Variational Estimate of Hydrogen Atom Ground State Energy

In summary, the conversation discusses obtaining a variational estimate for the ground state energy of the hydrogen atom using a trial function of the form \psi_T(r) = \text{exp } \left( - \alpha r^2 \right). The participants discuss the need to minimize the expectation value of the Hamiltonian, which includes both the kinetic and potential energy terms. The correct expression for the kinetic energy expectation value is given, but there is confusion about the radial Laplacian term and the integration process. The conversation ends with a request for clarification on how to proceed with the integrals.
  • #1
latentcorpse
1,444
0
Obtain a variational estimate of the ground state energy of the hydrogen atom by taking as a trial function [itex]\psi_T(r) = \text{exp } \left( - \alpha r^2 \right)[/itex]

How does your result compare with teh exact result?

You may assume that

[itex]\int_0^\infty \text{exp } \left( - b r^2 \right) dr = \frac{1}{2} \sqrt{ \frac{\pi}{b}}[/itex]

and that [itex] 1 \text{Ry} = \left( \frac{e^2}{4 \pi \epsilon_0} \right)^2 \frac{m}{2 \hbar^2}[/itex]

so i obviously need to minimise

[itex]E[r] = \frac{ \langle \psi \vline \hat{V} \vline \psi \rangle}{ \langle \psi \vline \psi \rangle}[/itex]

i think I'm getting the wrong answer because my V is wrong.

so i want a coulomb potential between a proton and a electron surely?

[itex]V=\frac{-e^2}{4 \pi \epsilon_0 r^2}[/itex]
clearly I'm going wrong since i have this [itex]r^{-2}[/itex] term that i don't know how to integrate (the only r dependence in hte given integral is in the exponent) and also i haven't used the rydberg thing (although i guess that might not be needed until the end of the question).

thanks for any help.
 
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  • #2
You need to minimize the expectation value of the whole Hamiltonian, not just the potential energy.

<H>=<T>+<V>

where <T> is the expectation value of the kinetic energy term, so you have to do two integrals. Furthermore, the Coulomb potential goes as 1/r not 1/r2.
 
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  • #3
okay but i still get integrals with r terms outside the exponential

[itex] \langle \psi | \hat{H} | \psi \rangle = - \int_V \frac{\hbar^2}{m} \text{exp} ( - 2 \alpha r^2) ( 2 \alpha r^2 - \alpha ) dV - \int_V \frac{e^2}{4 \pi \epsilon_0 r} \text{exp} ( - 2 \alpha r^2 )[/itex]

and the normalisation on the denominator works out at [itex]\sqrt{2} \pi^{\frac{3}{2}}[/itex]

any idea on how to proceed? thanks.
 
  • #4
Are you saying you don't know how to do the integrals? What is dV and what are your limits of integration?
 
  • #5
[itex]dV=r^2 \sin{\theta} dr d\theta d\phi[/itex]

so we can pull out a [itex]4 \pi[/itex] from the theta and phi integrals. then the r integral is from 0 to infinity since dV is all space.

so, for example how do we integrate

[itex]\int_0^\infty -\frac{4 \pi \hbar^2}{m} 2 \alpha r^2 \text{exp} ( - 2 \alpha r^2) dr[/itex]

the given formula doesn't apply here, does it?
 
  • #6
No it does not. However, before we go into that, can you write the complete expression for the integral giving the expectation value for the kinetic energy?
 
  • #7
ok.

[itex] \langle \psi | \hat{T} | \psi \rangle = \int_0^\infty \int_0^\pi \int_0^{2 \pi} e^{- \alpha r^2} -\frac{\hbar^2}{2m} (2 \alpha r^2 - \alpha) e^{- \alpha r^2} r^2 \sin{\theta} dr d \theta d \phi = -\frac{\hbar^2}{2m} \int_0^\infty \int_0^\pi \int_0^{2 \pi} r^2 (2 \alpha r^2 - \alpha) e^{- 2 \alpha r^2} \sin{\theta} dr d \theta d \phi [/itex]

i can't see why this is wrong...
 
  • #8
You are missing the normalization constant, but that can always be added later.
I think the radial Laplacian term is incorrect. Can you show how you derived it?
 

1. What is the Variational Estimate of Hydrogen Atom Ground State Energy?

The Variational Estimate of Hydrogen Atom Ground State Energy is a method used in quantum mechanics to approximate the lowest possible energy state of a hydrogen atom. It is based on the variational principle, which states that the energy of any trial wavefunction will always be greater than or equal to the true ground state energy.

2. How is the Variational Estimate of Hydrogen Atom Ground State Energy calculated?

The Variational Estimate of Hydrogen Atom Ground State Energy is calculated by using a trial wavefunction, which is a mathematical function that describes the electron's probability distribution in the atom. This function is then plugged into the Schrödinger equation, which is solved for the energy. The energy obtained from this calculation is an upper bound for the true ground state energy.

3. Why is the Variational Estimate of Hydrogen Atom Ground State Energy important?

The Variational Estimate of Hydrogen Atom Ground State Energy is important because it provides a way to approximate the ground state energy of a hydrogen atom, which is a fundamental concept in quantum mechanics. It also serves as a starting point for more advanced calculations and can be applied to other quantum systems besides the hydrogen atom.

4. How accurate is the Variational Estimate of Hydrogen Atom Ground State Energy?

The accuracy of the Variational Estimate of Hydrogen Atom Ground State Energy depends on the complexity of the trial wavefunction used. Generally, the more terms included in the wavefunction, the closer the estimated energy will be to the true value. However, it is important to note that this method can only provide an upper bound for the ground state energy, so the true energy may be lower than the estimated value.

5. What are some limitations of the Variational Estimate of Hydrogen Atom Ground State Energy?

One limitation of the Variational Estimate of Hydrogen Atom Ground State Energy is that it can only provide an upper bound for the true energy. This means that the estimated value may not be the exact ground state energy, but rather a value that is equal to or greater than the true energy. Additionally, the accuracy of the estimate depends on the complexity of the trial wavefunction used, so it may be challenging to obtain an accurate estimate for more complex quantum systems.

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