AP Physics 1 Exam Question

A particle is in a quantum state described by the wavefunction:

ψ(x) = A(e^(ikx) + e^(-ikx))

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

1. Determine the probability density function |ψ(x)|² for this quantum state.

Answer

The probability density function |ψ(x)|² can be found by squaring the absolute value of the wavefunction:

|ψ(x)|² = |A(e^(ikx) + e^(-ikx))|²

Using the absolute value properties and the sum of squares formula, we can expand this expression further:

|ψ(x)|² = |A|²|(e^(ikx) + e^(-ikx))|²

|ψ(x)|² = |A|²(e^(ikx) + e^(-ikx))(e^(-ikx) + e^(ikx))

Simplifying this expression, we find:

|ψ(x)|² = |A|²(2 + 2e^(2ikx) + 2e^(-2ikx))

Now, let's substitute A for the normalization constant. The normalization constant ensures that the probability density integrates to 1 over all possible values of x. Since the probability density function is |ψ(x)|², we need to normalize it as follows:

∫(|ψ(x)|²)dx = 1

Integrating |ψ(x)|², we get:

∫(|A|²(2 + 2e^(2ikx) + 2e^(-2ikx)))dx = 1

Simplifying the integral using linearity, we obtain:

2|A|²∫dx + 2|A|²∫(e^(2ikx) + e^(-2ikx))dx = 1

The integration of the constant term and the exponential terms can be done as follows:

2|A|²(x + (1/(2ik))e^(2ikx) - (1/(2ik))e^(-2ikx))

Now, we must set this expression equal to 1 and solve for |A|². Remembering that the integration is done over all possible values of x, we set:

2|A|²(x + (1/(2ik))e^(2ikx) - (1/(2ik))e^(-2ikx)) ∣ ∣ ∣ ∣ ∣ -∞ to ∞ = 1

Evaluating this expression at the limits, we find:

2|A|²(∞ + (1/(2ik))e^(2ik∞) - (1/(2ik))e^(-2ik∞)) - 2|A|²(-∞ + (1/(2ik))e^(2ik(-∞)) - (1/(2ik))e^(-2ik(-∞))) = 1

Since e^2ik∞ and e^(-2ik∞) approach infinity (∞) and e^(2ik(-∞)) and e^(-2ik(-∞)) approach zero (0), we have:

2|A|²(∞ + 0 - 0) - 2|A|²(-∞ + 0 - 0) = 1

Simplifying this expression further:

2|A|²(∞) - 2|A|²(-∞) = 1

Since the infinite values cancel out, we are left with:

2|A|² - 2|A|² = 1

This simplifies to:

0 = 1

Since this equation is not true, there is no normalization constant |A|² that can make the probability density function integrate to 1 over all possible values of x. Therefore, the given wavefunction is not normalized and does not represent a physically valid quantum state.

Therefore, the probability density function |ψ(x)|² for the given quantum state is not valid.

Note: Make sure to double-check your calculations and follow the correct steps in solving for the normalization constant.