Easy variational principle question that I can't integrate

In summary, the participants in this conversation discuss using a trial wavefunction to find an upper limit for the groundstate energy of a 1-D harmonic oscillator. They mention a Gaussian integral and the uncertainty principle, and also discuss normalizing the wave function and calculating the Hamiltonian.
  • #1
mrausum
45
0

Homework Statement



Use trial wavefunction exp(-bx^2) to get an upper limit for the groundstate energy of the 1-d harmonic oscillator

The Attempt at a Solution



This is always going to give an integral of x^2*exp(-x^2). How do you do it? :/
 
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  • #2
It's a Gaussian integral that comes out to be

[tex]\int_{-\infty}^{\infty} x^2 e^{-b x^2} = \frac{\sqrt{\pi}}{2b^{3/2}}[/tex]
 
  • #3
yeah, I know that from wolfram alpha too. But that can't be right, because then my upper limit for the ground energy would depend on 1/b^3/2, which would mean if b was very large it would violate the uncertainty principle.
 
  • #4
Did you normalize your wave test function?
 
  • #5
Feldoh said:
Did you normalize your wave test function?

Yeah that's the one, thanks.
 
  • #6
Wait..that still doesn't work. The norm.const = b^.5/(pi/2)^.5, which doesn't quite cancel with the b^3/2 in the denominator from the integral?
 
  • #7
Can you show me some of your work? When I do it I get the appropriate contributions from the kinetic and potential expectation values.
 
  • #8
sure:

[tex]<H> = <(b/\pi).exp(-bx^2)|H|(b/\pi).exp(-bx^2)>[/tex]


[tex]<H> = b/\pi\int^{inf}_{-inf}exp(-2bx^2).\left(\frac{2h^2x}{m}+\frac{1}{2}mw^2x^2\right)dx[/tex]
 
Last edited:
  • #9
The normalization constant is [tex]A = (\frac{2b}{\pi})^{1/4}[/tex]

From there the Hamiltonian of a 1D harmonic oscillator:

[tex]-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m \omega x^2[/tex]

I'm not sure if you all ready dealt with the second partial with respect to x do your expectation value of the Hamiltonian looks wrong so far.

Remember it's [tex]|A|^2\int_{-\infty}^{\infty} (e^{-b x^2})^* (-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m \omega x^2) e^{-b x^2} dx[/tex]
 

Related to Easy variational principle question that I can't integrate

1. What is the variational principle and how is it used in scientific research?

The variational principle is a mathematical concept that states that the true solution to a physical problem can be obtained by minimizing a specific functional. In scientific research, it is often used in theoretical physics and engineering to find the most accurate and efficient solutions to complex problems.

2. How does the variational principle relate to integration?

The variational principle involves finding the minimum value of a functional, which often involves solving an integral. This is because integrals can be used to represent the total energy or potential of a system, and minimizing this energy can lead to the most stable and accurate solution.

3. Can you explain the difference between easy and difficult variational principle problems?

Easy variational principle problems typically involve simple and well-defined systems with known equations and boundary conditions. Difficult problems, on the other hand, may involve complex systems with unknown variables and require more advanced mathematical techniques to solve.

4. What should I do if I can't integrate the variational principle for a certain problem?

If you are struggling to integrate the variational principle for a specific problem, it is best to seek help from a mathematics or physics expert. They can provide guidance and possibly suggest alternative approaches to solving the problem.

5. Are there any real-life applications of the variational principle?

Yes, the variational principle has many applications in various fields of science and engineering. It is used in quantum mechanics, fluid dynamics, and even in economics and finance to find optimal solutions. It is also the basis for many computational methods used in modern technology.

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