Unit 2: Uncertainty Relation and Properties of Wave Function

1. Heisenberg’s Uncertainty Relation and Applications

1.1 Mathematical Statement and Physical Meaning

Formulated by Werner Heisenberg in 1927, the Uncertainty Principle is a fundamental limit in quantum mechanics. It states that it is impossible to simultaneously measure both the exact position and exact momentum of a subatomic particle with absolute precision.

Heisenberg Uncertainty Principle:
Δx × Δp ≥ ℏ / 2 = h / 4π

Where:
• Δx = Uncertainty in position
• Δp = Uncertainty in momentum
• h = Planck’s constant (6.626 × 10^-34 J·s)
• ℏ = Reduced Planck’s constant = h / 2π = 1.054 × 10^-34 J·s

Other Forms of Uncertainty Relations:

• Energy-Time Uncertainty: ΔE × Δt ≥ ℏ / 2
• Angular Position-Angular Momentum Uncertainty: ΔL × Δθ ≥ ℏ / 2

Important Observation: The uncertainty relation is not a result of experimental inaccuracy or equipment limitations; it is an intrinsic property of wave-particle duality. A localized wave packet requires a superposition of many wavelengths, leading to spread in momentum.

1.2 Application A: Non-Existence of Electrons Within the Nucleus

We can use Heisenberg's Uncertainty Principle to prove that electrons cannot exist inside an atomic nucleus.

Step-by-step Proof:

1. Define Nuclear Size: The radius of an atomic nucleus is approximately r ≈ 10^-14 m. If an electron were present inside the nucleus, the maximum uncertainty in its position would be bounded by the nuclear diameter: Δx ≤ 2 × 10^-14 m.
2. Calculate Uncertainty in Momentum: According to the uncertainty principle:
   Δp ≥ ℏ / (2 Δx) = h / (4π Δx)
   Δp ≥ (6.626 × 10^-34) / (4 × 3.1416 × 2 × 10^-14) ≈ 2.63 × 10^-21 kg·m/s
3. Calculate Minimum Momentum and Energy: Since Δp represents uncertainty, the particle's momentum p must be at least of the order of Δp:
   p ≈ 2.63 × 10^-21 kg·m/s
   
   Because this momentum is extremely high, we must use the relativistic energy-momentum relation: E ≈ p × c (since kinetic energy dominates rest mass energy):
   E ≈ (2.63 × 10^-21 kg·m/s) × (3 × 10⁸ m/s) = 7.89 × 10^-13 J
   
   Converting Joules to Electron-Volts (1 eV = 1.6 × 10^-19 J):
   E ≈ (7.89 × 10^-13) / (1.6 × 10^-19) ≈ 20 MeV
4. Compare with Experimental Evidence: Experimental measurements during beta decay show that electrons emitted from radioactive nuclei have kinetic energies of only around 2 to 4 MeV. No electron bound within a nucleus with energy ~20 MeV or higher has ever been observed.

Conclusion: Because an electron confined to a nuclear dimension would require a minimum energy (~20 MeV) far exceeding observed atomic nuclear energies, electrons cannot reside inside the nucleus.

1.3 Application B: Radius and Ground State Energy of Hydrogen Atom

We can estimate the Bohr radius and ground state energy of a hydrogen atom using the uncertainty principle without solving Schrödinger's differential equation.

Step-by-step Derivation:

1. Position and Momentum Approximations: Let the average radius of the hydrogen atom be r. The uncertainty in position of the electron is Δx ≈ r. Therefore, uncertainty in momentum is Δp ≈ ℏ / r. We assume minimum momentum p ≈ Δp = ℏ / r.
2. Total Energy Expression: The total energy E of the electron is the sum of kinetic energy (K) and potential energy (V):
   E = K + V = p² / (2m) - e² / (4π ε₀ r)
   
   Substituting p ≈ ℏ / r:
   E = ℏ² / (2 m r²) - e² / (4π ε₀ r)
3. Finding Minimum Energy (Ground State): For ground state stability, energy E must be a minimum with respect to r. Setting dE/dr = 0:
   dE/dr = - ℏ² / (m r³) + e² / (4π ε₀ r²) = 0
   
   Solving for radius r (denoted as r₀, Bohr radius):
   ℏ² / (m r₀³) = e² / (4π ε₀ r₀²)
   r₀ = (4π ε₀ ℏ²) / (m e²) ≈ 0.53 Å (0.053 nm)
4. Calculate Ground State Energy: Substituting r₀ back into total energy expression:
   E₀ = ℏ² / [2 m (4π ε₀ ℏ² / m e²)²] - e² / [4π ε₀ (4π ε₀ ℏ² / m e²)]
   E₀ = - m e⁴ / (32 π² ε₀² ℏ²) ≈ -13.6 eV

Result: Ground state radius r₀ ≈ 0.53 Å and ground state energy E₀ = -13.6 eV.

2. Fundamental Postulates of Quantum Mechanics

Quantum mechanics is founded upon standard physical axioms known as postulates:

• Postulate 1 (State of a System): The physical state of any quantum system is completely specified by a wave function Ψ(r, t), which contains all accessible physical information about the system.
• Postulate 2 (Quantum Operators): Every physically observable quantity (e.g., position, momentum, energy) corresponds to a linear Hermitian operator in quantum mechanics.
  • Position Operator: x̂ = x
  • Linear Momentum Operator: p̂_x = -i ℏ (∂ / ∂x)
  • Total Energy (Hamiltonian) Operator: Ĥ = - (ℏ² / 2m) ∇² + V(r)
• Postulate 3 (Eigenvalues and Measurement): The only possible value obtained when measuring a physical observable A is one of the eigenvalues 'a' of the corresponding operator Â, defined by the eigenvalue equation: Â Ψ = a Ψ.
• Postulate 4 (Expectation Values): The average or expectation value ⟨A⟩ of an observable corresponding to operator  for a system in state Ψ is given by:
  ⟨A⟩ = ∫_-∞^∞ Ψ* Â Ψ dτ / ∫_-∞^∞ Ψ* Ψ dτ
• Postulate 5 (Time Evolution): The time development of a quantum system is governed by the Time-Dependent Schrödinger Equation:
  i ℏ (∂Ψ / ∂t) = Ĥ Ψ

3. Physical Interpretation and Boundary Conditions of Wave Functions

3.1 Born Interpretation of the Wave Function

In 1926, Max Born proposed the statistical probability interpretation of the wave function Ψ(x, y, z, t).

Born Probability Interpretation:
The wave function Ψ itself has no direct physical magnitude. However, the quantity |Ψ|² = Ψ* Ψ represents the probability density P(x, y, z, t).

The probability dP of finding the particle within an infinitesimal volume element dV = dx dy dz around position (x, y, z) at time t is:
dP = |Ψ|² dV = Ψ* Ψ dV

Where Ψ* is the complex conjugate of Ψ.

3.2 Boundary Conditions for Acceptable Wave Functions

For a wave function Ψ to represent a realistic, physically meaningful quantum state (termed a well-behaved wave function), it must satisfy four key boundary conditions:

1. Ψ must be single-valued: At any point in space and time, Ψ(x,t) must have only one value, preventing multiple simultaneous probability values for a single point.
2. Ψ must be continuous: The wave function Ψ(x,t) and its spatial first derivatives (∂Ψ/∂x, ∂Ψ/∂y, ∂Ψ/∂z) must be continuous everywhere across space boundaries (except where potential energy V becomes infinite).
3. Ψ must be finite everywhere: Ψ cannot become infinite at any point, as an infinite value would imply infinite probability density.
4. Ψ must be square-integrable: The integral of |Ψ|² over all space must be finite. As x → ±∞, Ψ must approach zero (Ψ → 0 as |x| → ∞).

4. Normalization and Orthogonality of Wave Functions

4.1 Normalization Condition

Since the total probability of finding a particle somewhere in all space must be 100% (or 1), any physically admissible wave function must satisfy the normalization condition.

Normalization Condition:
∫_-∞^∞ |Ψ(r, t)|² dV = ∫_-∞^∞ Ψ* Ψ dV = 1

If a wave function Ψ is unnormalized such that ∫ Ψ* Ψ dV = N ≠ 1, it can be normalized by multiplying by a constant normalization factor C = 1 / √N, so that Ψ_norm = C Ψ.

4.2 Orthogonality Condition

Two distinct wave functions Ψ_m and Ψ_n corresponding to two different non-degenerate energy eigenstates are orthogonal if their inner product over all space vanishes.

Orthogonality Condition:
∫_-∞^∞ Ψ_m* Ψ_n dV = 0 (for m ≠ n)

Orthonormality Condition: When wave functions are both normalized and orthogonal, they satisfy the unified orthonormality relation:

∫_-∞^∞ Ψ_m* Ψ_n dV = δ_mn

Where δ_mn is the Kronecker delta defined as:
• δ_mn = 1 if m = n
• δ_mn = 0 if m ≠ n

5. Probability Current Density and Equation of Continuity

5.1 Probability Density

Probability density P(r, t) represents the probability per unit volume of finding the particle at position r at time t:

P(r, t) = Ψ*(r, t) Ψ(r, t) = |Ψ(r, t)|²

5.2 Probability Current Density

Probability current density J represents the rate of flow of probability per unit area per unit time, directly analogous to electric current density in electromagnetism or mass flux in fluid mechanics.

Probability Current Density Formula:
J = (i ℏ / 2m) (Ψ ∇Ψ* - Ψ* ∇Ψ) = (ℏ / 2m i) (Ψ* ∇Ψ - Ψ ∇Ψ*)

In one dimension (1D):
J_x = (ℏ / 2m i) [ Ψ* (∂Ψ / ∂x) - Ψ (∂Ψ* / ∂x) ]

5.3 Derivation of Equation of Continuity

The conservation of total probability is mathematically expressed by the continuity equation.

Step-by-step Derivation:

1. Start with Time-Dependent Schrödinger Equation (TDSE):
   i ℏ (∂Ψ / ∂t) = - (ℏ² / 2m) ∇²Ψ + V Ψ     --- (Equation 1)
2. Take the Complex Conjugate of Equation 1:
   (Assuming potential V is real: V* = V)
   -i ℏ (∂Ψ* / ∂t) = - (ℏ² / 2m) ∇²Ψ* + V Ψ*     --- (Equation 2)
3. Multiply and Subtract Equations:
   Multiply Equation 1 by Ψ* from the left:
   i ℏ Ψ* (∂Ψ / ∂t) = - (ℏ² / 2m) Ψ* ∇²Ψ + V Ψ* Ψ     --- (Equation 3)
   
   Multiply Equation 2 by Ψ from the left:
   -i ℏ Ψ (∂Ψ* / ∂t) = - (ℏ² / 2m) Ψ ∇²Ψ* + V Ψ Ψ*     --- (Equation 4)
   
   Subtract Equation 4 from Equation 3:
   i ℏ [ Ψ* (∂Ψ / ∂t) + Ψ (∂Ψ* / ∂t) ] = - (ℏ² / 2m) [ Ψ* ∇²Ψ - Ψ ∇²Ψ* ]
4. Simplify Terms:
   The left side can be written as:
   i ℏ ∂(Ψ* Ψ) / ∂t = i ℏ (∂P / ∂t)
   
   The right side vector identity:
   Ψ* ∇²Ψ - Ψ ∇²Ψ* = ∇ · [ Ψ* ∇Ψ - Ψ ∇Ψ* ]
   
   Therefore:
   i ℏ (∂P / ∂t) = - (ℏ² / 2m) ∇ · [ Ψ* ∇Ψ - Ψ ∇Ψ* ]
5. Rearrange into Continuity Form:
   Divide both sides by i ℏ:
   ∂P / ∂t = - ∇ · [ (ℏ / 2m i) (Ψ* ∇Ψ - Ψ ∇Ψ*) ]
   
   Define J = (ℏ / 2m i) (Ψ* ∇Ψ - Ψ ∇Ψ*):
   ∂P / ∂t = - ∇ · J

Equation of Continuity:
∂P / ∂t + ∇ · J = 0

Physical Meaning: Any decrease in probability density inside a given volume element over time must be equal to the net outward flux of probability current density across its boundary surface. Probability is locally conserved.

6. Concept and Formula Summary Table