Schrodinger equation

The Schrodinger equation is a linear differential equation used in various fields of physics to describe the time evolution of quantum states. It is a fundamental aspect of quantum mechanics. The equation is named for its discoverer, Erwin Schrodinger.

Mathematical forms

General time-dependent form

The Schrodinger equation may generally be written

<math>i\hbar\frac{\partial}{\partial t}|\Psi\rangle=\hat H|\Psi\rangle</math>

where <math>i</math> is the imaginary unit,
<math>\hbar</math> is Planck's constant divided by <math>2\pi</math>,
<math>|\Psi\rangle</math> is the quantum mechanical state or wavefunction (expressed here in Dirac notation), and
<math>\hat H</math> is the Hamiltonian operator.

The left side of the equation describes how the wavefunction changes with time; the right side is related to its energy. For the simplest case of a particle of mass m moving in a one-dimensional potential V(x), the Schrodinger equation can be written

<math> -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V(x)\psi=i\hbar\frac{\partial \psi}{\partial t} </math>

Derivation

The quickest and easiest way to derive Schrodinger's equation is to understand the Hamiltonian operator in quantum mechanics. In classical mechanics, the total energy of a system is given by

<math> E = \frac{p^2}{2m} + V(x)</math>

where p is the momentum of the particle and V(x) is its potential energy. Applying the quantum mechanical operator for momentum:

<math> p = \frac{\hbar}{i}\frac{\partial}{\partial x} </math>

and subbing into the classical mechanical form for energy, we get the same Hamiltonian operator in quantum mechanics:

<math> \hat H = \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) </math>

from which Schrodinger's equation and the eigenvalue problem <math> \hat H\Psi = E\Psi</math> can be easily seen.

Eigenvalue problems

In many instances, steady-state solutions to the equation are of great interest. Physically, these solutions correspond to situations in which the wavefunction has a well-defined energy. The energy is then said to be an eigenvalue for the equation, and the wavefunction corresponding to that energy is called an eigenfunction or eigenstate. In such cases, the Schrodinger equation is time-independent and is often written

<math>E\psi=\hat H\psi</math>

Here, E is energy, H is once again the Hamiltonian operator, and <math>\psi</math> is the energy eigenstate for E.

One example of this type of eigenvalue problem is an electrons bound inside an atom.

Examples for the time-independent equation

Free particle in one dimension

In this case, <math>V(x)=0</math> and so we see that the solution to the Schrodinger equation must be

<math>\psi=Ae^{-ikx}</math>

with energy given by

<math>E=\frac{\hbar^2 k^2}{2m}</math>

Physically, this corresponds to a wave travelling with a momentum given by <math>\hbar k</math>, where k can in principle take any value.

Particle in a box

Consider a one-dimensional box of width a, where the potential energy is 0 inside the box and infinite outside of it. We can describe this potential by:

<math>U(x) = \begin{cases}

      \infty &\quad\text{if } x < 0\\
      0 &\quad\text{if } 0 \le x \le a \\
      \infty &\quad\text{if } x > a \ 

\end{cases} </math>

This is also known as "the infinite square well". This means that <math>\psi</math> must be zero outside the box. One can verify (by substituting into the Schrodinger equation) that

<math>\psi=\sin(kx)</math>

is a solution if <math>k=n\pi</math> where n is any integer. Thus, rather than the continuum of solutions for the free particle, for the particle in a box there is a set of discrete solutions with energies given by

<math>E_n=\frac{\hbar^2 k^2}{2m}=\frac{\hbar^2n^2\pi^2}{2m}</math>

Derivation

The particle cannot exits outside the box where the potential is infinite. Hence the wavefunction, <math>\psi</math>, here must be zero outside the box. Now let us consider inside the box. The Schrodinger equation inside the potential well becomes:

<math>- \frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi</math>

Rearranging this equation give <math>\frac{d^2 \psi}{dx^2} = - \frac{2mE}{\hbar^2} \psi</math>

This has the same form as the basic equation for simple harmonic motion. Hence we can give the solution as

<math>\psi (x) = A \cos{\omega x} + B \sin{\omega x}</math>

where <math>A</math> and <math>B</math> are unknown constants and <math>\omega</math> is equal to <math>\sqrt{\frac{2mE}{\hbar^2}}</math>. To ind the value of the the constants, we substitute in boundary conditions. An additional constraint is that wavefunctions must be continious (they cannot suddenly change value). Hence

<math>\psi (0) = 0 = A \cos{\omega 0} + B \sin{\omega 0} </math>

From this we can see that <math>A</math> must be zero. Using the continuous constraint again, we see that

<math>\psi (a) = 0 = B \sin{\omega a} </math>

The continuous constraint is only satisfied when <math>\omega a = n \pi</math> where <math>n</math> is an integer. We also note that <math>n=0</math> is not a solution as the wavefunction would be zero everywhere and so the probability of finding the particle is 0. We also discard solutions for <math>n < 0 </math> since these are not "new" solutions. As <math> \sin{-n x} =-\sin{x}</math>, these are just the negative of solutions with positive <math>n</math>. By setting <math>\omega a = n \pi</math>, we can find the energy <math>E_n</math> as:

<math>\sqrt{\frac{2m E_n}{\hbar^2}} = n \pi</math>

And so: <math>E_n=\frac{\hbar^2n^2\pi^2}{2m}</math>

You may wonder what happens about the constant <math>B</math>. Since the Schrodinger equation is linear, <math>B</math> can take any value and the resulting wavefunction would be a solution. Since <math>|\psi (x)|^2</math> represents the probability of finding the particle at position <math>x</math> and across all space, this probability must equal 1, we can solve for <math>B</math>. We do this by integrating <math>|\psi (x)|^2</math> from <math>-\infty</math> to <math> + \infty</math>. Therefore the full solution is:

<math>U(x) = \begin{cases}

      \sqrt{\frac{2}{a}} \sin{\omega x} &\quad\text{if } x < 0\\
      0 &\quad\text{otherwise} \\

\end{cases} </math>

where <math>\omega</math> is as above.