How do I find this expectation value?

[Kyrios]

Homework Statement

A hydrogen like ion (with one electron and a nucleus of charge Ze) is in the state
ψ = ψ_{2,0,0} - ψ_{2,1,0}

What's the expectation value of \hat{r} (position operator) as a function of Z?
Assuming origin at nucleus.

Homework Equations

for Z=1

< ψ | \hat{r} | ψ > = -3 \frac{4 π ε_0 \hbar^2}{m e^2} n_z

The Attempt at a Solution

Using the values for
ψ = ψ_{2,0,0} - ψ_{2,1,0}

I got

ψ = \frac{1}{4 \sqrt{2 π a_0 ^3}} e^{- \frac{r}{2 a_0}} ( 2 - \frac{r}{a_0} - \frac{r cos(\theta)}{a_0})

I wouldn't have a clue how to integrate this and I imagine there must be an easier way to find the expectation value.

 

[vela]

Homework Statement

A hydrogen like ion (with one electron and a nucleus of charge Ze) is in the state
ψ = ψ_{2,0,0} - ψ_{2,1,0}

What's the expectation value of \hat{r} (position operator) as a function of Z?
Assuming origin at nucleus.

Homework Equations

for Z=1

< ψ | \hat{r} | ψ > = -3 \frac{4 π ε_0 \hbar^2}{m e^2} n_z

What does $n_z$ represent? Is this result for the given $\psi$?

The Attempt at a Solution

Using the values for
ψ = ψ_{2,0,0} - ψ_{2,1,0}

I got

ψ = \frac{1}{4 \sqrt{2 π a_0 ^3}} e^{- \frac{r}{2 a_0}} ( 2 - \frac{r}{a_0} - \frac{r cos(\theta)}{a_0})

You're going to have to normalize the wave function before calculating the expectation value.

I wouldn't have a clue how to integrate this and I imagine there must be an easier way to find the expectation value.

Why not? It's an integral you should be able to do if you've taken basic calculus. Once you have the correct integral set up, it may look intimidating, but it's a straightforward to evaluate.

Initially, keep the integral in terms of $\psi_{200}$ and $\psi_{210}$. You'll be able to argue that some of the terms will vanish when integrated.

 

[Kyrios]

What does $n_z$ represent? Is this result for the given $\psi$?

It's the unit vector pointing in the z-direction. Yes, this is the Z=1 result for the given ψ.

You're going to have to normalize the wave function before calculating the expectation value.

The wave function is already normalised.


As for the integral, this is what I was doing initially:

< \hat{r} > = \int_{- \infty} ^ {\infty} ψ * \hat{r} ψ dV

< \hat{r} > = \int_{- \infty} ^ {\infty} ψ * ψ r^3 dr

< \hat{r} > = \int_{- \infty} ^ {\infty} \frac{1}{32 π a_0 ^3} e^{- \frac{r}{a_0}} ( 2 - \frac{r}{a_0} - \frac{r cos(\theta)}{a_0})^2 r^3 dr

What I've got here, even wolfram alpha can't integrate.

By keeping the ψ_{2,0,0} and ψ_{2,1,0} separate, do you mean

\hat{r} | ψ > = \hat{r} \psi_{200} - \hat{r} \psi_{210}

< \psi | \hat{r} | ψ > = ( \psi_{200} - \psi_{210} ) ( \hat{r} \psi_{200} - \hat{r} \psi_{210} )

< \psi | \hat{r} | ψ > = \psi_{200} \hat{r} \psi_{200} - \psi_{200} \hat{r} \psi_{210} - \psi_{210} \hat{r} \psi_{200} + \psi_{210} \hat{r} \psi_{210}

And then integrate each part separately and add them up?

 

[vela]

The wave function is already normalised.

No, it isn't. The individual wave functions are, but the linear combination isn't.

As for the integral, this is what I was doing initially:

< \hat{r} > = \int_{- \infty} ^ {\infty} ψ * \hat{r} ψ dV

< \hat{r} > = \int_{- \infty} ^ {\infty} ψ * ψ r^3 dr

< \hat{r} > = \int_{- \infty} ^ {\infty} \frac{1}{32 π a_0 ^3} e^{- \frac{r}{a_0}} ( 2 - \frac{r}{a_0} - \frac{r cos(\theta)}{a_0})^2 r^3 dr

What I've got here, even wolfram alpha can't integrate.

The limits and volume element aren't correct. In spherical coordinates, you want to calculate
$$\int_0^\infty \int_0^\pi \int_0^{2\pi} f(r,\theta,\phi)\,r^2\sin\theta\,d\phi\,d\theta\,dr$$ in general.

By keeping the ψ_{2,0,0} and ψ_{2,1,0} separate, do you mean

\hat{r} | ψ > = \hat{r} \psi_{200} - \hat{r} \psi_{210}

< \psi | \hat{r} | ψ > = ( \psi_{200} - \psi_{210} ) ( \hat{r} \psi_{200} - \hat{r} \psi_{210} )

< \psi | \hat{r} | ψ > = \psi_{200} \hat{r} \psi_{200} - \psi_{200} \hat{r} \psi_{210} - \psi_{210} \hat{r} \psi_{200} + \psi_{210} \hat{r} \psi_{210}

And then integrate each part separately and add them up?

Yes. Some of those terms you can argue will integrate to 0. (If you don't trust your physical intuition, you could also integrate them and verify that they vanish.)

 

[Kyrios]

Yes. Some of those terms you can argue will integrate to 0.

I think it is the < \psi_{200} | \hat{r} | \psi_{200} > and < \psi_{210} | \hat{r} | \psi_{210} > terms that will vanish, leaving
- < \psi_{200} | \hat{r} | \psi_{210} > - < \psi_{210} | \hat{r} | \psi_{200} >

which would give me.. -6 a_0 after integrating ?

 

[vela]

Looks good except that you neglected the effect of Z.

 

[Kyrios]

Looks good except that you neglected the effect of Z.

Ah yes, I guess that would just make it

\frac{-6 a_0}{Z}

since r is proportional to 1/Z ?

 

[vela]

I don't recall the Z dependence offhand, but what you said does sound familiar. It would probably be best if you looked at how the derivation changes if you use $Ze^2/r$ for the potential instead of just $e^2/r$ in the Schrodinger equation.

 

[DrClaude]

< ψ | \hat{r} | ψ > = -3 \frac{4 π ε_0 \hbar^2}{m e^2} n_z

That's not possible. The expectation value of $\hat{r}$, or average position, can't be negative.

It's the unit vector pointing in the z-direction.

It can't be. The expectation value of $\hat{r}$ is scalar, not a vector.

I think it is the < \psi_{200} | \hat{r} | \psi_{200} > and < \psi_{210} | \hat{r} | \psi_{210} > terms that will vanish

You just gave above the equation for $\langle \psi_{n l m} | \hat{r} | \psi_{n l m} \rangle$. Was it zero?

which would give me.. -6 a_0 after integrating ?

Again, you can't have a negative value.

 

[Kyrios]

That's not possible. The expectation value of $\hat{r}$, or average position, can't be negative.
It can't be. The expectation value of $\hat{r}$ is scalar, not a vector.

I'm just writing what it says on the sheet. $\hat{r}$ is written as the position operator, and I guess the electron is just along the negative z axis with respect to the nucleus. I don't think my professor would have made a mistake in his question (at least I rather hope not). I kind of see what you mean, but I've had negative expectation values before. Thanks anyway though.

 

[vela]

I think your professor meant the observable corresponding to the position vector $\vec{r}$ rather than the spherical coordinate $r$. At least, that's what I was assuming given what you had written.