Consider a particle with energy E bound to a finite square...

Consider a particle with energy $E$ bound to a finite square well of height $U$ and width $2 L$ situated on $-L \leq$ $x \leq+L .$ Because the potential energy is symmetric about the midpoint of the well, the stationary state waves will be either symmetric or antisymmetric about this point. (a) Show that for $E<U,$ the conditions for smooth joining of the interior and exterior waves lead to the following equation for the allowed energies of the symmetric waves: $$ k \tan k L=\alpha \quad \text { (symmetric case) } $$ where $\alpha=\sqrt{\left(2 m / \hbar^{2}\right)(U-E)}$ and $k=\sqrt{2 m E / \hbar^{2}}$ is the wavenumber of oscillation in the interior. (b) Show that the energy condition found in (a) can be rewritten as $$ k \sec k L=\frac{\sqrt{2 m U}}{\hbar} $$ Apply the result in this form to an electron trapped at a defect site in a crystal, modeling the defect as a square well of height 5 eV and width $0.2 \mathrm{nm}$. Solve the equation numerically to find the ground-state energy for the electron, accurate to ±0.001 eV.

Consider a particle with energy $E$ bound to a finite square well of height $U$ and width $2 L$ situated on $-L \leq$ $x \leq+L .$ Because the potential energy is symmetric about the midpoint of the well, the stationary state waves will be either symmetric or antisymmetric about this point. (a) Show that for $E<U,$ the conditions for smooth joining of the interior and exterior waves lead to the following equation for the allowed energies of the symmetric waves:
$$
k \tan k L=\alpha \quad \text { (symmetric case) }
$$
where $\alpha=\sqrt{\left(2 m / \hbar^{2}\right)(U-E)}$ and $k=\sqrt{2 m E / \hbar^{2}}$ is the
wavenumber of oscillation in the interior. (b) Show that the energy condition found in (a) can be rewritten as
$$
k \sec k L=\frac{\sqrt{2 m U}}{\hbar}
$$
Apply the result in this form to an electron trapped at a defect site in a crystal, modeling the defect as a square well of height 5 eV and width $0.2 \mathrm{nm}$. Solve the equation numerically to find the ground-state energy for the electron, accurate to ±0.001 eV.

Key Concepts

Numerical Methods for Solving Transcendental Equations

Since the energy quantization conditions in problems like the finite square well lead to transcendental equations, numerical methods such as the bisection method, Newton-Raphson method, or graphical techniques are employed to determine the energy eigenvalues with desired accuracy. These techniques are essential tools in quantum mechanics for obtaining approximate solutions when analytical methods fail.

Wavefunction Matching and Boundary Conditions

A critical aspect in solving quantum mechanical problems is ensuring the continuity of the wavefunction and its derivative at boundaries where the potential changes. In the case of the finite square well, matching the oscillatory interior solution with the exponentially decaying exterior solution at the well boundaries leads to conditions, often transcendental in nature, that determine the allowed energy levels.

Transcendental Equations in Energy Quantization

The smooth matching of the interior and exterior wavefunctions in a finite potential well typically results in transcendental equations that relate the wave numbers (or energies) inside and outside the well. These equations cannot be solved analytically in general and require numerical methods for finding the energy eigenvalues corresponding to bound states.

Quantum Bound States and Energy Quantization

In quantum systems with confining potentials such as the finite square well, only specific energy levels (bound states) are allowed. These quantized energy levels result from the requirement that the wavefunctions remain normalizable and satisfy appropriate boundary conditions at the interfaces between different potential regions.

Finite Square Well

The finite square well is a potential energy model used in quantum mechanics to study particles confined within a region by a potential barrier of finite height. In this model, a particle's wavefunction is oscillatory inside the well and decays exponentially outside, reflecting the fact that the particle has a nonzero probability of tunneling into the barrier. This model is fundamental for understanding bound states and tunneling phenomena.

Symmetry and Parity in Wavefunctions

For potentials that are spatially symmetric, such as the finite square well centered at the origin, the eigenfunctions of the Hamiltonian can be taken to have definite parity—either symmetric (even) or antisymmetric (odd). Exploiting this symmetry simplifies the analysis of boundary conditions and the derivation of energy quantization conditions.

Transcript

00:01 Wavelength allowed a bound particle are those of a typical standing wave which is given by lambda is equal to 2 l by n where l is the length of its home given that lambda is also equals to h by p we would have the momentum equals to n h by 2l and the kinetic energy equals to p squared by 2l thus it would be n square h square by 8m .2.

00:41 These are the actually the correct infinite well energies for the argument is perfectly valid when the potential energy is 0 inside the well and l is strictly constant but it is a pretty good guide to how the energy should go in other cases.

01:08 The length l allowed the wave should be roughly the region classically allowed to the particle which depends on the height of the total energy of total energy e relative to the potential energy potential energy given by u.

01:31 The wall is the classical turning point where there is no kinetic energy where e is equal to u the wall treating it as essentially one -dimensional problem apply these arguments to the hydrogen atom potential energy find the location of r of the classical turning point in terms of e use twice this distance for l twice of for finding the l and from this obtain an expression for the expected average in terms of e for the average potential use this value use this value at half the distance from the origin to the turning point in terms of e then write out the expected average total energy and solved for e.

02:53 The expression for electric potential, u is equal to minus 1 by 4 ,5, epsilon not, e squared by r.

03:01 Here, e is the magnitude of charge, how is the distance between two charges, epsilon not is the permittivity of free space.

03:08 The quantized kinetic energy, kinetic energy, for a particle of mass m in a box of length l is equal to n square, h squared by 8 ml squared here n is the principal quantum number and h being the plank constant the turning point for the system is at distance r r1 where there is no kinetic energy left and the total energy is just equals to the potential energy there is the expression for the total energy, e is equal to minus 1 by 4 pi epsilon not e squared by r1...

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