MODULE 7 | unit 7 solution of wave equation

PHY

MODULE 7

SOLUTION OF WAVE EQUATION

7.1 Solution of Stationary-State Schrodinger Equation for One Dimensional Problems–Particle inA Box:

Define the potential energy:

A particle in a 1D infinite potential well of dimension LL.

 The potential energy is 0 in the box (V=0 for 0<x<L) and drives to eternity at the walls of the box (V=∞ for x<0 or x>L).

 We accept the walls have infinite potential energy to confirm that the particle has zero chance of existence at the walls or outside the box.

 Undertaking significantly simplifies our later mathematical designs as we employ these boundary circumstances when resolving the Schrödinger Equation.

 Solution of the Schrödinger Equation:

The time-independent Schrödinger equation for a particle of mass (m) moving in one direction with energy (E) is

{\displaystyle i\hbar {\frac {\partial }{\partial t}}\psi (x,t)=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}\psi (x,t)+V(x)\psi (x,t),} $i\hbar {\frac {\partial }{\partial t}}\psi (x,t)=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}\psi (x,t)+V(x)\psi (x,t),$
{\displaystyle i\hbar {\frac {\partial }{\partial t}}\psi (x,t)=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}\psi (x,t)+V(x)\psi (x,t),}
{\displaystyle i\hbar {\frac {\partial }{\partial t}}\psi (x,t)=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}\psi (x,t)+V(x)\psi (x,t),} $i\hbar {\frac {\partial }{\partial t}}\psi (x,t)=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}\psi (x,t)+V(x)\psi (x,t),$

ℏ is the reduced Planck Constant ,here ℏ=h/2π
m is the mass of the particle
ψ(x) is the stationary time-independent wavefunction
V(x) is the potential energy as a function of position
E is the energy, a real number

7.2 The Delta Function Potential:

 Take a modest, attractive delta function potential and aspect for the bound states.

$\begin{displaymath}\bgroup\color{black} V(x)=-aV_0\delta(x)\egroup\end{displaymath}$

 These will have energy less than zero so the answers are

$\begin{displaymath}\bgroup\color{black} \psi(x)=\left\{ \matrix{Ae^{\kappa x} & x<0 \cr Ae^{-\kappa x} & x>0\cr} \right. \egroup\end{displaymath}$

Wherever

$\begin{displaymath}\bgroup\color{black} \kappa=\sqrt{-2mE\over\hbar^2} .\egroup\end{displaymath}$

 There are only double regions, above and below the delta function.

 We don't necessity to worry about the one point at x=0 -the two solutions will bout there.

 We have previously made the wave function nonstop at x=0 by using the similarcoefficient A for the answer in both sections.

 We now need to see the boundary state on the first derivative at x=0. Recall that the delta function causes a recognized gap in the first derivative.

$\begin{eqnarray*} \left.{d\psi\over dx}\right\vert _{+\epsilon}-\left.{d\psi\ove... ...kappa = -{2maV_0\over\hbar^2} \\ \kappa={maV_0\over\hbar^2} \\ \end{eqnarray*}$

 Placing in the formula for k in relations of the energy.

$\begin{eqnarray*} {-2mE\over\hbar^2}={m^2a^2V_0^2\over\hbar^4} \\ E=-{ma^2V_0^2\over 2\hbar^2} \\ \end{eqnarray*}$

7.3 Quantum Harmonic Oscillator:

The linear harmonic oscillator defines pulsations in molecules and their equivalents in solids, the phonons.Though the linear harmonic oscillator can signify justly non-elementary substances like a solid and a molecule, it delivers a gap into the most basic structure in the physical domain.

The Schrodinger Equation for a one –dimentional harmonic oscillator is

7.4 Scattering from A Potential Barrier and Tunnelling; Related Examples Like Alpha- Decay:

7.5 Particle in a three-dimensional box:

 Simplification of the consequences for a two-dimensional square box to a three-dimensional cubic box is direct.

 Meanwhile we living in a three-dimensional world, this simplification is ansignificant one, and we want to be able to consider about energy levels and wave functions in three dimensions.

 The potential energy $V(x,y,z)$ for the cubic box is well-defined to be 0 if $x\in[0,L]$ , $y\in[0,L]$ and $z\in[0,L]$ and infinite else.

 This resources that the wave function $\psi(x,y,z)$ needcontent six boundary circumstances $\psi(0,y,z) = 0$ , $\psi(x,0,z) = 0$ , $\psi(x,y,0) = 0$ , $\psi(L,y,z) = 0$ , $\psi(x,L,z) = 0$ and $\psi(x,y,L) = 0$ .

 We first note that the classical energy is the sum of three terms

$\begin{displaymath} {p_x^2 \over 2m} + {p_y^2 \over 2m} + {p_z^2 \over 2m} = E \end{displaymath}$

Where $p_x$ , $p_y$ and $p_z$ are the three mechanisms of the particle's momentum vector ${\bf p}$ . Thus, we can write the energy as

$\begin{displaymath} E = \varepsilon_x + \varepsilon_y + \varepsilon_z \end{displaymath}$

 Equivalent to the kinetic energy in the $x$ , $y$ and $z$ directions. Since the energy is a simple sum of energies for the $x$ , $y$ and $z$ directions, the wave function will be a invention of wave function forms for the one-dimensional box, and in order to fulfill the first three of the boundary conditions,

 we can take the $\sin$ functions:

$\begin{displaymath} \psi(x,y,z) = A\sin\left(\sqrt{2m\varepsilon_x \over \hbar^2... ...\right) \sin\left(\sqrt{2m\varepsilon_z \over \hbar^2}z\right) \end{displaymath}$

 As in the two-dimensional case, applying second three boundary conditions yields the allowed values of $\varepsilon_x$ , $\varepsilon_y$ and $\varepsilon_z$ , which now need three integers $n_x$ , $n_y$ and $n_z$ :

$\begin{displaymath} \varepsilon_{n_x} = {\hbar^2 \pi^2 \over 2mL^2}n_x^2 \;\;\;\... ...;\;\;\;\; \varepsilon_{n_z} = {\hbar^2 \pi^2 \over 2mL^2}n_z^2 \end{displaymath}$

So that the allowableprinciples of the total energy are

$\begin{displaymath} E_{n_xn_yn_z} = {\hbar^2 \pi^2 \over 2mL^2} \left(n_x^2 + n_y^2 + n_z^2\right) \end{displaymath}$

And the wave functions develop

$\begin{displaymath} \psi_{n_xn_yn_z}(x,y,z) = A\sin\left({n_x\pi x \over L}\righ... ...({n_y\pi y \over L}\right) \sin\left({n_z\pi z \over L}\right) \end{displaymath}$

The constant $A$ is resolute from the standardizationstate

$\displaystyle \int_0^L \int_0^L \int_0^L\vert\psi_{n_xn_yn_z}(x,y,z)\vert^2 dx dy dz$ $\textstyle =$ $\displaystyle 1$

$\displaystyle A^2 \int_0^L \sin^2\left({n_x\pi x \over L}\right)dx \int_0^L \si... ...ft({n_y\pi y \over L}\right)dy \int_0^L \sin^2\left({n_z\pi z \over L}\right)dz$ $\textstyle =$ $\displaystyle 1$

$\displaystyle A^2 {L \over 2}\cdot{L \over 2}\cdot{L \over 2}$ $\textstyle =$ $\displaystyle 1$

$\displaystyle A$ $\textstyle =$ $\displaystyle \left({2 \over L}\right)^{3/2}$

 Thus, the wave meanings are

$\begin{displaymath} \psi_{n_xn_yn_z}(x,y,z) = \left({2 \over L}\right)^{3/2} \si... ...({n_y\pi y \over L}\right) \sin\left({n_z\pi z \over L}\right) \end{displaymath}$

 As using the two-dimensional box, the three integers $n_x$ , $n_y$ and $n_z$ are limited to the usual numbers 1,2, 3, Thus, the lowest energy or ground-state energy is

$\begin{displaymath} E_{111} = {3\hbar^2 \pi^2 \over 2mL^2} \end{displaymath}$

 Imagining the wave functions is delicatesince of their high dimensionality. The most common method of imagining functions of three variables is the use of an isosurface.

 An isosurface of a function $f(x,y,z)$ is the whole set of points $x$ , $y$ , and $z$ for which $f(x,y,z) = C$ , where $C$ is a selected constant. Henceforth the name isosurface - the value of the function is the same at all points on the surface.

 For wave functions, where the sign can be positive or negative, it is valuable to base the value of $C$ not on the wave function significance but rather on the probability density $p_{n_xn_yn_z}(x,y,z) = \vert\psi_{n_xn_yn_z}(x,y,z)\vert^2$ .

 The numberunderdemonstrations two isosurfaces of the wave function $\psi _{123}(x,y,z)$ .

 The first happens at a probability density value of 0.64 and the other happens at 0.04. These values then suggest that the wave function can have a fixed positive or negative value beside the surface.

 In one case, $\psi_{123}(x,y,z) = \pm 0.8$ , while in the other, $\psi_{123}(x,y,z) = \pm 0.2$ .

7.6 Square-Well Potential

Consider a particle of mass $m$ and energy $E$ relating with the simple square potential well

$\begin{displaymath} V(x) = \left\{\begin{array}{lcl} -V_0&\mbox{\hspace{1cm}}&\m... ... x\leq a/2$}\ [0.5ex] 0&&\mbox{otherwise} \end{array}\right., \end{displaymath}$

Where $V_0>0$ .

Now, if E>0 then the particle is unbounded. Therefore, when the particle meets the well it is either reflected or transmitted. As is simply confirmed, the reflection and transmission probabilities are assumed and given by

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

$\displaystyle q^2$ $\textstyle =$ $\displaystyle \frac{2 m (E+V_0)}{\hbar^2}.$

Assume that E<0.

In this example, the particle is bounded (i.e., as ).

Now, it is simply got that independent solutions of Schrödinger's equation in the symmetric V(-x)= V(x) potential must be either completely symmetric

[i.e., (-x)=(x)], or completely anti-symmetric [i.e., (-x)=-(x)].

Also, the answers must fulfil the boundary condition

$\begin{displaymath} \psi\rightarrow 0\mbox{\hspace{2cm}as $\vert x\vert\rightarrow\infty$}. \end{displaymath}$

Let us, first of all, search for a completely symmetric solution. In the section to the left of the well x<-a/2 the solution of Schrödinger's equation which fulfills the boundary condition 0 and x - is

$\begin{displaymath} \psi(x) = A {\rm e}^{ k x}, \end{displaymath}$

Where

$\begin{displaymath} k^2 = \frac{2 m \vert E\vert}{\hbar^2}. \end{displaymath}$

By symmetry, the explanation in the region to the right of the well (i.e., x>a/2) is

$\begin{displaymath} \psi(x) = A {\rm e}^{-k x}. \end{displaymath}$

The solution inside the well (i.e., ) which fulfills the symmetry limit

(-x)=(x) is

$\begin{displaymath} \psi(x) = B \cos(q x), \end{displaymath}$

Where

$\begin{displaymath} q^2 = \frac{2 m (V_0+E)}{\hbar^2}. \end{displaymath}$

Now, we have supposed that .The constraint that (x) and its first derivative be constant at the edges of the well (i.e., at ) yields

$\begin{displaymath} k = q \tan(q a/2). \end{displaymath}$

Let Y=qa/2

It show that

$\begin{displaymath} E = E_0 y^2 - V_0, \end{displaymath}$

where

$\begin{displaymath} E_0 = \frac{2 \hbar^2}{m a^2}. \end{displaymath}$

Furthermore,

$\begin{displaymath} \frac{\sqrt{\lambda-y^2}}{y} = \tan y, \end{displaymath}$

With

$\begin{displaymath} \lambda = \frac{V_0}{E_0}. \end{displaymath}$

Now, y need lie in the range : i.e., $E$ must lie in the range .

$\begin{figure} \epsfysize =3in \centerline{\epsffile{Chapter05/fig07.eps}} \end{figure}$

Figure: The curves tany (solid) and $\sqrt {\lambda - y^2}/y$ (dashed), calculated for $\lambda = 1.5 \pi ^2$ . The latter curve takes the value $0$ when

Now, the solutions correspond to the connection of the curve with the curve tan y. Figure shows these two curves designed for a particular value of $\lambda$ . In this case, the curves meet twice, representing the presence of two completely symmetric bound states in the well. Also, it is obvious, from the figure, that as $\lambda$ rises (i.e., as the well becomes deeper) there are more and more bound states. However, it is also obvious that there is continually at least one completely symmetric bound state, no matter how small $\lambda$ becomes (i.e., no matter how shallow the well becomes). In the limit  (i.e., the limit in which the well becomes very deep), the solutions to Eq. (384) asymptote to the roots of tan y=.

This provides y=(2j-1)/2, where j is a positive integer, or

$\begin{displaymath} q = \frac{(2 j-1) \pi}{a}. \end{displaymath}$

These explanations are equal to the odd- $n$ infinite square well explanations stated by

$\begin{figure} \epsfysize =3in \centerline{\epsffile{Chapter05/fig08.eps}} \end{figure}$

Figure: The curves tan y (solid) and (dashed), calculated for $\lambda = 1.5 \pi ^2$ .

For the case of a completely anti-symmetric bound state, like examination to the above yields

$\begin{displaymath} -\frac{y}{\sqrt{\lambda-y^2}} = \tan y. \end{displaymath}$

The solutions of this equation correspond to the intersection of the curve tan y with the curve . Figure shows these two curves plotted for the same value of $\lambda$ as that used in Fig. In this case, the curves interconnect once, representing the being of a single completely anti-symmetric bound state in the well. It is, over, evident from the figure that as $\lambda$ increases (i.e., as the well becomes deeper) there are more and more bound states. However, it is also evident that when $\lambda$ becomes sufficiently small [i.e., ] then there is no completely anti-symmetric bound state. In other words, a very shallow potential well continuously keeps a completely symmetric bound state, but does not usually keep a completely anti-symmetric bound state.

In the boundary  (i.e., the limit in which the well becomes very deep), the solutions asymptote to the roots of tan y=0. This gives y=j, where j is a positive integer, or

$\begin{displaymath} q = \frac{2 j \pi}{a}. \end{displaymath}$

7.7 Linear Harmonic Oscillator

The Schrodinger Equation for a one –dimentional harmonic oscillator is

7.8Field-Ionization and Scanning Tunnelling Microscope:

7.9 Tunnelling In Semiconductor Structures

Tunneling is a quantum mechanical effect. A tunnel current occurs when electrons move through a barrier that they should not be able to cross. In classic terms, if you don't have enough energy to move "over" a barrier, you won't. However, in the world of quantum mechanics, electrons have wave properties. These waves do not end abruptly on a wall or barrier, but decrease rapidly. If the barrier is thin enough, the probability function can be extended to the next region via the barrier! Due to the low probability that an electron is on the other side of the barrier, given enough electrons, some will move and appear on the other side. When an electron moves through the barrier in this way, it is called a tunnel.

Quantum mechanics tells us that electrons have wave and particle properties. The construction of the tunnel is an undulating effect.

Schematic of electron wavefunction.

The image above shows that when an electron (the wave) hits a barrier, the wave does not stop abruptly, but gradually decreases exponentially. For a thick barrier, the wave does not pass.

The image below shows the scenario if the barrier is rather thin (about a nanometer). Part of the wave passes through, and therefore some electrons can appear on the other side of the barrier.

Due to the sharp decrease in the probability function across the barrier, the number of electrons that will actually form a tunnel depends heavily on the thickness of the barrier. The current through the barrier falls exponentially with the thickness of the barrier.

To extend this description to STM: The starting point of the electron is the tip or the sample, depending on the configuration of the instrument. The barrier is the gap (air, vacuum, liquid) and the second region is the other side, i.e. the tip or the sample, depending on the experimental configuration. By monitoring the current through the gap, we have excellent control over the distance of the tip-sample.

Diagram of the feedback loop of a Scanning Tunneling Microscope

Feedback loop and electron tunneling for scanning tunneling microscopy (STM).

Reference:

1. G. MAIN, “VIBRATIONS AND WAVES IN PHYSICS”, CAMBRIDGE UNIVERSITY PRESS,1993.

2. H. J. PAIN, “THE PHYSICS OF VIBRATIONS AND WAVES”, WILEY, 2006.

3. E. HECHT, “OPTICS”, PEARSON EDUCATION, 2008.

4. A. GHATAK, “OPTICS”, MCGRAW HILL EDUCATION, 2012.

5. O. SVELTO, “PRINCIPLES OF LASERS”, SPRINGER SCIENCE & BUSINESS MEDIA, 2010.

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