We want to solve the time independent Schrödinger equation
${\hat {H}}(x)\psi (x)=E\psi (x)$
$\left\lbrace {\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+V(x)\right\rbrace \psi (x)=E\psi (x)$ (Eq. 1)
for some specific case. We will consider a few different potential energies V(x) and see what the eigenvalues and eigenfunctions look like. We will also practice with some numerical exercises while making several observations on the behaviour of quantum particles. From now on, we may refer to the time independent Schrödinger equation as just the 'Schrödinger' equation.
Consider a particle of mass m which can only occupy the position between x=0 and x=L, and cannot escape from this portion of space. This is commonly known as the particle in a one dimensional box. A classical particle would go back and forth between the two boundaries. The potential energy for such a system can be written as
V(x) = 0 when 0<x<L; (Eq. 2)
V(x) = +∞ elsewhere
The wavefunction must be zero for x<0 and x>L. The wavefunction must also be continuous and so it must be that
ψ(0) = 0 and ψ(L) = 0 (Eq. 3)
These two conditions are known as boundary conditions in the theory of differential equations. Very often you can find many solutions to a differential equation but only a few would satisfy the boundary conditions.
If we consider only 0<x<L then the Schrödinger equation looks like
${\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\psi (x)}{\partial x^{2}}}=E\psi (x)$ or ${\frac {\partial ^{2}\psi (x)}{\partial x^{2}}}={\frac {2mE}{\hbar ^{2}}}\psi (x)$ (Eq. 4)
We need to find a function whose second derivative is proportional to the function itself and multiplied by a negative constant. As we know that
${\frac {\partial ^{2}\cos ax}{\partial x^{2}}}=a^{2}\cos ax$ and ${\frac {\partial ^{2}\sin ax}{\partial x^{2}}}=a^{2}\sin ax$ (Eq. 5)
it looks like a couple of possible solutions to Equation 1 are ψ(x) = cos(ax) or ψ(x) = sin(ax) where $a={\sqrt {\frac {2mE}{\hbar ^{2}}}}$. With differential equations, if you find two solutions, any linear combination of these solutions is still a solution. So the general solution of Equation 4 is
$\psi (x)=A\cos \left({\sqrt {\frac {2mE}{\hbar ^{2}}}}x\right)+B\sin \left({\sqrt {\frac {2mE}{\hbar ^{2}}}}x\right)$ (Eq. 6)
where A and B are constants that can take any values. Equation 6 is the solution if we ignore the boundary conditions (Equation 3). However, if we impose that ψ(0) = 0 we immediately see that it must be A = 0. If we impose ψ(L) = 0 we get $\psi (L)=B\sin \left({\sqrt {\frac {2mE}{\hbar ^{2}}}}L\right)=0$ (Eq. 7)
At this point, you need to remember that sin(y) = 0 if y = 0, ±π, ±2π,..., ±nπ and so it must be ${\sqrt {\frac {2mE}{\hbar ^{2}}}}L=n\pi$ where n = 1, 2,... (Eq. 8)
This is only possible if the energy E takes discrete values:
${\sqrt {\frac {2mE}{\hbar ^{2}}}}={\frac {n\pi }{L}}\to {\frac {2mE}{\hbar ^{2}}}={\frac {n^{2}\pi ^{2}}{L^{2}}}\to E={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}$ (Eq. 9)
The eigenvalues (energies) of the particle in the box Hamiltonian are therefore:
$E_{n}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}$ where n = 1, 2,..., (Eq. 10)
The index n is called the quantum number as it is a label of the energy level. The eigenfunctions (wavefunction) of the same Hamiltonian are:
$\psi _{n}(x)=B\sin \left({\sqrt {{\frac {2m}{\hbar ^{2}}}{\frac {n^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}}}x\right)=B\sin \left({\frac {n\pi x}{L}}\right)$ (Eq. 11)
Note that the wavefunctions are also labelled with the quantum number. B is any arbitrary constant. You must be able to draw these wavefunctions and the corresponding energy levels.
Probability and normalization[edit  edit source]
Any eigenfunction can be multiplied by any constant and it is still the same eigenfunction. It is customary to multiply them by a constant such that the integral of ψ_{n}(x)² over all space is unity. The wavefunction is said to be normalized if it has the following property:
$\int _{\infty }^{+\infty }\psi _{n}(x)^{2}\,dx=1$ (Eq. 12)
When a wavefunction is normalized, ψ_{n}(x)²dx is the probability of finding the particle in the interval between x and x+dx (Equation 2 Lesson 2)
In the case of the particle in the box, the normalized wavefunctions are:
$\psi _{n}(x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)$ (Eq. 13)
Some observations to keep in mind[edit  edit source]
The regions where the wavefunction is zero are called nodes (there are nodal points in 1D and nodal planes in 3D). Almost always the number of nodes increases with the energy. For the particle in the box, there are no nodes in the ground state (n = 1), 1 node for n = 2, 2 nodes for n = 3, etc.)
The energy levels are discrete because of the boundary condition (without them all the values of the energy were allowed  see question 8 below).
The particle in the box is a model system for all quantum mechanical systems. Whenever a particle is confined, discrete levels appear. It is possible to understand qualitatively many phenomena by just considering the particle in the box.
Exercise
 Calculate the lowest three energy levels of a particle of mass 10^{26} Kg in a box of length L = 10^{9} m.
 Calculate the lowest two energy levels (in eV) of an electron in a 2 Å long onedimensional box.
 Plot $\psi _{n}(x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)$ and its square ψ_{n}(x)^{2} for n=1,2,3,4 for 0<x<L
 State for which values of x the probability of finding the particle is maximum (for the one dimensional particle in the box) if the system is in state n=1, n=2, or n=3.
 Show that if ψ_{n}(x) is an eigenfunction of the Hamiltonian, Cψ_{n}(x) is also an eigenfunction (where C is any constant).
 Show that ${\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)$ is normalized. Is it always possible to normalize a wavefunction by multiplying it by an appropriate constant?
 If ψ_{n}(x) is normalized, ψ_{n}(x)*ψ_{n}(x)=ψ_{n}² is the probability density of finding the particle around x and $\int _{a}^{b}\psi _{n}(x)^{*}\psi _{n}(x)\,dx$ is the probability of finding the particle in the region between a<x<b (for any one dimensional system). Calculate the probability of finding a particle between x=0 and x=L/4 for a particle in a box in state n.
 A free particle is a particle without any interactions with potential energy V(x)=0 everywhere.
 Write the Hamiltonian for this system.
 Show that ψ(x) = e^{ikx} is an eigenfunction of this Hamiltonian.
 Find the eigenvalue corresponding to the eigenfunction e^{ikx}.
 Is the energy of the free particle quantized?
Solutions

 ${\frac {\pi ^{2}\hbar ^{2}}{2mL^{2}}}={\frac {(3.14)^{2}(1.06\cdot 10^{34})^{2}}{2\cdot 10^{26}\cdot (10^{9})^{2}}}=5.49\cdot 10^{23}{\mbox{J}}$
 $E_{1}={\frac {1^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}=$
 $E_{2}={\frac {2^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}=2.196\cdot 10^{24}$
 $E_{3}={\frac {3^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}=4.941\cdot 10^{24}$
 ${\frac {\pi ^{2}\hbar ^{2}}{2m_{e}L^{2}}}={\frac {(3.14)^{2}(1.06\cdot 10^{34})^{2}}{2\cdot 9.12\cdot 10^{31}\cdot (2\cdot 10^{10})^{2}}}=1.52\cdot 10^{18}{\mbox{J}}=9.5{\mbox{eV}}$ ; E_{1} = 9.5 eV ; E_{2} = 38 eV.
 Graph
 For n=1, x=L/2; for n=2, x=L/4 and 3L/4; for n=3, x=L/6 and L/2 and 5L/6
 ${\begin{matrix}{\hat {H}}(x)\psi (x)&=&E\psi (x)\\C{\hat {H}}(x)\psi (x)&=&CE\psi (x)\\{\hat {H}}(x)\left[C\psi (x)\right]&=&E\left[C\psi (x)\right]\end{matrix}}$
 Since you can multiply a wavefunction by any arbitrary constant (as seen in question 5) you can normalize any function such that the integral $\int \psi ^{*}\psi \,d\tau =Q$ can be calculated. The functions are normalized if they are multiplied by ${\frac {1}{\sqrt {Q}}}$
 The probability $\int _{a}^{b}\psi _{n}(x)^{*}\psi _{n}(x)\,dx$ for the ground state (n=1) and between 0 and L/4 (a=0, b=L/4) is given by $\int _{0}^{L/4}\psi _{1}(x)^{*}\psi _{1}(x)\,dx$ where $\psi _{1}={\sqrt {\frac {2}{L}}}\sin \left({\frac {\pi x}{L}}\right)$. The integral solved explicitly is: $\int _{0}^{L/4}\psi _{1}(x)^{*}\psi _{1}(x)\,dx={\frac {2}{L}}\int _{0}^{L/4}\sin \left({\frac {\pi x}{L}}\right)^{2}\,dx={\frac {2}{L}}\left[{\frac {1}{2}}x{\frac {L}{4\pi }}\sin {\frac {2\pi x}{L}}\right]={\frac {1}{4}}{\frac {1}{2\pi }}\sin {\frac {2\pi }{4}}\approx 0.091$

 ${\hat {H}}={\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}$
 ${\hat {H}}e^{ikx}={\frac {hbar^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}e^{ikx}={\frac {\hbar ^{2}k^{2}}{2m}}e^{ikx}$
 ${\frac {\hbar ^{2}k^{2}}{2m}}$
 No. Without boundary conditions k can take any values and so can the energy.


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