Shift in energy due to weak interaction

In summary, a particle is in a box of length L on x-axis. 0<x<L. The first order shift due to V(x) of the wavefunction is kx2/2.
  • #1
samee
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0

Homework Statement



A particle is in a box of length L on x-axis. 0<x<L. What is the first order shift due to V(x) of the wavefunction?

Homework Equations



H=p2/2m
Sin=sqrt(2/L)cos(npix/L)
En=n2pi2hbar2/2mL2
V(x)=kx2/2

The Attempt at a Solution



I'm super confused because the shift in the energy is usually determined by a sum, not an integral. Unless I calculate En(1) by taking;

En1=<n|H|n>
=<n|V|n>
=V * integral(2/Lcos2(nxpi/L)dx) with limits 0 to L

Is this the correct way to do this one? I didn't include k very well unless I put it in for V at the end...
 
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  • #2
You have to put V(x) inside the integral since it depends on x.
 
  • #3
So I set up my integral correctly? Thank god... So now I have;

En(1)=(k/2)*(2/L)*integral(x2*cos2(npix/L) dx) with limits 0 and L.

=k/L * integral(x2*cos2(npix/L) dx) with limits 0 and L.

Is this what the question is asking for? Also- How exactly do I solve this integral by hand? My calculator is giving me;

k/L*{180*(360npi2x cos(npix/L)+(n2pi4x2-64800L2)sin(npix/L)}/n3pi6
 
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  • #4
Your wavefunctions aren't right. If one end of the box is at x=0, you should have sines, not cosines.

I assume you're studying perturbation theory. You should be able to tell me what you're calculating.
 
  • #5
Yes, this is perturbation theory. I think that I'm finding the 1st order shift in the energy. The way the question asks though, "shift in energy of wavefunction due to k, to 1st order in k" sounds as if I should be shifting V(x) instead of the energy.

The question was written with cos in the wavefunction, but I can email my professor and ask if it's a typo... Or do you mean that instead of cos^2 in the integral I should have cos*sin?
 
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  • #6
In this problem, you're treating the spring potential as a perturbation. The shift in energy due to the perturbation can be expressed as a series in k. The problem is just asking you to find the first term in that series.

The solutions to the Schrodinger equation are linear combinations of sine and cosine. When you require that the wavefunction vanish at x=0, that forces the cosine term to disappear, so you're left with just sine.

I'd have guessed the intent was to have the well go from x=-L/2 to x=+L/2, so that the spring potential is centered in the well, but then you should have half the wavefunctions with cosine and the other half with sine.
 
  • #7
samee said:
Also- How exactly do I solve this integral by hand? My calculator is giving me;

k/L*{180*(360npi2x cos(npix/L)+(n2pi4x2-64800L2)sin(npix/L)}/n3pi6
First, use the trig identity cos^2 x = (1+cos 2x)/2. You can then integrate the first term. The second term will require repeated application of integration by parts.
 
  • #8
I'm not sure what to do now, that's how the question was written. I've emailed my prof. and hopefully will get a response in the morning explaining it. Sadly, this is due very soon. Is there no way to solve the problem as is? Or maybe I should say, were the wavefunction sine, then am I doing it correctly and will solving the integral be the only hurdle I need to jump?
 
  • #9
Yes, if the wavefunction was sine instead of cosine, it would be correct.
 
  • #10
Ha! You were correct! The wavefunction IS sine. Thank you so much! You're a lifesaver.
 

1. What is a shift in energy due to weak interaction?

A shift in energy due to weak interaction refers to the change in energy levels of particles caused by the weak nuclear force, one of the four fundamental forces in nature. This force is responsible for certain types of radioactive decay and plays a role in keeping the nucleus of an atom stable.

2. How does weak interaction affect the stability of atoms?

The weak nuclear force is responsible for the conversion of one type of subatomic particle (neutron) into another type (proton) and vice versa. This process, known as beta decay, helps to stabilize atoms by converting excess neutrons or protons into the correct ratio for a stable nucleus.

3. Can the weak interaction be observed in everyday life?

Although the effects of weak interaction are not easily observable in everyday life, it is a crucial force in maintaining the stability of matter. Without it, atoms would not be able to form and the universe as we know it would not exist.

4. How does the weak interaction differ from other fundamental forces?

The weak nuclear force is unique in that it is the only fundamental force that can change the type of particle it acts upon. It is also the shortest-ranged of the four forces, only acting within the nucleus of an atom.

5. What are some real-world applications of understanding the weak interaction?

Understanding the weak interaction has led to advancements in nuclear energy, medical imaging, and particle physics research. By studying the behavior of subatomic particles and their interactions, scientists can gain a better understanding of the fundamental components of the universe.

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