Optimizing a Wave Function

[Lunar_Lander]

"Optimizing" a Wave Function

Homework Statement

Consider a Hydrogen Atom, an electron in an attractive Coulomb potential of the form $V(r)=-\frac{e_0^2}{4\pi\epsilon_0r}$, where e₀ is the elementary charge. Assume the following wave function for the electron (with α>0):
$\psi(r)=Ae^{-\alpha r}$.

a) Determine the expectation value for the kinetic energy.
b) Determine the expectation value for the potential energy.
c) The expectation value of the total energy seems to be given by the sum of the results of a) and b). Plot the dependency of the kinetic, potential and total energy to α. Minimize the total energy in respect to α. If your calculations are correct, you will obtain the energy of the 1s state.

Homework Equations

$H=\frac{p^2}{2m}+V(r)$

The Attempt at a Solution

I am quite sure what to do in c), that it is required to calculate the derivative of the total energy function to alpha and then find the minimum of that. However, I need to get some help with a) and b). I know so far that the expectation value is calculated like this:

$<V>=<\psi|V|\psi>=\int_{-\infty}^{\infty} \psi\cdot V\cdot \psi~dr,$

and that in a), the Laplace operator needs to be applied in spherical coordinates. But how exactly is the integration done?

I would be grateful if someone could explain that, as I find the idea of integrating from negative to positive infinity somewhat puzzling.

 

[vela]

The integral is over all space, so what you should have is
$$\langle V \rangle = \int \psi^*(\vec{r}) V(\vec{r}) \psi(\vec{r})\,d^3\vec{r}$$In spherical coordinates, this becomes
$$\langle V \rangle = \int_0^\infty \int_0^\pi \int_0^{2\pi} \psi^*(\vec{r}) V(\vec{r}) \psi(\vec{r})\,r^2\sin\theta\,d\phi\,d\theta\,dr$$

 

[Lunar_Lander]

Thank you Vela! Would it maybe be possible for one of you to show me just how to correctly do this calculation, because my problem is that I usually need a good example and explanation on how to calculate complex things. When I look at calculations in books, those are usually good, but when I then try to apply the methods given there to problems, I usually don't get far.

I attempted though to do the integration you gave me, but I am not sure if it is correct (thus my request there).

$<V>=\int\int\int -\psi^2 r^2 \sin\theta \frac{e_0^2}{4\pi\epsilon_0 r}dr d\theta d\phi$
Then I did the integrations of θ and Phi and also put the values that are constants in front of the integral:
$-4\pi \frac{e_0^2}{4\pi\epsilon_0} \int \psi^2 r^2 \frac{1}{r}dr$
$-4\pi \frac{e_0^2}{4\pi\epsilon_0}\cdot A^2 \int \exp(-2\alpha\cdot r) r^2 \frac{1}{r}dr$

Is this even correct? I am feeling a bit lost, sorry.

 

[vela]

That looks fine so far. Simplify the integrand and then integrate.

 

[Lunar_Lander]

So, I am getting this:

$-4\pi\cdot A^2\frac{e_0^2}{4\pi\epsilon_0}\int \exp(-\alpha\cdot r)\cdot r~dr$
$-A^2\frac{e_0^2}{\epsilon_0}\int \exp(-\alpha\cdot r)\cdot r~dr$
and
$A^2\frac{e_0^2}{\epsilon_0}\cdot \frac{1-\exp(-\alpha\cdot r)\cdot (\alpha\cdot r+1)}{\alpha^2}$

Is this correct? If yes I'd continue on the kinetic energy, would this be:

$<T>=-\int\int\int r^2\cdot\sin(\theta)\cdot\psi^2\cdot\frac{\hbar}{2m}\Delta~dr~d\theta~d\phi$
?

 

[vela]

So, I am getting this:

$-4\pi\cdot A^2\frac{e_0^2}{4\pi\epsilon_0}\int \exp(-\alpha\cdot r)\cdot r~dr$
$-A^2\frac{e_0^2}{\epsilon_0}\int \exp(-\alpha\cdot r)\cdot r~dr$
and
$A^2\frac{e_0^2}{\epsilon_0}\cdot \frac{1-\exp(-\alpha\cdot r)\cdot (\alpha\cdot r+1)}{\alpha^2}$

Is this correct?

You need to plug the limits in. I didn't check to see if the antiderivative you came up with is correct.

If yes I'd continue on the kinetic energy, would this be:

$<T>=-\int\int\int r^2\cdot\sin(\theta)\cdot\psi^2\cdot\frac{\hbar}{2m}\Delta~dr~d\theta~d\phi$
?

No, that doesn't make sense. What is Δ supposed to be acting on?

 

[Lunar_Lander]

Sorry, I had once seen this notation ($T=\frac{\hbar}{2m}\Delta$) somewhere. Would it rather be:

$<T>=-\int\int\int r^2\cdot\sin(\theta)\cdot\psi^2\cdot\frac{\hbar}{2m}\Delta\psi~dr~d\theta~d\phi$ ?

I'll try the limits later on, but thanks for the help already :)!

 

[Steely Dan]

Sorry, I had once seen this notation ($T=\frac{\hbar}{2m}\Delta$) somewhere. Would it rather be:

$<T>=-\int\int\int r^2\cdot\sin(\theta)\cdot\psi^2\cdot\frac{\hbar}{2m}\Delta\psi~dr~d\theta~d\phi$ ?

I'll try the limits later on, but thanks for the help already :)!

You just have to be more careful with expectation value notation. That is, for an expectation value you sandwich the operator T between $\psi$ and its conjugate. So the kinetic energy operator (or equivalently, the momentum operator squared) is acting on $\psi$ only.

 

[Lunar_Lander]

I think I got <V>, the first solution I posted was from the indefinite integral.

$A^2\frac{e_0^2}{\epsilon_0}\int_0^{\infty}\exp(-2\alpha~r)r~dr$

$A^2\frac{e_0^2}{\epsilon_0}\cdot\frac{1}{4\alpha^2}=\frac{e_0^2\alpha}{4\pi\epsilon_0}$.

Also, A can be found by normalizing the original wave function:

$1=\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi}\psi^2r^2\sin(\theta)~dr~d\theta~d\varphi$

$1=\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi}A^2\exp(-2\alpha~r)r^2\sin(\theta)~dr~d\theta~d\varphi$

$1=2\pi\int_0^{\infty}\int_0^{\pi}A^2\exp(-2\alpha~r)r^2\sin(\theta)~dr~d\theta$

$1=4\pi\int_0^{\infty}A^2\exp(-2\alpha~r)r^2~dr$

$1=4\pi~A^2\frac{1}{4\alpha^3}$

$A=\sqrt{\frac{\alpha^3}{\pi}}$

 

[vela]

That looks right, except that you dropped the overall negative sign.

You should find the expected value for the kinetic energy is half that in magnitude. Did you come up with the correct expression for $\langle T \rangle$?