Problem Sheet: Week 11

Finite Square Well¶

1. In this question we will find the equation for the energy eigenvalues for the symmetric finite square well, i.e. the potential well such that

   $$V(x) = \begin{cases} 0 & \text{if } -\frac{a}{2} \leq x \leq \frac{a}{2}, \\ V_0 & \text{otherwise,} \end{cases}$$

   where $V_0$ specifies the height of the well, and $a$ the width. We will denote the region to the left of the well, such that $x < -a/2$ by region I, the region inside the well, such that $-a/2 \leq x \leq a/2$ by region II, and the region to the right of the well, such that $x > a/2$ by region III.

   We will only be interested in bound states, such that $E < V_0$.

   Since the potential is now symmetric about the origin, it is an even function. This means that that the energy eigenstates will be alternatively even and odd functions. In this question we will only be interested in the energy eigenvalues of the even solutions, i.e. every other energy, starting with the ground state.

   1. Write down the time-independent Schrödinger equation in region I, and solve it, to show that the permissible solution in this region is

      $$u_\rI(x) = Ae^{\zeta x},$$

      where $\zeta = \sqrt{2M(V_0 - E)}/\hbar$ and $A$ is a constant.

   2. Write down the time-independent Schrödinger equation in region II, and solve it, to show that an even solution in this region is

      $$u_\rII(x) = C\cos kx,$$

      where $k = \sqrt{2ME}/\hbar$ and $C$ is a constant.

   3. Using the fact that the wavefunction is even, show that the solution in region III is

      $$u_\rIII(x) = Ae^{-\zeta x}.$$

   4. Use the continuity of the wavefunction and its first derivative at the boundary between regions I and II to show that the following equation must hold, which is the quantisation condition

      $$\zeta = k \tan \frac{ka}{2}.$$

   5. Defining $k_0 = \sqrt{2MV_0}/\hbar$, such that $\zeta^2 = k_0^2 - k^2$, find a suitable pair of functions to sketch, such that the intersection points of these functions give the allowed values of $k$. Make a sketch, and indicate the intersection points.

Quantum Harmonic Oscillator¶

2. In this question we will derive the dimensionless form of the time-independent Schrödinger equation for the quantum harmonic oscillator.

   1. The three dimensionful quantities for the quantum harmonic are the mass of the particle $M$, the angular frequency ω (which specifies the spring constant $k = M\omega^2$), and $\hbar$. Write down the dimensions of each of these quantities.

   2. Find a combination of these quantities that has the dimension of length. That is, let

      $$a = M^\alpha \omega^\beta \hbar^\gamma,$$

      and find the exponents α, β and γ that give $a$ the dimension of length.

   3. Show that the time-independent Schrödinger equation for the harmonic oscillator becomes

      $$\frac{d^2 u}{dy^2} = (y^2-\mathcal{E})u(y),$$

      upon the substitutions $x = ya$ and $E = \frac{1}{2}\hbar \omega \mathcal{E}$.

3. In this question we will normalise the ground state and the first excited state of the quantum harmonic oscillator.

   1. Normalise the ground state wavefunction
   $$u_0(x) = A_0e^{-x^2/2a^2}.$$
   2. Normalise the first excited state wavefunction
   $$u_1(x) = A_1 \left(\frac{x}{a}\right) e^{-x^2/2a^2}.$$

4. In this question we will find the third excited state of the quantum harmonic oscillator. As in the notes, we will make a guess that the wavefunction is of the form

   $$u_3(y) = v(y)e^{-\frac{1}{2}y^2}.$$

   (11.1)
   1. Show that the function
   $$v(y) = A_3y(y-c)(y+c),$$

   where $A_3$ and $c$ are constants, is an odd function with 3 nodes (i.e. is zero precisely 3 times). Explain why this shows that it is a good guess for the form of $v(y)$ for the third excited state.

   2. Show that

      $$\frac{d^2 u_3}{dy^2} = A_3\Big(y^5 - (7+c^2)y^3 + (6 + 3c^2)y \Big)e^{-\frac{1}{2}y^2}.$$

   3. Using the time-independent Schrödinger equation from Problem 2, and equating the coefficients of the different powers of $y$, show that the following pair of equations must hold

      $$\begin{align*} -(7+c^2) &= - (\mathcal{E} + c^2),\\ 6+3c^2 &= \mathcal{E}c^2. \end{align*}$$

   4. Solve the equations from part (3) for $\mathcal{E}$ and $c$ to show that the third excited state of the quantum harmonic oscillator is

      $$u_3(x) = A_3\frac{x}{a}\left(\frac{x^2}{a^2} - \frac{3}{2}\right)e^{-x^2/2a^2},$$

      with corresponding energy eigenvalue

      $$E_3 = \frac{7}{2}\hbar \omega,$$

      after making appropriate substitutions for $y$ and $\mathcal{E}$.