigenvalues are orthogonal
- • Completeness: any state expands in eigenbasis
- • $|c_n|^2 = |\langle n|\psi\rangle|^2$ gives measurement probabilities
- • Discrete spectrum: bound states; Continuous: unbound states
Practice Problems
Problem 1:Find the eigenvalues and normalized eigenstates of the Pauli matrix $\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.
Solution:
Step 1: Set up the characteristic equation $\det(\sigma_z - \lambda I) = 0$:
$\det\begin{pmatrix} 1-\lambda & 0 \\ 0 & -1-\lambda \end{pmatrix} = (1-\lambda)(-1-\lambda) = 0$
Step 2: Solve: $-1 + \lambda^2 = 0$, giving $\lambda = \pm 1$.
Step 3: For $\lambda = +1$: $(\sigma_z - I)|\psi\rangle = 0$ gives $\begin{pmatrix} 0 & 0 \\ 0 & -2 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = 0$, so $b = 0$ and $|+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
Step 4: For $\lambda = -1$: $(\sigma_z + I)|\psi\rangle = 0$ gives $\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = 0$, so $a = 0$ and $|-\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.
Answer: Eigenvalues $\lambda = +1, -1$ with normalized eigenstates $|+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, $|-\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.
Problem 2:Prove that the eigenvalues of a Hermitian operator are real.
Solution:
Step 1: Let $\hat{A}$ be Hermitian ($\hat{A}^\dagger = \hat{A}$) with $\hat{A}|a\rangle = a|a\rangle$.
Step 2: Take the inner product: $\langle a|\hat{A}|a\rangle = a\langle a|a\rangle = a$ (assuming normalized).
Step 3: Take the conjugate: $\langle a|\hat{A}|a\rangle^* = \langle a|\hat{A}^\dagger|a\rangle = \langle a|\hat{A}|a\rangle$, so $a^* = a$.
Step 4: Since $a = a^*$, the eigenvalue $a$ must be real.
Answer: The Hermiticity condition $\hat{A}^\dagger = \hat{A}$ forces $a^* = a$, so all eigenvalues are real. This guarantees that measurement outcomes are real numbers.
Problem 3:A particle is in the state $|\psi\rangle = \frac{3}{5}|E_1\rangle + \frac{4}{5}|E_2\rangle$. Find the probability of measuring $E_1$ and the expectation value $\langle\hat{H}\rangle$.
Solution:
Step 1: Verify normalization: $|c_1|^2 + |c_2|^2 = (3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 1$. ✓
Step 2: Probability of measuring $E_1$: $P(E_1) = |c_1|^2 = (3/5)^2 = 9/25 = 0.36$.
Step 3: Expectation value: $\langle\hat{H}\rangle = |c_1|^2 E_1 + |c_2|^2 E_2 = \frac{9}{25}E_1 + \frac{16}{25}E_2$.
Answer: $P(E_1) = 9/25$ and $\langle\hat{H}\rangle = \frac{9E_1 + 16E_2}{25}$.
Problem 4:Evaluate the commutator $[\hat{x}, \hat{p}^2]$ using the canonical commutation relation $[\hat{x}, \hat{p}] = i\hbar$.
Solution:
Step 1: Use the identity $[\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{B}]\hat{C} + \hat{B}[\hat{A}, \hat{C}]$.
Step 2: Apply with $\hat{A} = \hat{x}$, $\hat{B} = \hat{C} = \hat{p}$:
$[\hat{x}, \hat{p}^2] = [\hat{x}, \hat{p}]\hat{p} + \hat{p}[\hat{x}, \hat{p}]$
Step 3: Substitute $[\hat{x}, \hat{p}] = i\hbar$:
$[\hat{x}, \hat{p}^2] = i\hbar\hat{p} + \hat{p}(i\hbar) = 2i\hbar\hat{p}$
Answer: $[\hat{x}, \hat{p}^2] = 2i\hbar\hat{p}$
Problem 5:Using $[\hat{x}, \hat{p}] = i\hbar$, derive the uncertainty relation $\Delta x\,\Delta p \geq \hbar/2$ for the harmonic oscillator ground state and show it is saturated.
Solution:
Step 1: The general uncertainty relation states $\Delta A\,\Delta B \geq \frac{1}{2}|\langle[\hat{A},\hat{B}]\rangle|$. For $\hat{x}$ and $\hat{p}$: $\Delta x\,\Delta p \geq \frac{1}{2}|i\hbar| = \frac{\hbar}{2}$.
Step 2: For the ground state $|0\rangle$ of the harmonic oscillator, $\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a} + \hat{a}^\dagger)$ and $\hat{p} = i\sqrt{\frac{m\omega\hbar}{2}}(\hat{a}^\dagger - \hat{a})$.
Step 3: Compute $\langle 0|\hat{x}^2|0\rangle = \frac{\hbar}{2m\omega}$ and $\langle 0|\hat{p}^2|0\rangle = \frac{m\omega\hbar}{2}$, with $\langle\hat{x}\rangle = \langle\hat{p}\rangle = 0$.
Step 4: Therefore $\Delta x = \sqrt{\frac{\hbar}{2m\omega}}$ and $\Delta p = \sqrt{\frac{m\omega\hbar}{2}}$.
Answer: $\Delta x\,\Delta p = \sqrt{\frac{\hbar}{2m\omega}} \cdot \sqrt{\frac{m\omega\hbar}{2}} = \frac{\hbar}{2}$. The ground state saturates the uncertainty bound, making it a minimum-uncertainty state.