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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\partial}{\partial t}|\Psi\rangle=\hat H|\Psi\rangle}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} is the imaginary unit,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar} is Planck's constant divided by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\pi} ,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} is the quantum mechanical state or wavefunction (expressed here in Dirac notation), and
${\hat {H}}$ 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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V(x)\psi=i\hbar\frac{\partial \psi}{\partial t} }

Note that solutions may be complex in nature.

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E = \frac{p^2}{2m} + V(x)}

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

$p={\frac {\hbar }{i}}{\frac {\partial }{\partial x}}$

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

${\hat {H}}={\frac {-\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+V(x)$

from which Schrodinger's equation and the eigenvalue problem ${\hat {H}}\Psi =E\Psi$ 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

$E\psi ={\hat {H}}\psi$

Here, E is energy, H is once again the Hamiltonian operator, and $\psi$ is the energy eigenstate for E.

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

Interpretation of the Wavefunction

The wavefunction is related to the probability of observing a particle at that location in space. More specifically, where the wavefunction has a greater value of $|\Psi |^{2}$ at one location, a, compared to another, b, then the particle is more likely to be observed at a than b.

Normalisation

Wavefunctions are said to be "normalisable" if the following integral exists:

$\int _{-\infty }^{\infty }|\Psi (x,t)|^{2}dx$

If that integral exists, then the wave function may be normalised by introducing a constant, $c$, such that

$\int _{-\infty }^{\infty }|c\Psi (x,t)|^{2}dx=|c|^{2}\int _{-\infty }^{\infty }|\Psi (x,t)|^{2}dx=1$

Note that $c$ may be complex. This will produce and new wavefunction, $c\Psi (x,t)$, which has the property that:

$P(a<x<b)=\int _{b}^{a}|c\Psi (x,t)|^{2}dx$

This is the probability of observing the particle between points a and b. For some wavefunctions, such as that of a free particle, the integral above does not exist and so they are said to be non-normalisable.

Examples for the time-independent equation

Free particle in one dimension

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

$\psi =Ae^{-ikx}$

with energy given by

$E={\frac {\hbar ^{2}k^{2}}{2m}}$

Physically, this corresponds to a wave travelling with a momentum given by $\hbar k$, 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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} }

This is also known as "the infinite square well". This means that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} must be zero outside the box. One can verify (by substituting into the Schrodinger equation) that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi=\sin(kx)}

is a solution if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=n\pi} 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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n=\frac{\hbar^2 k^2}{2m}=\frac{\hbar^2n^2\pi^2}{2m}}

Derivation

The particle cannot exits outside the box where the potential is infinite. Hence the wavefunction, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} , here must be zero outside the box. Now let us consider inside the box. The Schrodinger equation inside the potential well becomes:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle - \frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi}

Rearranging this equation give Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2 \psi}{dx^2} = - \frac{2mE}{\hbar^2} \psi}

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (x) = A \cos{\omega x} + B \sin{\omega x}}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B} are unknown constants and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega} is equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{\frac{2mE}{\hbar^2}}} . 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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (0) = 0 = A \cos{\omega 0} + B \sin{\omega 0} }

From this we can see that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} must be zero. Using the continuous constraint again, we see that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (a) = 0 = B \sin{\omega a} }

The continuous constraint is only satisfied when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega a = n \pi} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} is an integer. We also note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n=0} 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n < 0 } since these are not "new" solutions. As Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin{-n x} =-\sin{x}} , these are just the negative of solutions with positive Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} . By setting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega a = n \pi} , we can find the energy Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n} as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{\frac{2m E_n}{\hbar^2}} = n \pi}

And so: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n=\frac{\hbar^2n^2\pi^2}{2m}}

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

$U(x)={\begin{cases}{\sqrt {\frac {2}{a}}}\sin {\omega x}&\quad {\text{if }}x<0\\0&\quad {\text{otherwise}}\\\end{cases}}$

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega} is as above.

See Also

• Quantum Mechanics
• Differential equation