Infinitesimal Perturbation in a potential well

In summary: Yes. So, what about the numerators ##(<\psi^0_m|V|\psi^0_0>)^2##? Are any of these zero? Could they all be zero?There is nothing explicit in this theory which I think indicates that term being zero. So I would say it may/may not be 0?
  • #1
Baibhab Bose
34
3
Homework Statement
In an infinite potential well, two small constant perturbations \epsilon are added in the two opposite corners of the well in an infinitesimal extent (\delta). In what order of \epsilon the correction to the ground state energy would be? See the attached file.
Relevant Equations
<psi|Hp|psi>=E1
If I calculate ## <\psi^0|\epsilon|\psi^0>## and ## <\psi^0|-\epsilon|\psi^0>## separately and then add, the correction seems to be 0 since ##\epsilon## is a constant perturbation term.
SO how should I approach this? And how the Δ is relevant in this calculation?
 

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  • #2
Yes, the first-order correction is zero. Are you familiar with how to calculate the second-order correction to the energy?
 
  • #3
Baibhab Bose said:
If I calculate ## <\psi^0|\epsilon|\psi^0>## and ## <\psi^0|-\epsilon|\psi^0>## separately and then add, the correction seems to be 0 since ##\epsilon## is a constant perturbation term.
Careful with the notation. ## <\psi^0|\epsilon|\psi^0>## and ## <\psi^0|-\epsilon|\psi^0>## are not what you need to calculate for the first-order correction.

Rather, the expression you need to consider is ##<\psi^0|V(x)|\psi^0>##, where ##V(x)## is the perturbation:

##V(x) = -\epsilon## for ##0<x<\Delta##
##V(x) = + \epsilon## for ##a-\Delta < x < a##
##V(x) = 0##, otherwise.

(Here I assumed the origin x = 0 is at the left end of the well.)

For example, your expression ## <\psi^0|-\epsilon|\psi^0>## does not take into account that the perturbation ##-\epsilon## is only over the range ##0<x<\Delta##.
 
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  • #4
Yes, I know how 2nd order is calculated. $$ E^2_0=\sum_{n=/m} \frac{(<\psi_n|V|\psi_m>)^2}{E^0_n-E^0_m}$$
But I can't seem to practically calculate the correction in this case. Assuming ##\psi_n##s are ##\sqrt {\frac{2}{a}}sin(\frac{n\pi x}{a})## , what are the values of n's I'd have to limit to?
 
  • #5
Baibhab Bose said:
$$ E^2_0=\sum_{n=/m} \frac{(<\psi_n|V|\psi_m>)^2}{E^0_n-E^0_m}$$
The right-hand side is not quite correct. There is nothing on your right-hand side that indicates that you are getting the second-order correction to the ground state. Also, be careful that you have the terms in the denominator in the correct order.

You will not have to work out the values of the matrix elements in order to answer the question. You will just need to consider whether or not the right-hand side overall is positive, negative, or zero.
 
  • #6
Oh yes, so the correct form would be ## E_0^2=\sum_{n\neq m} \frac {(<\psi_m^0|V|\psi_0^0>)^2}{E^0_0-E^0_m}## for the ground state.
So in this question we just need the order of ##\epsilon## which, by speculation, would be ##\epsilon^2## since the 2nd order correction depends on square of the perturbation term!
But how do we ensure its sign?
 
  • #7
Baibhab Bose said:
Oh yes, so the correct form would be ## E_0^2=\sum_{n\neq m} \frac {(<\psi_m^0|V|\psi_0^0>)^2}{E^0_0-E^0_m}## for the ground state.
This still needs a little fixing up. Your summation notation indicates that ##n## is the summation index, but the index ##n## does not appear in the expression following the summation symbol. Instead, it looks like maybe your expression is for ##m## as the summation index. Anyway, for whatever symbol you take to be the summation index, what particular value of this index is to be excluded in the summation?

So in this question we just need the order of ##\epsilon## which, by speculation, would be ##\epsilon^2## since the 2nd order correction depends on square of the perturbation term!
But how do we ensure its sign?
How does the value of ##E^0_0## compare to any other ##E^0_m##?
 
  • #8
TSny said:
This still needs a little fixing up. Your summation notation indicates that ##n## is the summation index, but the index ##n## does not appear in the expression following the summation symbol. Instead, it looks like maybe your expression is for ##m## as the summation index. Anyway, for whatever symbol you take to be the summation index, what particular value of this index is to be excluded in the summation?

How does the value of ##E^0_0## compare to any other ##E^0_m##?
Oh, my bad !
## E_0^2=\sum_{m\neq 0} \frac{(<\psi^0_m|V|\psi^0_0>)^2}{E_0^0-E^0_m}##
Now,
##E_0^0##= unperturbed Ground state Wavefunction's Energy.
##E^0_m##=Unperturbed energy eigenvalue of the higher state wavefunctions, since m index represents states higher than the ground state, ##E_0^0-E^0_m## should be negative, right?
 
  • #9
Baibhab Bose said:
Oh, my bad !
## E_0^2=\sum_{m\neq 0} \frac{(<\psi^0_m|V|\psi^0_0>)^2}{E_0^0-E^0_m}##
Now,
##E_0^0##= unperturbed Ground state Wavefunction's Energy.
##E^0_m##=Unperturbed energy eigenvalue of the higher state wavefunctions, since m index represents states higher than the ground state, ##E_0^0-E^0_m## should be negative, right?
Yes.
So, what about the numerators ##(<\psi^0_m|V|\psi^0_0>)^2##? Are any of these zero? Could they all be zero?
 
  • #10
There is nothing explicit in this theory which I think indicates that term being zero. So I would say it may/may not be 0?
 
  • #11
Baibhab Bose said:
There is nothing explicit in this theory which I think indicates that term being zero. So I would say it may/may not be 0?
Knowing the form of the perturbation and the form of the unperturbed energy eigenstates, you should be able to tell if a particular matrix element ##\langle \psi^0_m|V|\psi^0_0 \rangle## is zero. Hint: Think in terms of even or odd functions about the midpoint of the well.
 

Related to Infinitesimal Perturbation in a potential well

1. What is an infinitesimal perturbation in a potential well?

An infinitesimal perturbation in a potential well refers to a small change or disturbance in the potential energy of a system. This perturbation can be caused by external factors such as external forces or internal factors such as fluctuations in the system itself.

2. How does an infinitesimal perturbation affect the behavior of a system?

The effect of an infinitesimal perturbation depends on the magnitude and direction of the change in potential energy. In some cases, it can cause the system to deviate from its equilibrium state and lead to oscillations or even chaos. In other cases, the system may return to its original state once the perturbation is removed.

3. What are some real-world examples of infinitesimal perturbations in potential wells?

Infinitesimal perturbations can be seen in various systems, such as a pendulum experiencing small fluctuations in air resistance, a planet orbiting a star experiencing gravitational disturbances from other planets, or a chemical reaction being influenced by slight changes in temperature or pressure.

4. How do scientists study infinitesimal perturbations in potential wells?

Scientists use mathematical models and simulations to study the effects of infinitesimal perturbations in potential wells. These models can help predict the behavior of a system under different perturbations and provide insights into the underlying dynamics of the system.

5. Can infinitesimal perturbations be beneficial?

Yes, infinitesimal perturbations can be beneficial in certain cases. For example, they can help stabilize a system that is on the verge of instability, or they can be used to control and manipulate the behavior of a system for practical purposes. However, in some cases, they can also have negative effects and cause unintended consequences.

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