Free Particle Model Trigonometry Practice Problems

The free particle model in quantum mechanics describes the behavior of a particle that is not subjected to any external forces or potentials. Understanding this model is crucial for grasping more complex quantum systems. To solidify your understanding, working through trigonometry practice problems related to the free particle model can be incredibly beneficial. Trigonometry plays a vital role in analyzing wave functions and probabilities associated with free particles, particularly when dealing with momentum and position representations.

The Essence of the Free Particle Model

The free particle model simplifies the quantum world by eliminating any interactions or constraints on the particle. This means the potential energy, V(x), is zero everywhere. Consequently, the time-independent Schrdinger equation reduces to a simpler form:

(-ħ²/2m) d²/dx² ψ(x) = E ψ(x)

where:

• ħ is the reduced Planck constant
• m is the mass of the particle
• ψ(x) is the wave function of the particle
• E is the energy of the particle

The solutions to this equation are plane waves, which can be expressed using trigonometric functions (sine and cosine) or complex exponentials. These solutions represent the probability amplitude of finding the particle at a particular position, x. The momentum of the particle, p, is related to its energy by E = p²/2m, a crucial relationship for solving related problems.

Why Trigonometry Matters

Trigonometry is fundamental in the free particle model for several reasons:

1. Representing Wave Functions: Plane waves, the solutions to the free particle Schrdinger equation, can be expressed using trigonometric functions: ψ(x) = A cos(kx) + B sin(kx), where k is the wave number related to the momentum by p = ħk, and A and B are constants.
2. Superposition: Quantum mechanics allows for the superposition of states. Combining multiple plane waves with different wave numbers results in more complex wave functions that can be analyzed using trigonometric identities and principles.
3. Normalization: Wave functions must be normalized, meaning the integral of the probability density (|ψ(x)|²) over all space must equal 1. Trigonometric integrals often arise during the normalization process.
4. Fourier Analysis: The momentum representation of a wave function can be obtained through Fourier analysis, which relies heavily on trigonometric functions and integrals.
5. Boundary Conditions: When the free particle is confined to a finite region (particle in a box), boundary conditions imposed on the wave function lead to quantized energy levels and trigonometric solutions that satisfy these conditions.

Trigonometry Practice Problems: Free Particle Model

Let's delve into a series of practice problems that combine the free particle model with trigonometry, enhancing your understanding and problem-solving skills.

Problem 1: Wave Function and Probability Density

A free particle of mass m has a wave function given by:

ψ(x) = A cos(kx)

where A is a normalization constant and k is the wave number.

a) Determine the probability density |ψ(x)|².

b) If the particle is confined to the region -L/2 ≤ x ≤ L/2, find the normalization constant A.

c) Calculate the probability of finding the particle in the region 0 ≤ x ≤ L/4.

Solution:

a) The probability density is the square of the absolute value of the wave function:

|ψ(x)|² = (A cos(kx))² = A² cos²(kx)

b) To find the normalization constant A, we need to ensure that the integral of the probability density over the given region is equal to 1:

∫[-L/2, L/2] |ψ(x)|² dx = ∫[-L/2, L/2] A² cos²(kx) dx = 1

Using the trigonometric identity cos²(kx) = (1 + cos(2kx))/2, we get:

A² ∫[-L/2, L/2] (1 + cos(2kx))/2 dx = 1

A² [x/2 + sin(2kx)/(4k)] evaluated from -L/2 to L/2 = 1

A² [(L/4 + sin(kL)/(4k)) - (-L/4 + sin(-kL)/(4k))] = 1

A² [L/2 + sin(kL)/(2k)] = 1

Therefore, A² = 1 / [L/2 + sin(kL)/(2k)]

A = √{1 / [L/2 + sin(kL)/(2k)]}

c) To calculate the probability of finding the particle in the region 0 ≤ x ≤ L/4, we integrate the probability density over this region:

P = ∫[0, L/4] |ψ(x)|² dx = ∫[0, L/4] A² cos²(kx) dx

Using the same trigonometric identity, we get:

P = A² ∫[0, L/4] (1 + cos(2kx))/2 dx

P = A² [x/2 + sin(2kx)/(4k)] evaluated from 0 to L/4

P = A² [(L/8 + sin(kL/2)/(4k)) - (0 + 0)]

P = A² [L/8 + sin(kL/2)/(4k)]

Substituting the value of A² from part (b):

P = {1 / [L/2 + sin(kL)/(2k)]} * [L/8 + sin(kL/2)/(4k)]

Problem 2: Superposition of Plane Waves

A free particle is described by a wave function that is a superposition of two plane waves:

ψ(x) = C [sin(k₁x) + sin(k₂x)]

where C is a constant, k₁ and k₂ are wave numbers.

a) Using trigonometric identities, rewrite the wave function in terms of cosine and sine functions with the average and difference of the wave numbers.

b) Determine the probability density |ψ(x)|².

c) Analyze the behavior of the probability density when k₁ and k₂ are close to each other (k₁ ≈ k₂).

Solution:

a) Using the trigonometric identity sin(a) + sin(b) = 2 sin((a+b)/2) cos((a-b)/2), we can rewrite the wave function:

ψ(x) = C [2 sin((k₁+k₂)/2 * x) cos((k₁-k₂)/2 * x)]

Let k_avg = (k₁+k₂)/2 and Δk = (k₁-k₂)/2

Then, ψ(x) = 2C sin(k_avg * x) cos(Δk * x)

b) The probability density is:

|ψ(x)|² = (2C sin(k_avg * x) cos(Δk * x))²

|ψ(x)|² = 4C² sin²(k_avg * x) cos²(Δk * x)

c) When k₁ and k₂ are close to each other, Δk is small. Therefore, cos(Δk * x) varies slowly compared to sin(k_avg * x). The term cos²(Δk * x) acts as an envelope modulating the faster oscillations of sin²(k_avg * x). This creates a pattern of alternating regions of high and low probability density, representing wave packets. The particle is more likely to be found in regions where cos²(Δk * x) is close to 1.

Problem 3: Momentum and Wave Number

A free particle has a well-defined momentum p. Its wave function is given by:

ψ(x) = B e^(ikx)

where B is a constant and k is the wave number.

a) Express the wave number k in terms of the momentum p and the reduced Planck constant ħ.

b) If the wave function is given instead by ψ(x) = B cos(kx), what are the possible values of the momentum p?

c) Explain the physical difference between the wave functions ψ(x) = B e^(ikx) and ψ(x) = B cos(kx) in terms of momentum.

Solution:

a) The relationship between momentum and wave number is given by:

p = ħk

Therefore, k = p/ħ

b) If the wave function is ψ(x) = B cos(kx), we can express it as a superposition of two plane waves:

cos(kx) = (e^(ikx) + e^(-ikx))/2

This implies that the particle can have two possible momenta:

p₁ = ħk and p₂ = -ħk

The particle can be moving in either the positive or negative x-direction.

c) The wave function ψ(x) = B e^(ikx) represents a particle with a definite momentum p = ħk moving in the positive x-direction. It's an eigenstate of the momentum operator.

The wave function ψ(x) = B cos(kx) represents a particle in a superposition of two momentum eigenstates. It is equally likely to have momentum ħk (moving in the positive x-direction) or momentum -ħk (moving in the negative x-direction). This state does not have a definite momentum.

Problem 4: Particle in a Finite Region

Consider a free particle confined to a region 0 ≤ x ≤ L. The wave function must be zero at the boundaries (ψ(0) = 0 and ψ(L) = 0).

a) Show that the general solution to the Schrdinger equation in this region can be written as:

ψ(x) = D sin(nxπ/L)

where n is an integer and D is a constant.

b) Determine the possible energy levels E for the particle in terms of n, ħ, m, and L.

c) Find the normalization constant D.

Solution:

a) The general solution to the free particle Schrdinger equation is:

ψ(x) = A cos(kx) + B sin(kx)

Applying the boundary condition ψ(0) = 0:

ψ(0) = A cos(0) + B sin(0) = A = 0

So, the wave function reduces to:

ψ(x) = B sin(kx)

Applying the boundary condition ψ(L) = 0:

ψ(L) = B sin(kL) = 0

For a non-trivial solution (B ≠ 0), we must have:

sin(kL) = 0

This implies that kL = nπ, where n is an integer (n = 1, 2, 3, ...). Therefore, k = nπ/L

Substituting this value of k into the wave function:

ψ(x) = B sin(nπx/L)

We can replace B with D to match the given form:

ψ(x) = D sin(nπx/L)

b) The energy E is related to the wave number k by:

E = ħ²k²/2m

Substituting k = nπ/L, we get:

E = ħ²(nπ/L)² / 2m

E = n²π²ħ² / (2mL²)

These are the quantized energy levels for the particle in the box.

c) To find the normalization constant D, we need to ensure that:

∫[0, L] |ψ(x)|² dx = 1

∫[0, L] D² sin²(nπx/L) dx = 1

Using the trigonometric identity sin²(θ) = (1 - cos(2θ))/2, we have:

D² ∫[0, L] (1 - cos(2nπx/L))/2 dx = 1

D² [x/2 - (L/(4nπ)) sin(2nπx/L)] evaluated from 0 to L = 1

D² [L/2 - (L/(4nπ)) sin(2nπ) - (0 - 0)] = 1

Since sin(2nπ) = 0, we have:

D² (L/2) = 1

Therefore, D² = 2/L

D = √(2/L)

Problem 5: Time-Dependent Wave Function

A free particle's wave function at time t = 0 is given by:

ψ(x, 0) = G cos(k₀x)

where G is a constant and k₀ is the initial wave number.

a) Determine the time-dependent wave function ψ(x, t).

b) Describe how the probability density |ψ(x, t)|² changes with time.

Solution:

a) The time-dependent wave function for a free particle is obtained by multiplying each momentum eigenstate by its corresponding time evolution factor:

ψ(x, t) = ψ(x, 0) * e^(-iEt/ħ)

Since cos(k₀x) = (e^(ik₀x) + e^(-ik₀x))/2, we can write ψ(x, 0) as a superposition of two plane waves with wave numbers k₀ and -k₀. The corresponding energies are E = ħ²k₀²/2m for both.

Thus, ψ(x, t) = G * [(e^(ik₀x) * e^(-i(ħ²k₀²/2m)t/ħ) + e^(-ik₀x) * e^(-i(ħ²k₀²/2m)t/ħ) ] /2

ψ(x, t) = (G/2) * [e^(i(k₀x - (ħk₀²/2m)t)) + e^(-i(k₀x + (ħk₀²/2m)t))]

Using Euler's formula, we can rewrite this as:

ψ(x, t) = G cos(k₀x - (ħk₀²/2m)t) if we only consider the wave moving in the positive x direction. However, the initial condition contains both positive and negative k, so it should be:

ψ(x, t) = (G/2) [cos(k₀x - (ħk₀²/2m)t) + cos(k₀x + (ħk₀²/2m)t)]

Using the trigonometric identity cos(a) + cos(b) = 2 cos((a+b)/2)cos((a-b)/2), we get:

ψ(x, t) = G cos(k₀x) cos((ħk₀²/2m)t)

b) The probability density is:

|ψ(x, t)|² = [G cos(k₀x) cos((ħk₀²/2m)t)]²

|ψ(x, t)|² = G² cos²(k₀x) cos²((ħk₀²/2m)t)

The probability density oscillates in both space (due to the cos²(k₀x) term) and time (due to the cos²((ħk₀²/2m)t) term). The spatial oscillations are determined by the initial wave number k₀, while the temporal oscillations are determined by the energy E = ħ²k₀²/2m. The particle's probability density is not static; it changes periodically with time. The envelope of the wave will remain the same, modulated in time.

Advanced Concepts and Trigonometry

Beyond these fundamental problems, trigonometry plays a vital role in more advanced concepts within the free particle model and quantum mechanics in general:

• Wave Packets and Group Velocity: Constructing wave packets involves superposing plane waves with a range of wave numbers. The group velocity, which represents the velocity of the wave packet's envelope, can be calculated using trigonometric relationships and derivatives.
• Uncertainty Principle: The Heisenberg uncertainty principle states that the product of the uncertainties in position and momentum is bounded below. Trigonometric functions and Fourier analysis are used to analyze the spread of wave functions in both position and momentum space to verify this principle.
• Scattering Theory: When considering scattering from a potential barrier or well, the wave function in the free region can be expressed as a superposition of incident, reflected, and transmitted waves. Trigonometric functions are used to determine the amplitudes of these waves based on boundary conditions at the potential.
• Quantum Computing: Qubits, the basic units of quantum information, can be represented using trigonometric functions on the Bloch sphere. Quantum gates, which manipulate qubits, often involve trigonometric transformations.

Conclusion

Mastering the free particle model and its associated trigonometry is essential for understanding the fundamentals of quantum mechanics. The practice problems presented here provide a solid foundation for tackling more complex quantum systems. By understanding the relationship between wave functions, momentum, energy, and probability, you can develop a deeper appreciation for the quantum world. Remember to always relate the mathematical formalism to the physical interpretation, and don't hesitate to explore further applications of trigonometry in quantum mechanics. The journey into the quantum realm is filled with mathematical beauty, and trigonometry is a key to unlocking its secrets.