• Support PF! Buy your school textbooks, materials and every day products Here!

How do I find this expectation value?

  • Thread starter Kyrios
  • Start date
  • #1
28
0

Homework Statement


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

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


Homework Equations



for Z=1

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


The Attempt at a Solution



Using the values for
[tex] ψ = ψ_{2,0,0} - ψ_{2,1,0} [/tex]

I got

[tex] 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}) [/tex]

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

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257

Homework Statement


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

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


Homework Equations



for Z=1

[tex] < ψ | \hat{r} | ψ > = -3 \frac{4 π ε_0 \hbar^2}{m e^2} n_z[/tex]
What does ##n_z## represent? Is this result for the given ##\psi##?


The Attempt at a Solution



Using the values for
[tex] ψ = ψ_{2,0,0} - ψ_{2,1,0} [/tex]

I got

[tex] 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}) [/tex]
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.
 
  • #3
28
0
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:

[tex] < \hat{r} > = \int_{- \infty} ^ {\infty} ψ * \hat{r} ψ dV [/tex]

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

[tex] < \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 [/tex]

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

[tex] \hat{r} | ψ > = \hat{r} \psi_{200} - \hat{r} \psi_{210} [/tex]

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

[tex] < \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} [/tex]

And then integrate each part separately and add them up?
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
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:

[tex] < \hat{r} > = \int_{- \infty} ^ {\infty} ψ * \hat{r} ψ dV [/tex]

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

[tex] < \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 [/tex]

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

[tex] \hat{r} | ψ > = \hat{r} \psi_{200} - \hat{r} \psi_{210} [/tex]

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

[tex] < \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} [/tex]

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.)
 
  • #5
28
0
Yes. Some of those terms you can argue will integrate to 0.
I think it is the [tex] < \psi_{200} | \hat{r} | \psi_{200} > and < \psi_{210} | \hat{r} | \psi_{210} > [/tex] terms that will vanish, leaving
[tex] - < \psi_{200} | \hat{r} | \psi_{210} > - < \psi_{210} | \hat{r} | \psi_{200} > [/tex]

which would give me.. -6 a_0 after integrating ?
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
Looks good except that you neglected the effect of Z.
 
  • #7
28
0
Looks good except that you neglected the effect of Z.
Ah yes, I guess that would just make it

[tex] \frac{-6 a_0}{Z} [/tex]

since r is proportional to 1/Z ?
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
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.
 
  • Like
Likes 1 person
  • #9
DrClaude
Mentor
7,271
3,428
[tex] < ψ | \hat{r} | ψ > = -3 \frac{4 π ε_0 \hbar^2}{m e^2} n_z[/tex]
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 [tex] < \psi_{200} | \hat{r} | \psi_{200} > and < \psi_{210} | \hat{r} | \psi_{210} > [/tex] 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.
 
  • #10
28
0
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.
 
  • #11
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
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.
 

Related Threads on How do I find this expectation value?

Replies
11
Views
726
Replies
4
Views
1K
  • Last Post
Replies
1
Views
889
  • Last Post
Replies
2
Views
763
Replies
1
Views
1K
Replies
3
Views
2K
Replies
1
Views
942
Replies
6
Views
1K
Replies
31
Views
933
Replies
2
Views
2K
Top