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r2 - 23 Jul 2007 - 21:49:34 - ThanhTruongYou are here: TWiki >

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Particle in One-Dimensional Box

Description: A particle of mass m moving freely along the X-axis between x = 0 and x = L, i.e. the particle is subjected to the potential

(insert figure)

$V(x) = \left \lbrace \begin{array}{l} \infty \; \; \; for \: x &lt; 0 \; \; \; (Region \: I) \\0 \; \; \; \; \; 0 \leq x \leq L \; \; (Region \: II)\\ \infty \; \; \;for \: x &gt; L \; \; \; (Region \: III)\end{array} \right.$

Note: Classically, one may relate this problem to a frictionless bead moving along a string between two points.

The general time-independent Schrödinger equation for one-dimensional system is given by

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

where $\psi (x)$ is the eigenfunction and E is the eigenvalue.

To solve this problem, we examine the wavefunction in three separate regions defined above.

From classical mechanics we know that the particle would not be in regions I and III. In quantum mechanics we need to examine the behavior of the wavefunction in those regions.

$\frac{ d^2 \psi }{ dx^2} - \lim_{ V \rightarrow \infty } \frac{ 2mV(x)}{ \hbar ^2} \psi (x) = - \frac{ 2mE}{ \hbar ^2} \psi (x)$

From Postulate 1, $\psi (x)$ must be finite. Hence, for a finite E, the right-hand side of the above equation is finite yet the left-hand side is infinite unless $\psi (x) = 0$ in these regions.

$\psi ^I (x) = \psi ^{III} (x) = 0$

In Region II, since V = 0, we need to worry on the kinetic energy term. The Schrödinger equation becomes

$\frac{ d^2 \psi }{ dx^2} = - \frac{ 2mE}{ \hbar ^2} \psi (x) = - k^2 \psi (x)$

where $k^2 = \frac{ 2mE}{ \hbar ^2}$ .

The general solution to the above ordinary second-order differential equation is

$\psi ^{II} (x) = A \sin (kx) + B\cos (kx)$

Also from Postulate 1, $\psi$ must be continuous and thus we must impose the following boundary conditions:

$\psi ^I (0) = \psi ^{II} (0) = 0$

$\psi ^{II} (L) = \psi ^{III} (L) = 0$

From $\psi ^{II} (0) = 0 \rightarrow B = 0 \; \; \; \Rightarrow \; \; \psi^{II} (x) = A \sin (kx)$

From $\psi ^{II} (L) = 0 \rightarrow \sin (kL) = 0$

This means $kL = n\pi$ where n = 0, 1, 2, 3,...

However, we need to rule out the case of n=0 because it implies $\psi = 0$ everywhere, or the particle does not exist. Thus, the only acceptable values of n are 1, 2, 3, ...

From $k^2 = \frac{ 2mE}{ \hbar ^2}$ and $kL = n\pi$ , we obtain

$E_n = \frac{ h^2n^2}{8mL^2 } \; \; \; \; n = 1, 2, 3, ...$

where is the $n^{th}$ energy level and n is the quantum number. Thus, the energy quantization arises from the boundary conditions.

To determine the constant A of the wavefunction, we use the normalization condition

$\int_{ - \infty }^{ \infty } \psi^* \psi dx = A^2 \int_{ 0}^{ L} \sin^2 (kx)dx = 1$

$\Rightarrow \; \; \frac{1}{2} A^2 L = 1 \; \; \Rightarrow \; \; A = \sqrt{ \frac{ 2}{ L} }$

The final solution to the one-dimensional partical in the box is

$E_n = \frac{ h^2n^2}{8mL^2 } \; \; \; \; n = 1, 2, 3, ...$

$\psi_n (x) = \sqrt{ \frac{ 2}{ L}} \sin \left ( \frac {n\pi x} {L} \right )$

(insert figure)

Observations and new concepts

Zero-point energy: From classical mechanics, the lowest energy for this system is zero, namely the particle at rest. However, the results show that in QM the lowest energy is not zero,

$E_1 = \frac {h^2}{8mL^2} \; \; \; \; \leftarrow \; \; the \:zero-point \: energy$

A possible explanation: Since $\psi^I = \psi^{III} = 0$ and $\psi^{II}$ must not be zero everywhere, the continuity condition of the wavefunction force it to curve and the curvature of the wavefunction, $\frac {d^2 \psi}{dx^2}$ , implies the non-zero kinetic energy even at the lowest level.

Probability density: From classical mechanics, the probability of finding the particle between dx around a location between 0 and L is uniform. However, in QM such a probability is given by

$\psi^* \psi dx = \frac{ 2}{ L} \sin^2 \left (\frac{ n \pi x_0}{ L} \right ) dx$

and is not uniform.

The Corresponding Principle: Notice that $\psi^2$ becomes more uniform as n increases and reflects the classical results. Thus, as the quantum number increases to infinity, quantum mechanics approaches the classical mechanics limits.

(insert figure)

Also note that $E_n \propto \frac{ 1}{ L^2}$ thus as the side of the box L gets larger the degree of energy quantization decreases and the system approaches the classical regime when the energy spacing $\Delta E &lt; kT$ .

-- ThanhTruong - 23 Jul 2007

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