Stationary Solutions & Energy levels
Solving the 1D Schrodinger equation
The Schrodinger equation was given by:
\[
i \hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} + V(x,t) \Psi(x,t)
\]
- The complex wave function \(\Psi(x, t)\) is not observable
- Potential energy: \(V \rightarrow V(x,y,z,t)\)
- \(\hbar = \frac{h}{2\pi} = 1.055 \times 10^{-34}\) J s
- Probability to find particle in \(x\) at time \(t\) given by \(|\Psi(x, t)|^2\):
\[
P(x \in [a,b]) = \int_a^b |\Psi(x, t)|^2 dx
\]
Solving the 1D Schrodinger equation
- Wave function \(\Psi(x, t)\) defines \(|\Psi(x, t)|^2\)
- How to calculate \(\Psi(x, t=0)\)?
- Evolution in time of \(\Psi(x, t)\)?
\[
\begin{aligned}
i \hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2}\\
\\
\quad + V(x,t) \Psi(x,t)\quad\\
\end{aligned}
\]
How do we solve for given \(V(x,t)\) ?
Solving the 1D Schrodinger equation
\[
i \hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} + V(x,t) \Psi(x,t)
\]
How do we solve this equation for given \(V(x,t)\) ?
- Assume \(V(x, t)\) independent of time: \(V(x) \leftarrow V(x, t)\)
- Solve by separation of the variables \(\Psi(x,t) = \psi(x) \phi(t)\)
\[
i \hbar \frac{\partial (\psi(x)\phi(t))}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2( \psi(x)\phi(t))}{\partial x^2} + V(x) \psi(x)\phi(t)
\]
Solving the 1D Schrodinger equation
\[
i \hbar \frac{\partial (\psi(x)\phi(t))}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2( \psi(x)\phi(t))}{\partial x^2} + V(x) \psi(x)\phi(t)
\]
\[
\Rightarrow i \hbar \psi(x) \frac{\partial \phi(t)}{\partial t} = -\phi(t)\frac{\hbar^2}{2m} \frac{\partial^2\psi(x)}{\partial x^2} + V(x) \psi(x)\phi(t)
\]
Divide the equation by \(\Psi(x, t) = \psi(x) \phi(t)\)
\[
\Rightarrow i \hbar \frac{1}{\phi(t)} \frac{\partial \phi(t)}{\partial t} = -\frac{\hbar^2}{2m}\frac{1}{\psi(x)} \frac{\partial^2\psi(x)}{\partial x^2} + V(x)
\]
\(\longrightarrow\) the left hand side depends only on \(x\) and the right hand side only on \(t\).
\[
\Rightarrow i \hbar \frac{1}{\phi(t)} \frac{\partial \phi(t)}{\partial t} = -\frac{\hbar^2}{2m}\frac{1}{\psi(x)} \frac{\partial^2\psi(x)}{\partial x^2} + V(x) = \textrm{constant } E
\]
Time-dependence & Stationary equation
\[
i \hbar \frac{1}{\phi(t)} \frac{\partial \phi(t)}{\partial t} = -\frac{\hbar^2}{2m}\frac{1}{\psi(x)} \frac{\partial^2\psi(x)}{\partial x^2} + V(x) = E
\]
\(\longrightarrow\) System of 2 ordinary differential equations:
\[
\left\{
\begin{aligned}
{\color{blue}{-\frac{\hbar^2}{2m} \frac{1}{\psi(x)}\frac{d^2\psi(x)}{d x^2} + V(x) }} &{\color{blue}{= E}}\\
{\color{red}{i\hbar\frac{1}{\phi(t)}\frac{d \phi(t)}{d t}}} &{\color{red}{= E}}\\
\end{aligned}
\right.
\]
IF we can solve both equations \(\Longrightarrow\) \(\Psi(x,t) = \color{blue}{\psi(x)} \color{red}{\phi(t)}\) is a solution
Time evolution
- Solving the equation for \(\phi(t)\)
\[
{\color{red}{i\hbar\frac{1}{\phi(t)}\frac{d \phi(t)}{d t}}} {\color{red}{= E}} \quad\Rightarrow\quad
\frac{d \phi(t)}{d t}=-\frac{i}{\hbar}E \phi(t)
\]
1st order differential equation with general solution:
\[
\phi(t) = C \exp(- i E t /\hbar)
\]
Full solution of the form (C is absorbed):
\[
\Psi(x,t) = \psi(x) \phi(t) = \psi(x) \exp(- i E t /\hbar)
\]
Notice that the probability \(|\Psi(x, t)|^2 = |\psi(x)|^2\) is independent of \(t\)
Time-independent equation
Time-independent Schrodinger equation (TISE):
\[
\color{blue}{-\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{d x^2} + V(x) \psi(x) = E \psi(x)}
\]
Or we can write
\[
\hat{H} \psi = E \psi \quad \textrm{with Hamiltonian }\,\, \hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{d x^2} + V(x)
\]
The expectation value of \(\hat{H}\) is:
\[
\langle \hat{H} \rangle = \int \Psi^* \hat{H} \Psi dx = \int \Psi^* E \Psi dx = E \int |\Psi|^2 dx = E \int |\psi|^2 dx = E
\]
General Solution of the TDSE
- From the theory of differential equations:
- The general solution is a linear superposition of solutions \(\{\psi_n(x)\} = \psi_1(x), \psi_2(x), \psi_3(x), \dots\)
- Independent solutions
- Separate energies \(\{E_n\}\) for corresponding \(\{\psi_n(x)\}\)
- Solutions form an infinite and complete basis
\[
\Psi(x, t) = \sum_{n = 1}^\infty c_n \psi_n(x) \, e^{-iE_n t/\hbar}
\]
Notice: General probability \(|\Psi(x, t)|^2\) does depend on time
General Solution of the TDSE
\[
\Psi(x, t) = \sum_{n = 1}^\infty c_n \psi_n(x) \, e^{-iE_n t/\hbar}
\]
One can proof that \(|c_n|^2\) is the probability to measure energy as \(E_n\) (Griffith’s Chapter 3):
\[
\langle \hat{H} \rangle = \int \Psi^* \hat{H} \Psi dx = \sum_{n = 1}^\infty |c_n|^2 E_n \quad \textrm{ and } \quad \sum_{n = 1}^\infty |c_n|^2 = 1
\]
Potential energy function V(x)
\[
\hat{H}\psi(x) = -\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{d x^2} + V(x) \psi(x) = E \psi(x)
\]
- Potential energy \(V(x)\) is linked to force \(F = - \frac{\partial V}{\partial x}\)
\(\Longrightarrow\) if \(V(x)\) is a constant corresponds to zero force
\(\Longrightarrow\) A linear \(V(x)\) corresponds to a constant force
\(\Longrightarrow\) A parabolic \(V(x)\) corresponds to a linear force (like a spring)
Square potential energy well
Infinite well
- Inside the well a particle can exist
- Outside the well the potential is infinite
\[
\left\{
\begin{aligned}
V(x < 0) & = \infty \\
V(0 < x < L) & = 0 \\
V(x > L) & = \infty \\
\end{aligned}
\right.
\]
- Task: solve the stationary Schrodinger equation for \(V(x)\)
\[
-\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{d x^2} + V(x) \psi(x) = E \psi(x)
\]
Infinite well: Solution in the well
- Particles outside would have infinite energy
- Wave function \(\psi(x)\) should be zero outside
- Assume \(\psi(0) = \psi(L) = 0\) \(\longleftarrow\) \(\psi(x)\) ctu
- Inside the well \(V(x) = 0\):
\[
-\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{d x^2} = E \psi(x)
\]
General solution:
\[
\psi(x) = A \cos(k x) + B \sin(k x)
\]
Infinite well: Solution in the well
\[
-\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{d x^2} = E \psi(x)
\]
\[
\psi(x) = A \cos(k x) + B \sin(k x)
\]
- with \(A\) and \(B\) complex numbers
- \(k = \sqrt{2mE / \hbar^2}\) a complex number
Apply BC’s \(\psi(0) = \psi(L) = 0\):
\[
\psi(0) = 0 \quad\Rightarrow\quad A = 0 \\
\]
\[
\psi(x) = B \sin(k\,x) \\
\]
Infinite well: Energies
\[
\psi(x) = B \sin(k\,x) \\
\]
Apply the other BC: \(\quad\psi(L) = 0\):
\[
k_n = \sqrt{2mE_n/\hbar^2} = n \pi / L \\
\]
\[
\Longrightarrow \left\{
\begin{aligned}
\psi_n(x) &= A_n \sin\left(\frac{n\pi x}{L}\right)\\
E_n &= \frac{\hbar^2 k_n^2}{2m} = \frac{\hbar^2}{2m} \left( \frac{n\pi}{L} \right)^2
\end{aligned}
\right.
\]
Infinite well
\[
\psi(x) = B \sin(k\,x) \\
\]
Apply the other BC: \(\quad\psi(L) = 0\):
\[
k_n = \sqrt{2mE_n/\hbar^2} = n \pi / L \\
\]
\[
\Longrightarrow \left\{
\begin{aligned}
\psi_n(x) &= A_n \sin\left(\frac{n\pi x}{L}\right)\\
E_n &= \frac{\hbar^2 k_n^2}{2m} = \frac{\hbar^2}{2m} \left( \frac{n\pi}{L} \right)^2
\end{aligned}
\right.
\]
Infinite well: Normalization
\[
\begin{aligned}
\psi_n(x) &= A_n \sin\left(\frac{n\pi x}{L}\right)\\
\end{aligned}
\]
- Obtain \(A_n\) from normalization \(\int|\psi|^2 = 1\)
\[
1 = \int_0^L |A_n|^2 \left|\sin\left(\frac{n\pi x}{L}\right)\right|^2 dx = \frac{|A_n|^2 L}{2}
\]
\[
\Longrightarrow \quad |A_n|^2 = \frac{2}{L} \Rightarrow |A_n| = \sqrt{\frac{2}{L}}
\]
\[
\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)
\]
Infinite well: summary
\[
\left\{
\begin{aligned}
\psi_n(x) &= \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\\
&\\
E_n &= \frac{\hbar^2 k_n^2}{2m} = \frac{\hbar^2}{2m} \left( \frac{n\pi}{L} \right)^2\\
&\\
n &= 1, 2, 3, 4, \dots\\
\end{aligned}
\right.
\]
Plot shows the wave function (\(\psi\), grey), probability (\(|\psi|^2\), color) for first 3 eigenstates
Eigenenergies and eigenstates
\[
\begin{aligned}
\textrm{Eigenstates} \quad &\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\\
&\\
\textrm{Eigenenergies} \quad &E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{\hbar^2}{2m} \left( \frac{n\pi}{L} \right)^2\\
&\\
n &= 1, 2, 3, 4, \dots\\
\end{aligned}
\]
- Lowest state \(n = 1\) we call ground state
- Higher states \(n > 1\) are excited states
- Parity of wave functions is either:
- Even (\(n = 1, 3, 5, \dots\))
- Odd (\(n = 2, 4, 6, \dots\))
Properties
\[
\begin{aligned}
\textrm{Eigenstates} \quad &\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\\
\end{aligned}
\]
The eigenstates are orthonormal:
\[
\int \psi_m(x)^*\, \psi_n(x) dx = \delta_{nm}
\]
Eigenstates form a complete basis
Every \(f(x)\) we can expand as a series:
\[
f(x) = \sum_{n=1}^\infty c_n \psi_n(x) = \sqrt{\frac{2}{L}} \sum_{n=1}^\infty c_n \sin\left(\frac{n\pi x}{L}\right)
\]
Properties of stationary eigenstates
\[
\begin{aligned}
\psi_n \textrm{ are orthonormal}\quad & \int \psi_m(x)^* \, \psi_n(x) \, dx = \delta_{mn}\\
\psi_n \textrm{ form a complete basis}\quad & f(x) = \sum_{n = 1}^\infty c_n \psi_n(x) \qquad \forall f(x)\\
\textrm{Coefficients $c_n$ are given by}\quad & c_n = \int \psi_n(x)^* \, f(x) \, dx\\
\end{aligned}
\]
Proof of last property:
\[
\begin{aligned}
\int \psi_m(x)^* \, f(x) \, dx & = \int \psi_n(x)^* \, \sum_{n = 1}^\infty c_n \psi_n(x) \, dx\\
& = \sum_{n = 1}^\infty c_n \int \psi_m(x)^* \, \psi_n(x) \, dx = \sum_{n = 1}^\infty c_n \delta_{mn} = c_m
\end{aligned}
\]
Stationary solution of the TISE
For the infinite well
\[
\psi(x) = \sqrt{\frac{2}{L}} \sum_{n = 1}^\infty c_n \sin\left(\frac{n\pi}{L} x\right)
\]
Example state: \[
\left\{
\begin{aligned}
& c_1 = 4/5,\\
& c_2 = \sqrt{1 - c_1^2} = 3/5,\\
& n > 2 \longrightarrow c_n = 0\\
\end{aligned}
\right.
\]
- How does the wave function (\(\psi\), color) and the probability (\(|\psi|^2\), gray) look?
- What if we let time evolve?
Infinite well: solution of the TDSE
Adding time evolution
\[
\begin{aligned}
\Psi(x, t) & = \sum_{n = 1}^\infty c_n \psi_n(x) \, e^{-iE_n t/\hbar}\\
& \textrm{with }\sum_{n = 1}^\infty |c_n|^2 = 1\\
\end{aligned}
\]
Coefficients \(|c_n|^2\) give the probability to measure energy as \(E_n\):
\[
\langle \hat{H} \rangle = \int \Psi^* \hat{H} \Psi dx = \sum_{n = 1}^\infty |c_n|^2 E_n
\]
But \(\langle \hat{x} \rangle = \int x \Psi^* \Psi \, dx\) is not constant!
Expand a function in eigenstates
- Suppose we have a certain wave function
\[
f(x) = A\,((L/2)^4 - (x-L/2)^4), \,\, \textrm{with } x \in [0, L]
\]
- Normalization constant \(A = \sqrt{\frac{64}{45} \left(\frac{L}{2}\right)^9}\)
- Since \(f(0) = f(L) = 0\) we can expand \(f(x)\) in eigenstates of the infinite well
\[
\begin{aligned}
f(x) &= \sqrt{\frac{2}{L}} \sum_{n = 1}^\infty c_n \sin\left(\frac{n\pi}{L} x\right)\\
&\textrm{with} \quad c_n = \int_0^L \psi_n(x)^* \, f(x) \, dx\\
\end{aligned}
\]
A more complex example
![]()
Ammonia molecule has two possible geometries
- The ammonia molecule \(NH_3\) has two possible geometries
- Experiments tell that \(NH_3\) flips between states
- Possible by quantum tunneling
