AP Physics 2 Exam Question

A particle of mass $\text{m}$ is in a one-dimensional infinite potential well with width $\text{L}$. Initially, the particle is in its ground state, $\phi_1(x)$, which has a probability density function given by:

$$\Psi_1(x) = \frac{2}{L} \sin\left(\frac{\pi x}{L}\right)$$

where $\Psi_1(x)$ is the wave function and $x$ is the position of the particle.

a) Calculate the normalization constant, $A$, for the wave function.

b) Calculate the probability that measuring the particle's position yields a value between $\frac{L}{4}$ and $\frac{L}{2}$.

c) At time $t = 0$, the particle's wave function is altered such that it is described by the following superposition:

$$\Psi(x,0) = \frac{\sqrt{2}}{L} \sin\left(\frac{2\pi x}{L}\right) + \frac{\sqrt{3}}{2}\phi_1(x)$$

Calculate the probability that measuring the particle's position at $t = 0$ yields a value between $\frac{L}{4}$ and $\frac{L}{2}$.

d) Calculate the expectation value of the energy of the particle at $t = 0$ given its altered wave function.

e) Discuss the physical interpretation of the wave function alteration in part (c).

Provide all calculations, formulas used, and detailed explanations for each part.

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Answer:

a) In order to calculate the normalization constant $A$, we need to ensure that the total probability (integral of the probability density function) is equal to 1.

The probability density function $\Psi_1(x)$ is given by:

$$\Psi_1(x) = \frac{2}{L} \sin\left(\frac{\pi x}{L}\right)$$

To calculate the normalization constant, we integrate the square of the wave function and set it equal to 1:

$$1 = \int_{0}^{L} |A \Psi_1(x)|^2 \, dx$$

$$1 = \int_{0}^{L} |A|^2 \left(\frac{2}{L} \sin\left(\frac{\pi x}{L}\right)\right)^2 \, dx$$

$$1 = \frac{4|A|^2}{L^2} \int_{0}^{L} \sin^2\left(\frac{\pi x}{L}\right) \, dx$$

Using the trigonometric identity $\sin^2(\theta) = \frac{1}{2}(1 - \cos(2\theta))$, we can simplify the integral:

$$1 = \frac{4|A|^2}{L^2} \int_{0}^{L} \frac{1}{2}\left(1 - \cos\left(\frac{2\pi x}{L}\right)\right) \, dx$$

Simplifying further:

$$1 = \frac{2|A|^2}{L^2} \left[x - \frac{L}{2\pi}\sin\left(\frac{2\pi x}{L}\right) \right]_{0}^{L}$$

Substituting the limits of integration:

$$1 = \frac{2|A|^2}{L^2} \left[L - \frac{L}{2\pi}\sin\left(2\pi\right)\right]$$

Since $\sin(2\pi) = 0$:

$$1 = 2|A|^2$$

Solving for $|A|^2$: [ |A|^2 = \frac{1}{2} ]

Taking the square root: [ |A| = \frac{1}{\sqrt{2}} ]

Therefore, the normalization constant is $A = \frac{1}{\sqrt{2}}$.

b) The probability of measuring the particle's position between $L/4$ and $L/2$ can be calculated by integrating the absolute square of the wave function between those limits:

$$P = \int_{L/4}^{L/2} \left|\frac{1}{\sqrt{2}} \frac{2}{L} \sin\left(\frac{\pi x}{L}\right)\right|^2 \, dx$$

Simplifying:

$$P = \frac{1}{2} \int_{L/4}^{L/2} \left(\frac{2}{L}\right)^2 \sin^2\left(\frac{\pi x}{L}\right) \, dx$$

Using the trigonometric identity $\sin^2(\theta) = \frac{1}{2}(1 - \cos(2\theta))$:

$$P = \frac{1}{2} \int_{L/4}^{L/2} \frac{1}{L^2}\left(1 - \cos\left(\frac{2\pi x}{L}\right)\right) \, dx$$

Simplifying further:

$$P = \frac{1}{2L^2} \left[x - \frac{L}{2\pi}\sin\left(\frac{2\pi x}{L}\right) \right]_{L/4}^{L/2}$$

Substituting the limits of integration:

$$P = \frac{1}{2L^2} \left[\frac{L}{2} - \frac{L}{4} - \frac{L}{2}\sin\left(\pi\right) + \frac{L}{4}\sin\left(\frac{\pi}{2}\right)\right]$$

Since $\sin(\pi) = 0$ and $\sin(\pi/2) = 1$:

$$P = \frac{1}{2L^2} \left[\frac{L}{2} - \frac{L}{4} + \frac{L}{4}\right] = \frac{1}{2L^2} \left[\frac{L}{2}\right] = \frac{1}{4}$$

Therefore, the probability of measuring the particle's position between $L/4$ and $L/2$ is $\frac{1}{4}$.

c) We are given the altered wave function at $t=0$:

$$\Psi(x,0) = \frac{\sqrt{2}}{L} \sin\left(\frac{2\pi x}{L}\right) + \frac{\sqrt{3}}{2}\phi_1(x)$$

To calculate the probability of measuring the particle's position between $L/4$ and $L/2$ at $t=0$, we need to integrate the absolute square of the wave function between those limits:

$$P = \int_{L/4}^{L/2} |\Psi(x,0)|^2 \, dx$$

Simplifying:

$$P = \int_{L/4}^{L/2} \left|\frac{\sqrt{2}}{L} \sin\left(\frac{2\pi x}{L}\right) + \frac{\sqrt{3}}{2}\phi_1(x)\right|^2 \, dx$$

Expanding the square:

$$P = \int_{L/4}^{L/2} \left(\frac{\sqrt{2}}{L}\right)^2 \sin^2\left(\frac{2\pi x}{L}\right) \, dx + \int_{L/4}^{L/2} \left(\frac{\sqrt{3}}{2}\right)^2 |\phi_1(x)|^2 \, dx + 2\int_{L/4}^{L/2} \left(\frac{\sqrt{2}}{L}\right) \left(\frac{\sqrt{3}}{2}\phi_1(x)\right) \sin\left(\frac{2\pi x}{L}\right) \, dx$$

Using the trigonometric identity $\sin^2(\theta) = \frac{1}{2}(1 - \cos(2\theta))$:

$$P = \int_{L/4}^{L/2} \frac{1}{2L^2} \left(1 - \cos\left(\frac{4\pi x}{L}\right)\right) \, dx + \frac{3}{4} \int_{L/4}^{L/2} |\phi_1(x)|^2 \, dx + \sqrt{\frac{3}{2}} \int_{L/4}^{L/2} \frac{1}{L}\phi_1(x) \sin\left(\frac{2\pi x}{L}\right) \, dx$$

The first integral simplifies to:

$$\int_{L/4}^{L/2} \frac{1}{2L^2} \left(1 - \cos\left(\frac{4\pi x}{L}\right)\right) \, dx = \frac{1}{2L^2} \left[x - \frac{L}{4\pi}\sin\left(\frac{4\pi x}{L}\right)\right]_{L/4}^{L/2}$$

Substituting the limits of integration:

$$\frac{1}{2L^2} \left[\frac{L}{2} - \frac{L}{4} - \frac{L}{4} + \frac{L}{4\pi}\sin\left(\frac{2\pi}{2}\right) - \frac{L}{4\pi}\sin\left(\frac{2\pi}{4}\right) \right]$$

Since $\sin(2\pi) = 0$ and $\sin\left(\frac{\pi}{2}\right) = 1$:

$$\frac{1}{2L^2} \left[\frac{L}{4\pi}\right]$$

Simplifying:

$$\frac{1}{8\pi L^2}$$

The second integral is the normalization integral for $\phi_1(x)$. Since $\phi_1(x)$ is the ground state wave function, it is already normalized, hence the integral evaluates to 1:

$$\int_{L/4}^{L/2} |\phi_1(x)|^2 \, dx = 1$$

The third integral simplifies to:

$$\sqrt{\frac{3}{2}} \int_{L/4}^{L/2} \frac{1}{L}\phi_1(x) \sin\left(\frac{2\pi x}{L}\right) \, dx = \sqrt{\frac{3}{2}} \left[\frac{1}{L} \int_{L/4}^{L/2} \phi_1(x) \sin\left(\frac{2\pi x}{L}\right) \, dx\right]$$

Since $\phi_1(x)$ is an eigenfunction of the Hamiltonian operator, it is orthogonal to the sine term (which corresponds to an eigenfunction of the position operator). Therefore, the integral is zero:

$$\frac{1}{L} \int_{L/4}^{L/2} \phi_1(x) \sin\left(\frac{2\pi x}{L}\right) \, dx = 0$$

Thus, the third integral evaluates to zero.

The overall probability is then:

$$P = \frac{1}{8\pi L^2} + 1 + 0 = \frac{1}{8\pi L^2} + 1$$

Therefore, the probability of measuring the particle's position between $L/4$ and $L/2$ at $t=0$ is $\frac{1}{8\pi L^2} + 1$.

d) The expectation value of the energy, $\langle E \rangle$, can be calculated using the time-independent Schrödinger equation and the wave function at $t=0$:

$$\Psi(x,t) = \sum_{n=1}^{\infty} c_n \phi_n(x) e^{-iE_nt/\hbar}$$

where $c_n$ are the coefficients of the expansion, $\phi_n(x)$ are the stationary states, $E_n$ are the eigenvalues, and $\hbar$ is the reduced Planck's constant.

Taking the inner product of $\Psi(x,t)$ with the Hamiltonian operator, then dividing by the inner product of $\Psi(x,t)$ with itself, we obtain the expectation value of the energy:

$$\langle E \rangle = \frac{\langle \Psi(x,t) | \hat{H} | \Psi(x,t) \rangle}{\langle \Psi(x,t) | \Psi(x,t) \rangle}$$

Substituting the altered wave function at $t=0$:

$$\Psi(x,0) = \frac{\sqrt{2}}{L} \sin\left(\frac{2\pi x}{L}\right) + \frac{\sqrt{3}}{2}\phi_1(x)$$

and recognizing that $\phi_1(x)$ is an eigenstate of the Hamiltonian operator with eigenvalue $E_1$, the expectation value becomes:

$$\langle E \rangle = \frac{\left|\langle \Psi(x,0) | \hat{H} | \Psi(x,0) \rangle\right|}{\langle \Psi(x,0) | \Psi(x,0) \rangle}$$

Substituting the eigenvalue equation for $\phi_1(x)$, we have:

$$\langle E \rangle = \frac{\left|\langle \Psi(x,0) | E_1 | \Psi(x,0) \rangle\right|}{\langle \Psi(x,0) | \Psi(x,0) \rangle}$$

Since $\phi_1(x)$ is normalized, $\langle \Psi(x,0) | \Psi(x,0) \rangle$ is equal to 1. Also, $E_1$ can be factored out of the inner product because it is just a constant. Thus, we obtain:

$$\langle E \rangle = E_1 = \frac{\pi^2 \hbar^2}{2mL^2}$$

Therefore, the expectation value of the energy of the particle at $t=0$ given its altered wave function is $\frac{\pi^2 \hbar^2}{2mL^2}$.

e) The altered wave function at $t=0$ is a superposition of the ground state $\phi_1(x)$ and another stationary state with double the frequency, $\frac{\sqrt{2}}{L} \sin\left(\frac{2\pi x}{L}\right)$. This means the particle now has a non-zero amplitude to be found in the excited state $\frac{\sqrt{2}}{L} \sin\left(\frac{2\pi x}{L}\right)$. As time progresses, the wave function will evolve, causing the particle to transition between the ground state and the excited state periodically.

The physical interpretation of this alteration is that it introduces a higher energy component to the particle's state, leading to oscillatory behavior and potential changes in observable properties, such as the position and energy of the particle. This alteration reflects the inherent probabilistic nature of quantum physics, where particles can exist in superpositions of states and exhibit wave-like behavior.