← Quantum Mechanics
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Topic 1 of 8

Wave Functions & Probability

In quantum mechanics, the state of a particle is described by a wave function ψ(x,t). The probability of finding the particle between x and x+dx is |ψ|²dx. This Born interpretation is the foundation of all quantum predictions.

Key Concepts

  • Wave function ψ(x,t) encodes all information about a quantum state
  • Probability density: P(x) = |ψ(x,t)|²
  • Normalization: ∫|ψ|²dx = 1 over all space
  • Expectation value: ⟨x⟩ = ∫x|ψ|²dx
  • Wave functions must be continuous, normalizable, and single-valued

Key Equations

Probability density
P(x,t)=∣ψ(x,t)∣2P(x,t) = |\psi(x,t)|^2
Normalization
∫−∞∞∣ψ∣2 dx=1\int_{-\infty}^{\infty} |\psi|^2\,dx = 1
Expectation value of x
⟨x⟩=∫−∞∞x∣ψ∣2 dx\langle x \rangle = \int_{-\infty}^{\infty} x|\psi|^2\,dx
Expectation of operator
⟨Q⟩=∫ψ∗Q^ψ dx\langle Q \rangle = \int \psi^* \hat{Q} \psi\,dx
Worked Example

Example Problem

Problem

A particle has wave function ψ(x) = A e^{-x²/a²} for a = 2.0 nm. Find A such that ψ is normalized.

Solution

Normalization: A² ∫e^{-2x²/a²}dx = 1. The Gaussian integral gives √(πa²/2). So A² = √(2/πa²) = √(2/(π×4×10⁻¹⁸)). A = (2/πa²)^(1/4) = (2/(π×4nm²))^(1/4) ≈ (0.159/nm²)^(1/4) ≈ 0.632 nm^(-1/2).

Practice

Exercises

20 problems
1 of 20

A particle has wave function ψ(x) = A for 0 ≤ x ≤ L = 3.0 nm. Drag the slider to find the normalization constant A so that ∫₀ᴸ |ψ|² dx = 1.

∫|ψ|²dx =
0.270 / 1.000
A = 0.3000 nm⁻¹/²
2 of 20

For the normalized box wave function (L = 3.0 nm), drag the boundary to find the probability of the particle being in 0 ≤ x ≤ 1.0 nm.

Drag the yellow handle to set the right boundary at 1.0 nm, then submit the probability.

P(0 → 1.50 nm) = 0.5000
3 of 20

A particle is in state ψ(x) = A sin(πx/L) for 0≤x≤L. Find A in terms of L. For L=1.0 nm, A in nm^(-1/2) is:

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4 of 20

For ψ = √(2/L) sin(πx/L) with L=1.0 nm, find ⟨x⟩ in nm.

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5 of 20

A wave function is ψ(x) = A e^{-|x|/a} with a=1.0 nm. Find A in nm^(-1/2).

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6 of 20

For ψ(x) = (1/√a)e^{-|x|/a} with a=1.0 nm, find ⟨x²⟩ in nm².

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7 of 20

A particle has P(x<0) = 0.3. What is P(x≥0)?

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8 of 20

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9 of 20

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10 of 20

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11 of 20

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12 of 20

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13 of 20

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14 of 20

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15 of 20

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16 of 20

Find the normalization constant A>0A > 0 for ψ(x)=Ax(1−x)\psi(x) = Ax(1-x) defined on [0,1][0,1] (and zero elsewhere). Use the equation editor below to enter your exact answer.

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17 of 20

For ψ1(x)=2/Lsin⁡(πx/L)\psi_1(x) = \sqrt{2/L}\sin(\pi x/L) on [0,L][0,L], compute the expectation value ⟨x⟩\langle x \rangle. Enter your answer as a multiple of LL (use LL as a symbol; we set L=1L=1 to check).

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18 of 20

For any real-valued normalizable wave function ψ(x)\psi(x), show that ⟨p⟩=0\langle p \rangle = 0. Enter the numerical value of ⟨p⟩\langle p \rangle.

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19 of 20

For the ground state of the infinite square well ψ1=2/Lsin⁡(πx/L)\psi_1 = \sqrt{2/L}\sin(\pi x/L) on [0,L][0,L], compute ⟨x2⟩\langle x^2 \rangle. Enter your answer (with L=1L=1, so enter a pure number).

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20 of 20

For the state ∣ψ⟩=12(∣0⟩+∣1⟩)|\psi\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle), the density matrix is ρ=∣ψ⟩⟨ψ∣\rho = |\psi\rangle\langle\psi|. Compute the off-diagonal element ρ01=⟨0∣ρ∣1⟩\rho_{01} = \langle 0|\rho|1\rangle. Enter the numerical value.

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Key Takeaways

  • The wave function ψ contains all physical information about a quantum state
  • Probability is found from the modulus squared |ψ|²
  • Physical wave functions must be normalized to unit total probability
  • Expectation values are averages weighted by the probability density