The Schrödinger Equation

In quantum mechanics, the state of a system is defined by a mathematical function %$\Psi$%, called wavefunction.  %$\Psi$% is a function of the coordinates of all particles in the system and is also a function of time.

For simplicity, let examine the case where  %$\Psi$% is independent of time first.

Time-Independent Schrödinger Equation 

For a particle of mass m moving in an one-dimensional potential V(x) with an energy E, the wavefunction %$\psi$% of the particle satisfies the time-independent  Schrödinger Equation:

%MATHMODE{-\frac{\hbar^2}{2m} \frac{ { d}^{ 2} \psi }{ d { x}^{ 2} } + V(x) \psi = E \psi }%

 where  %$\hbar$% is %$\frac{ h}{ 2 \pi }$%. It can be rewritten as

%MATHMODE{ \left[ -\frac{\hbar^2}{2m} \frac{ { d}^{ 2} }{ d { x}^{ 2} } + V(x) \right ]\psi = E \psi}%

We can abbreviate the mathematical operation in the square bracket as

%MATHMODE{\hat H = -\frac{\hbar^2}{2m} \frac{ { d}^{ 2} }{ d { x}^{ 2} } + V(x)}%

%$\hat H$% is known as the Hamiltonian Operator.  In the general three-dimensional case, the Hamiltonian operator has the form:

%MATHMODE{\hat H = -\frac{\hbar^2}{2m} \left ( \frac{ { d}^{ 2} }{ d { x}^{ 2} } + \frac{ { d}^{ 2} }{ d { y}^{ 2} } + \frac{ { d}^{ 2} }{ d { z}^{ 2} }\right ) + V(\vec r) = -\frac{\hbar^2}{2m} { \nabla }^{ 2} + V(\vec r)}%

We will discuss more about operators in the next chapter.

The Schrödinger Equation then has a simpler form:

%MATHMODE{\hat H \psi = E \psi}%

Let take a closer look at the time-independent Schrödinger equation for a special case, a free particle where V(x) = 0.  The Schrödinger equation then is given by:

%MATHMODE{-\frac{\hbar^2}{2m} \frac{ { d}^{ 2} \psi }{ d { x}^{ 2} } = E \psi }% or %MATHMODE{\frac{ { d}^{ 2} \psi }{ d { x}^{ 2}} = - \frac{ 2mE}{ { \hbar}^{ 2} } \psi}%

The general solution to this ordinary second-order differential equation has the form

%MATHMODE{ \psi = { e}^{ ikx} = cos(kx) + isin(kx)}%

where the wave vector k is

%MATHMODE{k = \sqrt{ \frac{ 2mE}{ { \hbar }^{ 2} } } \; \; \; or \; \; \; E = \frac{ { k}^{ 2} { \hbar }^{ 2} }{ 2m}}%

 Since V=0, the particle only possesses the kinetic energy, and thus

%MATHMODE{E = \frac{ { k}^{ 2} { \hbar }^{ 2} }{ 2m} = \frac{ { p}^{ 2} }{ 2m} }%

Consequently,

%MATHMODE{p = \hbar k}%

Since the wavefunction %$\psi$% oscillates with %$ \lambda = \frac{ 2 \pi }{ k}$%, the above equation yields the de Broglie's relation

%MATHMODE{\lambda = \frac{ h}{ p}}%

Time-dependent Schrödinger Equation

The state of the system, %$\Psi$%, changes in time according to the time-dependent Schrödinger Equation:

%MATHMODE{\hat H \Psi = i \hbar \frac{ \partial \Psi }{ \partial t}}%

where %$i = \sqrt { -1}$%. Note that %$\Psi$% is used to denote the time-dependent whereas %$\psi$% for the time-independent wavefunction.

To solve the time-dependent Schrödinger Equation, we use the separation of variable technique.

First, assume %$\Psi (x,t) = \psi (x) \xi (t)$%, then substitute it into the above equation to yield

%MATHMODE{\hat H \left[ \psi (x) \xi (t)\right ] = i \hbar \frac{ \partial \left[ \psi (x) \xi (t)\right ] }{ \partial t}}%

%MATHMODE{ \xi (t) \hat H \psi (x) = i \hbar \psi (x) \frac{ d \xi (t)}{ dt} }%

Divide both sides by %$\psi(x) \xi (t)$%

%MATHMODE{ \frac{ 1}{ \psi (x)} \hat H \psi (x) = \frac{ i \hbar }{ \xi (t)} \frac{ d\xi(t)}{ dt}}%

Since the left side of the above equation depends only on x while the right side only on t, in order for it to be true for all x and t, it must be equal to some constant E. From the left-side, we obtain

%$\frac{ 1}{ \psi (x)} \hat H \psi (x) = E \; \; \; \; or \; \; \; \; \hat H \psi = E \psi \;\; \; \leftarrow}$% the time-independent Schrodinger equation

From the right-side,

%MATHMODE{ \frac{ i \hbar }{ \xi(t)} \frac{ d \xi(t)}{ dt} = E \; \; \; \; or \; \; \; \; \frac{ d \xi(t)}{ dt} = - \frac{ iE}{ \hbar } \xi(t) }%

This is a ordinary first-order differential equation where solution has the form

%MATHMODE{\xi(t) = C { e}^{ - \frac{ iEt}{ \hbar } }}%

where C is a constant. Let take C=1, then

%MATHMODE{\Psi (x,t) = \psi (x) { e}^{ - \frac{ iEt}{ \hbar } }}%

This means that to obtain the time-dependent wavefunction, we need to solve the time-independent Schrödinger for %$\psi$% and E, then the time-dependent wavefunction can be obtained from the above equation.

Significance of the Wavefunction

If the amplitude of the wavefunction of a particle is %$\psi$% at some point %$\vec r$% in space, then the probability of finding the particle in an infinitesimal volume dv = dxdydz at the point %$\vec r$%  is proportional to %${\psi}^{*}\psi dv$%. %${\psi}^{*}\psi $% is called the probability density.

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Note that if %$\psi$% is the solution of the Schrödinger equation, then any %$N\psi$% wavefunction, where N is a constant, is also a solution of that equation. Since the probability of finding the particle somewhere is space is 1, we can find a constant N, called the Normalization Constant , such that

%MATHMODE{ \int {(N\psi)}^{ *} (N\psi) d \tau = 1}%

The integral is over all space accessible to the particle.

In Cartesian coordinate

%MATHMODE{ \int d \tau = \int_{ - \infty }^{ \infty } dx \; \int_{ - \infty }^{ \infty } dy \; \int_{ - \infty }^{ \infty } dz}%

In spherical polar coordinate

%MATHMODE{ \int d \tau = \int_{ 0 }^{ \infty } { r}^{ 2}dr \; \int_{ 0}^{ \pi } sin \theta d \theta \; \int_{ 0}^{ 2 \pi } d \phi }%

Finally, in order for the wavefunction to have any physical interpretation, it must

1. be continuous,
2. have continuous first-derivative
3. be single-valued,
4. be square integrable, namely %$\int {\psi}^{*}\psi d\tau$% is finite.

(insert figures)

-- ThanhTruong - 12 Jul 2007