Flag Modern Physics> A small dip in the potential energy is in...
question mark

A small dip in the potential energy is introduced into the one dimentional square well problem ( infinite potential well) A. Write a general expression for he first order corection to the ground state energy.B.Evaluate the energy corection for a( length of the dip) = L/10. with n=1 I'll draw an aprox figure . | < ------------------------------------------L---------------------------------------->|| <--------a-------->| _____________| | | ^ | | | e ||_________________| |_______________________|

Amit Saxena , 11 Years ago
Grade upto college level
anser 1 Answers
Askiitians Tutor Team

Last Activity: 16 Days ago

In quantum mechanics, when we introduce a small perturbation to a system, we can use perturbation theory to find how this affects the energy levels of the system. In the case of a one-dimensional infinite potential well, the ground state energy can be modified by a small dip in potential energy. Let’s break down the steps to derive the first-order correction to the ground state energy and evaluate it for a specific case.

Understanding the Infinite Potential Well

In an infinite potential well, the potential energy V(x) is defined as:

  • V(x)=0 for 0<x<L
  • V(x)= otherwise

The wave functions for the ground state (n=1) in this well are given by:

ψ1(x)=2Lsin(πxL)

The corresponding energy for the ground state is:

E1=2π22mL2

Introducing the Perturbation

Now, let’s introduce a small dip in the potential energy, V(x), which we can represent as:

V(x)=V0 for 0<x<a (where a=L10)

Outside this region, the potential remains zero. The first-order correction to the energy can be calculated using the formula:

En(1)=ψn|V|ψn

For the ground state (n=1), this becomes:

E1(1)=0Lψ1(x)V(x)ψ1(x)dx

Calculating the First-Order Energy Correction

Substituting the wave function and the potential into the integral, we have:

E1(1)=0aψ1(x)(V0)ψ1(x)dx

Since ψ1(x)=2Lsin(πxL), we can write:

E1(1)=V00L10(2Lsin(πxL))2dx

Now, simplifying the integral:

E1(1)=V02L0L10sin2(πxL)dx

Evaluating the Integral

To evaluate the integral, we can use the identity:

sin2(x)=1cos(2x)2

Thus, we have:

0L10sin2(πxL)dx=12[L10L2πsin(2π10)]

Substituting this back into our expression for E1(1), we can find the first-order correction to the ground state energy.

Final Expression

After performing the calculations, we can summarize the first-order energy correction as:

E1(1)=V0L(L220L24π2sin(2π10))

Now, substituting a=L10 and evaluating the integral will give you the specific numerical value for the energy correction based on the chosen V0.

This approach illustrates how perturbation theory allows us to analyze small changes in a quantum system and understand their impact on energy levels. If you have any specific values for V0 or further questions about the calculations, feel free to ask!

star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments