AP Physics 2 Exam Question: Quantum Phenomena

Consider a particle of mass $m$ in a one-dimensional potential energy well with a width of $L$. The potential energy inside the well is given by the equation:

$$V(x) = \begin{cases} 0, & \text{if } 0 \leq x \leq L \\ \infty, & \text{otherwise} \end{cases}$$

The particle's wave function inside the well, $\psi(x)$, can be represented as a linear combination of the ground state and the first excited state as follows:

$$\psi(x) = A\left(\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right) + B\ \sqrt{\frac{2}{L}}\ \sin\left(\frac{2\pi x}{L}\right)\right)$$

Where $A$ and $B$ are normalization constants.

a) Determine the possible values of the energy, $E$, of the particle in terms of the fundamental constant $\hbar$ and the particle's mass $m$.

b) Calculate the normalization constant $A$.

c) Calculate the normalization constant $B$.

d) Determine the probability density, $P(x)$, for finding the particle at position $x$.

e) Calculate the average position, $\langle x \rangle$, of the particle in the well.

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Answer:

a) The energy, $E$, can be determined by solving the time-independent Schrödinger equation:

$$-\frac{{\hbar^2}}{{2m}}\frac{{d^2\psi}}{{dx^2}} + V(x)\psi = E\psi$$

Inside the well, the potential energy is zero, so the Schrödinger equation simplifies to:

$$-\frac{{\hbar^2}}{{2m}}\frac{{d^2\psi}}{{dx^2}} = E\psi$$

We can start by taking the second derivative of the wave function:

$$\frac{{d^2\psi}}{{dx^2}} = \frac{{d^2}}{{dx^2}}\left(A\left(\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right) + B\ \sqrt{\frac{2}{L}}\ \sin\left(\frac{2\pi x}{L}\right)\right)\right)$$

Simplifying this expression, we have:

$$\frac{{d^2\psi}}{{dx^2}} = \frac{{d}}{{dx}}\left(A\left(\sqrt{\frac{2}{L}}\ \frac{{\pi}}{{L}}\cos\left(\frac{\pi x}{L}\right) + B\ \sqrt{\frac{2}{L}}\ \frac{{2\pi}}{{L}}\cos\left(\frac{2\pi x}{L}\right)\right)\right)$$

$$\frac{{d^2\psi}}{{dx^2}} = -A\left(\sqrt{\frac{2}{L}}\ \frac{{\pi^2}}{{L^2}}\sin\left(\frac{\pi x}{L}\right) + B\ \sqrt{\frac{2}{L}}\ \frac{{4\pi^2}}{{L^2}}\sin\left(\frac{2\pi x}{L}\right)\right)$$

Substituting this back into the Schrödinger equation, we get:

$$-\frac{{\hbar^2}}{{2m}}\left(-A\left(\sqrt{\frac{2}{L}}\ \frac{{\pi^2}}{{L^2}}\sin\left(\frac{\pi x}{L}\right) + B\ \sqrt{\frac{2}{L}}\ \frac{{4\pi^2}}{{L^2}}\sin\left(\frac{2\pi x}{L}\right)\right)\right) = E\left(A\left(\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right) + B\ \sqrt{\frac{2}{L}}\ \sin\left(\frac{2\pi x}{L}\right)\right)\right)$$

Simplifying this expression, we find:

$$\frac{{\hbar^2}}{{2m}}\left(\frac{{\pi^2}}{{L^2}}A \sin\left(\frac{\pi x}{L}\right) + \frac{{4\pi^2}}{{L^2}}B \sin\left(\frac{2\pi x}{L}\right)\right) = E\left(A\left(\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right) + B\ \sqrt{\frac{2}{L}}\ \sin\left(\frac{2\pi x}{L}\right)\right)\right)$$

Dividing both sides by $\psi$, we obtain:

$$\frac{{\hbar^2}}{{2m}}\left(\frac{{\pi^2}}{{L^2}}A + \frac{{4\pi^2}}{{L^2}}B\right) = E$$

Therefore, the possible values of the energy, $E$, are given by:

$$\boxed{E = \frac{{\hbar^2}}{{2m}}\left(\frac{{\pi^2}}{{L^2}}A + \frac{{4\pi^2}}{{L^2}}B\right)}$$

b) To calculate the normalization constant $A$, we need to integrate the square of the absolute value of the wave function over all space and set it equal to 1 (since the wave function must be normalized). The integral is as follows:

$$\int_{0}^{L} |\psi(x)|^2\ dx = \int_{0}^{L} \left[A\left(\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right) + B\ \sqrt{\frac{2}{L}}\ \sin\left(\frac{2\pi x}{L}\right)\right)\right]^2\ dx$$

Simplifying and evaluating the integral, we find:

$$A^2 \left(\frac{2}{L}\ \frac{L}{2}\right) = A^2$$

Since the integral equals 1, we have:

$$A^2 = 1$$

Taking the square root of both sides, we get:

$$A = 1$$

Therefore, the normalization constant $A$ is equal to 1.

c) To calculate the normalization constant $B$, we can use the same approach as in part (b). The integral is as follows:

$$\int_{0}^{L} |\psi(x)|^2\ dx = \int_{0}^{L} \left[A\left(\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right) + B\ \sqrt{\frac{2}{L}}\ \sin\left(\frac{2\pi x}{L}\right)\right)\right]^2\ dx$$

Expanding and evaluating the integral, we have:

$$A^2 \int_{0}^{L} \left(\frac{2}{L}\ \sin^2\left(\frac{\pi x}{L}\right) + 2B\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right) \sqrt{\frac{2}{L}}\ \sin\left(\frac{2\pi x}{L}\right) + B^2\frac{2}{L}\ \sin^2\left(\frac{2\pi x}{L}\right)\right)\ dx$$

Simplifying this expression and evaluating the integral, we find:

$$A^2 \left(\frac{2}{L}\ \frac{L}{2} + 2B\sqrt{\frac{2}{L}}\ 0 \times 1 + B^2\frac{2}{L}\ \frac{L}{2}\right) = A^2 + B^2$$

Since the integral equals 1, we have:

$$A^2 + B^2 = 1$$

Substituting the value of $A$ (from part b) into this equation, we obtain:

$$1 + B^2 = 1$$

Simplifying, we find:

$$B^2 = 0$$

Taking the square root of both sides, we get:

$$B = 0$$

Therefore, the normalization constant $B$ is equal to 0.

d) The probability density, $P(x)$, for finding the particle at position $x$ can be calculated as the square of the absolute value of the wave function:

$$P(x) = |\psi(x)|^2 = \left|1\left(\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right) + 0\ \sqrt{\frac{2}{L}}\ \sin\left(\frac{2\pi x}{L}\right)\right)\right|^2$$

Simplifying and squaring the terms, we find:

$$P(x) = \left(\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right)\right)^2$$

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

Therefore, the probability density $P(x)$ for finding the particle at position $x$ is given by:

$$\boxed{P(x) = \frac{2}{L}\ \sin^2\left(\frac{\pi x}{L}\right)}$$

e) The average position, $\langle x \rangle$, of the particle in the well can be calculated using the following formula:

$$\langle x \rangle = \int_{0}^{L} x\ |\psi(x)|^2\ dx$$

Substituting the wave function and evaluating the integral, we have:

$$\langle x \rangle = \int_{0}^{L} x\ \left(1\left(\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right) + 0\ \sqrt{\frac{2}{L}}\ \sin\left(\frac{2\pi x}{L}\right)\right)\right)^2\ dx$$

Simplifying, we find:

$$\langle x \rangle = \int_{0}^{L} x\left(\sqrt{\frac{2}{L}}\ \sin\left(\frac{\pi x}{L}\right)\right)^2\ dx$$

$$\langle x \rangle = \int_{0}^{L} x\left(\frac{2}{L}\ \sin^2\left(\frac{\pi x}{L}\right)\right)\ dx$$

Expanding and evaluating the integral, we obtain:

$$\langle x \rangle = \frac{2}{L}\int_{0}^{L} x\ \sin^2\left(\frac{\pi x}{L}\right)\ dx$$

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

$$\langle x \rangle = \frac{2}{L}\int_{0}^{L} x\left(\frac{1}{2}\left(1 - \cos\left(\frac{2\pi x}{L}\right)\right)\right)\ dx$$

Distributing and evaluating the integral, we find:

$$\langle x \rangle = \frac{1}{L}\left(\int_{0}^{L} x\ dx - \int_{0}^{L} x\cos\left(\frac{2\pi x}{L}\right)\ dx\right)$$

The first integral is straightforward:

$$\int_{0}^{L} x\ dx = \left[\frac{1}{2}x^2\right]_{0}^L = \frac{1}{2}(L^2 - 0^2) = \frac{1}{2}L^2$$

For the second integral, we use integration by parts with $u = x$ and $dv = \cos\left(\frac{2\pi x}{L}\right)\ dx$:

$$du = dx, \quad v = \frac{L}{2\pi}\sin\left(\frac{2\pi x}{L}\right)$$

Applying the integration by parts formula, we have:

$$\int_{0}^{L} x\cos\left(\frac{2\pi x}{L}\right)\ dx = \left[\frac{L}{2\pi}x\sin\left(\frac{2\pi x}{L}\right)\right]_{0}^L - \int_{0}^{L} \frac{L}{2\pi}\sin\left(\frac{2\pi x}{L}\right)\ dx$$

The first term evaluates to zero at both integration limits:

$$\frac{L}{2\pi}L\sin\left(\frac{2\pi L}{L}\right) - \frac{L}{2\pi}0\sin\left(\frac{2\pi 0}{L}\right) = 0$$

Simplifying the second term, we obtain:

$$-\frac{L}{2\pi}\int_{0}^{L} \sin\left(\frac{2\pi x}{L}\right)\ dx = -\frac{L}{2\pi}\left[-\frac{L}{2\pi}\cos\left(\frac{2\pi x}{L}\right)\right]_{0}^L$$

Again, the first term evaluates to zero at both integration limits:

$$-\frac{L}{2\pi}\left[-\frac{L}{2\pi}\cos\left(\frac{2\pi L}{L}\right)\right] + \frac{L}{2\pi}\left[-\frac{L}{2\pi}\cos\left(\frac{2\pi 0}{L}\right)\right] = \frac{L^2}{2\pi}\left(\frac{1}{\pi} - \frac{1}{\pi}\right) = 0$$

Substituting these results back into the expression for $\langle x \rangle$, we get:

$$\langle x \rangle = \frac{1}{L}\left(\frac{1}{2}L^2 - 0\right) = \frac{1}{2}L$$

Therefore, the average position $\langle x \rangle$ of the particle in the well is given by:

$$\boxed{\langle x \rangle = \frac{1}{2}L}$$

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Note: The solutions provided are based on the given information and assumptions. If additional information or context is provided, the solutions may need to be adjusted accordingly.