# Homework Help: Max probability of finding an electron in 3dz2

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1. Oct 2, 2016

### Titan97

1. The problem statement, all variables and given/known data
The radial wavefunction for $3d_{z^2}$ orbital is
$$R=N\sigma^2e^{-\frac{\sigma}{3}}(3\cos^2\theta-1)$$
$$\sigma=\frac{r}{a_0}$$
Find r and θ for which the probabiity of finding the electron is maximum

2. Relevant equations
None

3. The attempt at a solution
$R^2$ gives the probability density. and $4\pi r^2dr\cdot R^2$ gives the probability at $(r,r+dr)$. How can I find a particular $r$? Should I use $4\pi r^2\cdot R^2$ as the probability at $r$? but that doesnt make any sense to me.

Last edited: Oct 3, 2016
2. Oct 2, 2016

### kuruman

You have a function of r (or σ) and you need to maximize it. How do you maximize (or minimize) a function?

Last edited: Oct 2, 2016
3. Oct 2, 2016

### haruspex

Not sure how to read the expression for R. What exactly in the exponent of e? Looks like you did not bracket it correctly in the LaTeX.

4. Oct 2, 2016

### haruspex

I think you have not understood the difficulty.
Consider a probability density function over the plane, expressed in polar coordinates. If it asks for the most likely r, you would take into account the whole area in the annulus (r, r+dr). But that might not be the same r as you would find for the peak local density at a single (r, θ) point.
Titan97, I would take the single point view, so no 4πr2 factor.

5. Oct 2, 2016

### kuruman

Perhaps I misunderstood the question. It is asking "Find r and θ for which the probability of finding the electron is maximum". I interpreted "probability of finding the electron" as ψ*ψ dV which is dimensionless not as the probability density ψ*ψ which has dimensions. It looks like the language is unclear because "probability" should be followed either by "within volume element dV" or by "density".

6. Oct 2, 2016

### haruspex

An area element can be thought of as consisting of a dr axis and an rdθ axis. If you read the question as finding the likelihood peak in respect of r, that would involve integrating rdθ before differentiating wrt r. Thus there is a bias towards greater r. Your interpretations (which I support) maximise wrt r and θ independently, so no initial integration step.

7. Oct 3, 2016

### Titan97

@haruspex I cant do a dr times the probability density. I need the probability as a function of r and theta. If I just take |R|^2, that will give me the probability density.

8. Oct 3, 2016

### haruspex

I did not suggest that. I wrote that if you wanted the probability of r being in a certain range you would have to integrate over the annulus. In the plane polar illustration I described for kuruman, that would be integrating with rdθ. Is that not the point of your post?

9. Oct 4, 2016

### kuruman

Yes, there is a bias towards greater r and we agree that one needs to maximize independently. We also agree that the probability of finding the electron within element dV is given by dP = const. R2(r) (Y20)2(dr)(r sinθ dθ)(r dφ)
In the expression above, I separated the volume element dV into the three "axes" as suggested by haruspex. At this point, haruspex suggests (I think) that one should proceed by maximizing the radial part R2(r) separately from the angular part (Y20)2 to find values for r and θ. This is maximizing the probability distribution, not the probability and I agree that, if the problem asked to maximize the probability, that's what one should do.

The problem is asking "Find r and θ for which the probability of finding the electron is maximum", which I find ambiguously incomplete, like asking "find the work done on the charge when moved from point A to point B in an electric field." Done by what force? The agent who moves the charge or the electric force? Here, I find myself asking what probability? Let me explain how I see it. Starting with what we all agree on, dP = const. R2(r) (Y20)2(dr)(r sinθ dθ)(r dφ), the probability that the electron is found within area at element dA = (dr)(r sinθ dθ) and any azimuthal angle φ is obtained by integrating over the azimuthal angle to get
dP = const1 R2(r) (Y20)2r (dr)(r sinθ dθ)
At this point, if we wish to maximize independently, we can find the probability of finding the electron at any angle θ and any angle φ by integrating over θ to get
dP = const2 R2(r) r2 (dr)
This can be interpreted two ways
1. Leave it as is so that dP = const2 {r2 R2(r)} dr which is the probability of finding the electron at any angle φ and θ and radially between r and r + dr. What gets maximized is what's between angular brackets.
2. "Borrow" 4π from the constant and write this as dP = const3 {R2(r)} A dr which is the probability of finding the electron anywhere within a spherical shell of area A = 4πr2 and thickness dr. As before, what gets maximized is what's between angular brackets and this maximization gives the same answer as maximizing the radial probability distribution.

What if we integrated the radial part first? Then, we would get
dP = const4 (Y20)2(sinθ dθ)
If we want to maximize the probability of finding the electron between θ and θ+dθ, we maximize sinθ (Y20)2.
If we want to maximize the probability of finding the electron between cosθ and cos(θ+dθ), we maximize (Y20)2 because sinθ dθ = d(cosθ) (never mind the negative sign; it's there to show that cosθ decreases with increasing θ).

This is how I see it.

10. Oct 4, 2016

### BvU

I'm not convinced integration is intended by the exercise composer
Probability is probability density times $dV$. Higher $\Psi^*\Psi$ means a higher probability (for equal $dV$).
It would be different if the exercise asked for the most probable r or the most probable $\theta$.

11. Oct 5, 2016

### kuruman

I stand corrected. Thanks.

12. Oct 10, 2016

### Titan97

A higher psi*psi means a higher probability density. Not a higher probability for different dV. Am I correct?

And, After reading some articles, i found that $4\pi r^2 R^2$ (R is the radial wave function) gives the probability.

13. Oct 11, 2016

### BvU

Of course, provided you mean $\ |d^3V| \$.
No. Be more precise and more explicit. As you should know by now, $\Psi^* \Psi$ is a probability density. Your expression has the wrong dimension for a probability.

14. Oct 11, 2016

### Titan97

@BvU I argued this with my professor and so did others in my class. But he told this gives the probability even if it has a dimension.(I couldn't argue more as i trusted his statement becuase his education level is far higher than mine). I even got marks for using the above formula.

I used $$\frac{\text{d}r^2R^2}{\text{d}r}=0$$ to get the answer.

15. Oct 11, 2016

### BvU

It is wise not too argue too much with a teacher. But you don't want to take a probability with a dimension for granted either -- dificult.

See Haru in post #4.

I maintain that $\Psi^* \Psi \;d^3 V$ is a probability to e.g. find a particle in that $d^3 V$
and thereby the highest probability is where $\Psi^* \Psi$ is maximum

$4\pi r^2 dr$ is the volume of a shell between $r$ and $r+dr$ so that $R(r)^2 \;4\pi r^2 \;dr\ \$ (with $R(r)$ the radial part of the wave function $\Psi$) is the probability to find the particle in such a shell. But only for spherically symmetric wave functions where you can separately integrate over $\theta$ and $\phi$.

16. Oct 13, 2016

### Titan97

@BvU high probabilty density does not mean high probability. How can you say psi*psi has to be maximum for probability to be maximum?

17. Oct 13, 2016

### BvU

How does one compare probabilities ? I would say per equal volume. we've been through this in #12 and #13. In #10 I wonder about the intentions of the exercise composer; did you render the exact formulation of the exercise ?

(Because asking for the most probable $r$ or the most probable $r^2$ is asking for something else than asking for the " $r$ and $\theta$ " where the probability of finding the electron is maximum !)