Free particle: propagating waves
\[
-\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} = E \psi(x), \quad \textrm{ as } V(x) = 0
\]
\[
\Rightarrow \quad \frac{d^2 \psi(x)}{dx^2} = - k^2 \psi(x), \quad \textrm{ with } k = \frac{\sqrt{2 m E}}{\hbar}
\]
- Solutions are unconstrained: all energy values
- Similar to a very wide well (with infinite walls)
\[
\psi(x) = A e^{ikx} + B e^{-ikx}
\]
Free particle solutions
\[
\psi(x) = A e^{ikx} + B e^{-ikx}
\]
- Problem 1: Not normalizable
- Problem 2: Velocity is half of classical velocity
Problem 1: Not normalizable
\[
\psi(x) = A e^{ikx} + B e^{-ikx}
\]
\[
\begin{aligned}
|\psi(x)|^2 &= \psi(x)^* \, \psi(x)\\
&= (A^* e^{-ikx} + B^* e^{ikx}) (A e^{ikx} + B e^{-ikx})\\
&= |A|^2 + |B|^2 + (A B^* e^{i2kx} + A^* B e^{-i2kx})\\
&= |A|^2 + |B|^2 + \Re\left(A B^* e^{i2kx}\right)\\
\Rightarrow \int_{-\infty}^{+\infty} |\psi(x)|^2 dx &= (|A|^2 + |B|^2) \, \infty + \int_{-\infty}^{+\infty} \Re\left(A B^* e^{i2kx}\right) dx\\
\end{aligned}
\]
The last integral is bounded (oscillating between finite values)
The total integral \(\int |\psi|^2 dx \longrightarrow +\infty\), and therefore doesn’t exist.
Problem 2: Velocity too slow
We can rewrite \(\Psi(x, t) = \psi(x) e^{-iEt/\hbar}\) with \(E = \frac{\hbar^2 k^2}{2m}\):
\[
\begin{aligned}
\Psi(x, t) &= A e^{ikx - iEt/\hbar} + B e^{-ikx - iEt/\hbar}\\
&= A e^{ikx - iEt/\hbar} + B e^{-ikx - iEt/\hbar}\\
&= A e^{ik(x - \frac{\hbar k}{2m}t)} + B e^{-ik(x - \frac{\hbar k}{2m}t)}\\
\end{aligned}
\]
- This is a function of \(x \pm v t\), a “wave” moving with velocity \(v\)
- The velocity is \(v_\textrm{quantum} = \frac{\hbar k}{2m} = \sqrt{\frac{E}{2m}}\)
- Classically the velocity \(v_\textrm{classical} = \sqrt{\frac{2 E}{m}}\) because \(E = \frac{1}{2}m v^2\)
- The classical velocity is twice the one according to quantum mechanics!
Wave packet
- Superposition of propagating waves is normalizable
- Solves the velocity problem as well!
- Is consistent with uncertainty principle
- Fourier’s trick to find coefficients
Wave packet
- Superposition of waves
- Momentum around main \(\sum_{n=-N}^N k_0 \pm n\delta k\)
- Plots from top to bottom \(N = 1, 2, 4, 8\)
\[
\psi(x) = \sum_{n=-N}^N e^{i (k_0 + n\delta k) x}
\]
Wave packet
Phase velocity \(v\)
\[
v = \frac{\omega}{k}
\]
Group velocity:
\[
v_g = \frac{d\omega}{dk}
\]
\[
\Psi(x, t) = \sum_{n=-N}^N e^{i (k_0 + n\delta k) x}
\]
Phase and group velocity
\[
\begin{aligned}
\Psi(x, t) &= \frac{1}{\sqrt{2\pi}}\int \phi(k) e^{i (k x - \omega t)}\\
\end{aligned}
\]
Assume \(\phi(k)\) around \(k_0\)
\[
\begin{aligned}
\Psi(x, t) &\approx \frac{1}{\sqrt{2\pi}}\int \phi(k_0 + s) e^{i ((k_0 +s) x - (\omega_0 + s\omega_0') t)}\\
&\approx \frac{1}{\sqrt{2\pi}} e^{i (k_0 x - \omega_0 t)} \int \phi(k_0 + s) e^{i s(x - \omega_0' t)}\\
\end{aligned}
\]
Finite square well
Finite square potential well
Bound states \(0 < E < V_0\)
Scattering states \(E > V_0\)
Schrodinger equation:
\[
- \frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} = \left(E - V(x)\right) \, \psi(x)
\]
Finite well: bound states
For bound states \(0 < E < V_0\)
\[
\textrm{inside: }\quad \psi(-a < x < a) = C \sin(\lambda x) + D \cos(\lambda x) \qquad \lambda = \frac{\sqrt{2 m E}}{\hbar}
\]
\[
\textrm{outside:} \quad
\begin{aligned}
\psi(x < -a) &= A e^{-\kappa x} + B e^{\kappa x} = B e^{\kappa x}\\
\psi(x > a) &= F e^{-\kappa x} + G e^{\kappa x} = F e^{-\kappa x}
\end{aligned}
\qquad \kappa = \frac{\sqrt{2 m (V_0 - E)}}{\hbar}
\]
Finite well: bound states
Since \(\psi(x)\) is continuous in \(x = -a\) and \(x = a\):
\[
\begin{aligned}
B e^{-\kappa a} &= -C \sin(\lambda a) + D \cos(\lambda a)\\
F e^{-\kappa a} &= C \sin(\lambda a) + D \cos(\lambda a)\\
\end{aligned} \qquad \psi(x) \textrm{ continuous in } x = -a, a
\]
\[
\begin{aligned}
B \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) + D \sin(\lambda a)\\
-F \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) - \lambda D \sin(\lambda a)\\
\end{aligned} \qquad \frac{d\psi(x)}{dx} \textrm{ continuous in } x = -a, a
\]
Finite well: bound states
\[
\begin{aligned}
B e^{-\kappa a} &= -C \sin(\lambda a) + D \cos(\lambda a)\\
F e^{-\kappa a} &= C \sin(\lambda a) + D \cos(\lambda a)\\
\end{aligned} \qquad \psi(x) \textrm{ continuous in } x = -a, a
\]
\[
\begin{aligned}
B \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) + \lambda D \sin(\lambda a)\\
-F \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) - \lambda D \sin(\lambda a)\\
\end{aligned} \qquad \frac{d\psi(x)}{dx} \textrm{ continuous in } x = -a, a
\]
Add the first two equations and subtract the others:
\[
\begin{aligned}
(B + F) e^{-\kappa a} &= 2 D \cos{\lambda a}\\
(B + F) \frac{\kappa}{\lambda} e^{-\kappa a} &= 2 D \sin{\lambda a}\\
\end{aligned}
\]
And divide them \(\Leftrightarrow B \neq -F\)
\[
\begin{aligned}
\frac{\kappa}{\lambda} &= \tan{\lambda a}\\
\end{aligned}
\]
Finite well: bound states
\[
\begin{aligned}
B e^{-\kappa a} &= -C \sin(\lambda a) + D \cos(\lambda a)\\
F e^{-\kappa a} &= C \sin(\lambda a) + D \cos(\lambda a)\\
\end{aligned} \qquad \psi(x) \textrm{ continuous in } x = -a, a
\]
\[
\begin{aligned}
B \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) + \lambda D \sin(\lambda a)\\
-F \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) - \lambda D \sin(\lambda a)\\
\end{aligned} \qquad \frac{d\psi(x)}{dx} \textrm{ continuous in } x = -a, a
\]
Now subtract the first two equations and add the others:
\[
\begin{aligned}
(B - F) e^{-\kappa a} &= -2 C \sin{\lambda a}\\
(B - F) \frac{\kappa}{\lambda} e^{-\kappa a} &= 2 C \cos{\lambda a}\\
\end{aligned}
\]
And divide them \(\Leftrightarrow B \neq F\)
\[
\begin{aligned}
\frac{\kappa}{\lambda} &= -\cot{\lambda a} \qquad \textrm{ if } B \neq F\\
\frac{\kappa}{\lambda} &= \tan{\lambda a} \qquad \textrm{ if } B \neq -F \textrm{ from before} \\
\end{aligned}
\]
Finite well: bound states
\[
\begin{aligned}
\frac{\kappa}{\lambda} &= -\cot{\lambda a} \qquad \textrm{ if } B \neq F\\
\frac{\kappa}{\lambda} &= \tan{\lambda a} \qquad \textrm{ if } B \neq -F \textrm{ from before} \\
\end{aligned}
\]
- The relations are inconsistent!
- Only possible if one of them is invalid: \(B = F\) or \(B = -F\)
\[
B = F \quad \Rightarrow \quad
\begin{aligned}
(B - F) e^{-\kappa a} &= -2 C \sin{\lambda a}\\
(B - F) \frac{\kappa}{\lambda} e^{-\kappa a} &= 2 C \cos{\lambda a}\\
\end{aligned} \quad \Rightarrow C = 0, \psi(x) = D \cos(\lambda x)
\]
\[
B = -F \quad \Rightarrow \quad
\begin{aligned}
(B + F) e^{-\kappa a} &= 2 D \cos{\lambda a}\\
(B + F) \frac{\kappa}{\lambda} e^{-\kappa a} &= 2 D \sin{\lambda a}\\
\end{aligned}
\quad \Rightarrow D = 0, \psi(x) = C \sin(\lambda x)
\]
Finite well: bound states (wave function)
Symmetric solutions
\[
\left\{
\begin{aligned}
\psi(x < -a) &= B e^{\kappa x}\\
\psi(-a < x < a) &= D \cos(\lambda x)\\
\psi(x > a) &= B e^{-\kappa x}\\
\end{aligned}
\right.
\]
with \(F = B \quad \& \quad \kappa = \lambda \tan \lambda a\)
Asymmetric solutions
\[
\left\{
\begin{aligned}
\psi(x < -a) &= B e^{\kappa x}\\
\psi(-a < x < a) &= C \sin(\lambda x)\\
\psi(x > a) &= -B e^{-\kappa x}\\
\end{aligned}
\right.
\]
with \(F = -B \quad \& \quad \kappa = -\lambda \cot \lambda a\)
Finite well: bound states
Symmetric solutions
\[
\left\{
\begin{aligned}
\psi(x < -a) &= B e^{\kappa x}\\
\psi(-a < x < a) &= D \cos(\lambda x)\\
\psi(x > a) &= B e^{-\kappa x}\\
\end{aligned}
\right.
\]
with \(F = B \quad \& \quad \kappa = \lambda \tan \lambda a\)
Asymmetric solutions
\[
\left\{
\begin{aligned}
\psi(x < -a) &= B e^{\kappa x}\\
\psi(-a < x < a) &= C \sin(\lambda x)\\
\psi(x > a) &= -B e^{-\kappa x}\\
\end{aligned}
\right.
\]
with \(F = -B \quad \& \quad \kappa = -\lambda \cot \lambda a\)
Let’s first obtain the energy from \(\kappa\) and \(\lambda\):
\[
\kappa^2 = \frac{2m}{\hbar^2}(V_0 - E), \qquad \lambda^2 = \frac{2m}{\hbar^2}(E)
\]
Define \(z = \lambda \, a\) and \(z_0 = a\, \frac{\sqrt{2mV_0}}{\hbar}\)
\[
\Longrightarrow \quad \kappa^2 + \lambda^2 = \frac{2m}{\hbar^2}(V_0) = z_0^2/a^2 \quad \Longrightarrow \quad \kappa^2 a^2 = z_0^2 - z^2
\]
Finite well: bound states
Symmetric solutions
\[
\left\{
\begin{aligned}
\psi(x < -a) &= B e^{\kappa x}\\
\psi(-a < x < a) &= D \cos(\lambda x)\\
\psi(x > a) &= B e^{-\kappa x}\\
\end{aligned}
\right.
\]
with \(F = B \quad \& \quad \kappa = \lambda \tan \lambda a\)
\[
\Longrightarrow \tan z = \sqrt{\frac{z_0^2}{z^2} - 1}
\]
Asymmetric solutions
\[
\left\{
\begin{aligned}
\psi(x < -a) &= B e^{\kappa x}\\
\psi(-a < x < a) &= C \sin(\lambda x)\\
\psi(x > a) &= -B e^{-\kappa x}\\
\end{aligned}
\right.
\]
with \(F = -B \quad \& \quad \kappa = -\lambda \cot \lambda a\)
\[
\Longrightarrow -\cot z = \sqrt{\frac{z_0^2}{z^2} - 1}
\]
where we made use of:
\[
\kappa^2 a^2 = z_0^2 - z^2 \Longrightarrow \kappa/\lambda = \sqrt{z_0^2/z^2 - 1}
\]
Finite well: E for symmetric bound states
Symmetric solutions
\[
\left\{
\begin{aligned}
\psi(x < -a) &= B e^{\kappa x}\\
\psi(-a < x < a) &= D \cos(\lambda x)\\
\psi(x > a) &= B e^{-\kappa x}\\
\end{aligned}
\right.
\]
with \(F = B \quad \& \quad \kappa = \lambda \tan \lambda a\)
\[
\Longrightarrow \tan z = \sqrt{\frac{z_0^2}{z^2} - 1}
\]
Finite well: E for asymmetric bound states
Asymmetric solutions
\[
\left\{
\begin{aligned}
\psi(x < -a) &= B e^{\kappa x}\\
\psi(-a < x < a) &= C \sin(\lambda x)\\
\psi(x > a) &= -B e^{-\kappa x}\\
\end{aligned}
\right.
\]
with \(F = -B \quad \& \quad \kappa = -\lambda \cot \lambda a\)
\[
\Longrightarrow -\cot z = \sqrt{\frac{z_0^2}{z^2} - 1}
\]
Asymmetric energies approximately “shifted” to the right
Finite well: graphical energy-levels
- Symmetric and asymmetric solutions
- Energy levels at intersections of red curve with blue/green curves.
- Energies close to infinite well energies (vertical asymptotes):
\[
E_n = \frac{\hbar^2 z^2}{2m a^2} \lessapprox E^\infty_n = \frac{\hbar^2}{2m}\frac{n^2\pi^2}{(2a)^2}
\]
- \(z_0\) is intersection of red curve with x-axis
- Increasing \(z_0 \, \Longrightarrow \,\) more energy-levels
- Decreasing \(z_0 \, \Longrightarrow \,\) less energy-levels
- \(z_0 = \frac{\sqrt{2mV_0}}{\hbar} \, a \propto \sqrt{V_0} a\)
Finite well: numerical solutions
Numerically solving for \(z\):
\[
\begin{aligned}
\tan z &= \sqrt{z_0^2/z^2 - 1},\\
-\cot z &= \sqrt{z_0^2/z^2 - 1}\\
\end{aligned}
\]
Gives us:
- Energy levels \(E_n = \frac{\hbar^2\lambda^2}{2m} = \frac{\hbar^2 z^2}{2m a^2}\)
- Values for \(C\) or \(D\) as function of \(B\):
\[
\begin{aligned}
B &= D \cos(\lambda a) e^{\kappa a}\\
B &= -C \sin(\lambda a) e^{\kappa a}\\
\end{aligned}
\]
- Normalization \(\longrightarrow\) final unknown \(B\)
