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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:

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

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

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

We can abbreviate the mathematical operation in the square bracket as

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

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

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

We will discuss more about operators in the next chapter.

The Schrödinger Equation then has a simpler form:

$\hat H \psi = E \psi$

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:

$-\frac{\hbar^2}{2m} \frac{ { d}^{ 2} \psi }{ d { x}^{ 2} } = E \psi$

$\frac{ { d}^{ 2} \psi }{ d { x}^{ 2}} = - \frac{ 2mE}{ { \hbar}^{ 2} } \psi$

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

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

where the wave vector k is

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

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

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

Consequently,

$p = \hbar k$

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

$\lambda = \frac{ h}{ p}$

Time-dependent Schrödinger Equation

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

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

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

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

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

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

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

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,

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

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

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

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

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

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.

(insert figure)

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

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

The integral is over all space accessible to the particle.

In Cartesian coordinate

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

In spherical polar coordinate

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

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

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

(insert figures)

-- ThanhTruong - 12 Jul 2007

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