Did I use the correct identities in my variation method calculations?

In summary, the conversation is about determining the value of the appropriate variational integral W for a one dimensional quartic oscillator with the potential function V(x) = cx^4. The participant has used the trial function e^(-alpha(x^2)/2) and has attached a picture of their work. They are questioning if their answer is correct because they had to use identities (listed as Identity 1 and Identity 2) in their calculation. They used the phi*Vphi/phi*phi equation and simplified it to 0.0625C (pi/(alpha^5))^(1/2) after using Identity 1 and Identity 2.
  • #1
Emily Smith
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<Moderator's note: Moved from a technical forum and thus no template.>

Problem:
One dimensional quartic oscillator, V(x) = cx^4 (c is a constant)
Use the trial function e^(-aplha(x^2)/2) to determine the value of the appropriate variational integral W.

I've attached a picture of my work.

I feel that my answer is not correct, because I ended up using identities (listed below) at the end because I was grasping for straws. So my question is: did I mess up by using the identities?

Identity 1 = x^n e^bx^2 = (2n!)/(2^(2n+1)n!)(pi/(b^2n+1))^1/2
Identity 2 = e^-bx^2 = 1/2 (pi/b)^1/2

I used the phi* V phi / phi*phi equation. Started with phi *(V)phi, e^(-aplha(x^2)/2)(Cx^4)e^(-aplha(x^2)/2) dx
I pulled the C out since it was constant and simplified to e^(-aplha(x^2))(x^4)
From there I used the Identity 1 and got 0.0625C (pi/(aplha^5))^(1/2).

For phi*phi I used Identity 2 and got sqrt(pi/alpha)/2

Finally I divided hi* V phi by phi*phi which gave me 0.125C(sqrt(pi/(alpha^5))/ sqrt(pi/alpha)
IMG_20181001_204034.jpg
IMG_20181001_204034.jpg
 

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  • #2
Please type in what you've done, but the least, rotate your pictures.
 

Related to Did I use the correct identities in my variation method calculations?

What is the variation method?

The variation method is a mathematical technique used to approximate the ground state energy of a quantum mechanical system. It involves variationally optimizing a trial wavefunction to find the lowest possible energy for a given system.

Why is the variation method important?

The variation method is important because it allows scientists to approximate the energy of complex quantum systems that cannot be solved analytically. It is also a useful tool for understanding the behavior of different physical systems.

How does the variation method work?

The variation method works by starting with a trial wavefunction, which is an educated guess for the ground state wavefunction of a system. This wavefunction is then varied by adjusting a set of parameters, and the resulting energy is calculated. This process is repeated until the energy is minimized, providing an approximation for the ground state energy of the system.

What are the limitations of the variation method?

The variation method is limited by the accuracy of the trial wavefunction chosen. If the trial wavefunction is not a good representation of the actual ground state, the resulting energy approximation will also be inaccurate. Additionally, the variation method can be computationally intensive and may not be feasible for very large or complex systems.

What are some applications of the variation method?

The variation method has many applications in quantum mechanics, including calculating the energy levels of atoms and molecules, studying the properties of solids and liquids, and predicting the behavior of complex chemical reactions. It is also used in many fields of physics, such as nuclear physics and condensed matter physics.

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