# Probability density function,normalize and expectation values

1. Aug 4, 2012

### zhillyz

1. The problem statement, all variables and given/known data

The probablity density function of the $n-state$ of an electron is proportional to

$f$$n$$=$$(\frac{rz}{a_{0}})^{2n}$$e^ \frac{-2Zr}{\large na_{0}}$

show that the expectation value of the potential energy of the electron in
the $n-th$ quantum state of the hydrogen atoms is;

$\frac{{-Z^2}{e^2}}{\large a_{0} n^2}$

For that do

(a) derive normalisation factor of the function. (5 marks)
(b) calculate the expectation value for $<1/r>$. (5 marks).
(c) calculate the expectation value for $< V >$. (5 marks).

You may use Mathematica or Mathcad for this exercise

2. Relevant equations

Normalization procedure

Take a function $f(x)$ and the Normalizing Constant as $N$ then

$N^2∫f(x)*f(x) dx = 1$
and
$N = \frac{1}{\sqrt{\large\int_o^\infty f(x)*f(x)}}$

3. The attempt at a solution

Put equation into mathematica then square it, then integrate it, then square root it and then divide function by answer?
If integrated from negative to positive infinity it should equal zero?

Wondering if my logic is okay before I move onto expectation values?

2. Aug 4, 2012

### cepheid

Staff Emeritus
Yeah, your logic for normalization seems fine. The only thing that I don't understand is the bit in red, which doesn't make sense to me. fn is a function of r, the radial coordinate of the electron right? Well, r can't be less than 0 anyway, since it is a distance. So I don't think you can integrate the function over negative r, but even if you could, I don't understand why you claim that the result would be 0.

3. Aug 4, 2012

### gabbagabbahey

Keep in mind that this isn't a purely one-dimensional problem:

$$1=N^2\int_{ \text{all space} }|f^2(\mathbf{r})|d^3 \mathbf{r}$$

What is the volume differential in spherical coordinates?

4. Aug 5, 2012

### zhillyz

The zero was a typo I meant 1. I have only briefly been exposed to spherical coordinates about a year ago so not sure how I shall use them.

From researching online though I can see that;

$dV = r^2sin\theta \,dr\, d\theta \,d\varphi$.

and

$\int_{\varphi = 0}^{2\pi} \int_{\theta = 0}^\pi \int_{r = 0}^\infty f(r,\theta,\varphi) r^2sin\theta \; dr \; d\theta \; d\varphi$.

I shall see what I can find on using these when I come home.

5. Aug 5, 2012

### zhillyz

So I am not entirely sure how I am getting on here, but I think the point in the question stating that MathCad or Mathematica could be used was as the integral is difficult, and as I have never used spherical coordinates or Mathematica I don't know if what Im doing is correct.

Basically I took great effort with the input of my function $'f$$n$$'$ and got the software to square the function which gave me the function below back;

$f$$n$$=\frac{n^2 rZ^4\exp\frac{-4Zr}{an}}{a^4}$

but then when I integrate it with respect to r from 0 to infinity I get the exact same function back with and infinity symbol on the end.

Also once I have got the correct integral I plan to square root it and divide 1 by the answer to have my Normalization constant which I would then multiply the initial function with to normalize it does this seem correct?

Thank you in advance for any help.

6. Aug 5, 2012

### gabbagabbahey

First off, I think I erred in my previous post (and failed to catch your error). You need to distinguish between wavefunction and probability density.

For an un-normalized wavefunction $\Psi(\mathbf{r})$, the normalization constant is given by

$$N^2\int_{ \text{all space} } | \Psi(\mathbf{r}) |^2 d^3 \mathbf{r} = 1$$

How is the probability density $f_n(r)$ related to to the wavefunction?

7. Aug 5, 2012

### zhillyz

Ahh, the probability density function is the wave function multiplied by its complex conjugate and integrated from one point to another?

probability density function = $f$$n$$= \int \varphi * \varphi \, dr$

So normalization factor is merely;

$\frac{1}{\large\sqrt{f_n}}$

Does this seem correct?

8. Aug 5, 2012

### gabbagabbahey

No, the probability density is just the wavefunction multiplied by its complex conjugate (not integrated - it's a density after all) $f_n(r)=| \Psi(\mathbf{r}) |^2$ so,

$$N^2\int_{ \text{all space} } | \Psi(\mathbf{r}) |^2 d^3 \mathbf{r} = N^2\int_{ \text{all space} } f_n(r) d^3 \mathbf{r} =1$$

You will still need to integrate (do the angles first, and don't forget the factor of r^2 from d^3r when doing the radial integration). If you have trouble getting Mathematica to spit out a useable answer, post your Mathematica input in code tags so we can see what you are doing wrong.

9. Aug 5, 2012

### zhillyz

Damn, thought I had got it easy there for a second :) okay 2minutes I will see what I can do.

10. Aug 5, 2012

### zhillyz

Right, so I have ignored the Mathematica part of this question so far till I get my head round spherical coordinate integration. The way I see it just now what I need to do is this;

$\int_0^{\pi}\int_0^{2\pi}\int_0^{\infty} f_n\; {d^3}r$

${d^3}r = dV = {r^2}sin\varphi dr d\theta d\varphi$

So in $f_n$ there is no term dependent on$\phi$or$\theta$ so we can evaluate as follows;

$\int_0^{\pi} sin\varphi \,d\varphi *\int_0^{2\pi} 1 \,d\theta * \int_0^\infty f_n dr$

which equates to

$(-cos \pi + cos 0)*(2\pi)*\int_0^{\infty} f_n(r)*r^2 dr$

and then I need to get into Mathematica yes?

11. Aug 5, 2012

### zhillyz

Integrate[(((rZ)/(a))^(2 n))*(Exp[(-2 rZ)/an])*(r^2), {r, 0, Infinity}]

Integrate::idiv: Integral of r^2 does not converge on {0,\[Infinity]}. >>

12. Aug 5, 2012

### gabbagabbahey

Mathematica isn't human, it will interpret "rZ" as a single variable and "r Z" as the product of two variables

13. Aug 5, 2012

### zhillyz

Ah okay thank you, and the post before that seem okay to you?

14. Aug 5, 2012

### zhillyz

In[8]:= Integrate[(((r Z)/(a))^(2 n))*(Exp[(-2 r Z)/a n])*(r^2), {r,
0, Infinity}]

Out[8]= ConditionalExpression[(
2^(-3 - 2 n) a^3 (Z/a)^(2 n) ((n Z)/a)^(-2 n) Gamma[3 + 2 n])/(
n^3 Z^3), Re[(n Z)/a] > 0 && Re[n] > -(3/2)]

15. Aug 5, 2012

### gabbagabbahey

Seems fine to me (although I would rewrite -cos∏+cos0 as 2 ).

Whenever you are computing a multidimensional integral (surface, volume, etc.) as long as the integrand is not a function of more than one variable, and the variables of integration are independant over whatever region you are integrating over, you can rearrange the integrals in that manner/

One thing you should take note of is that the $N$ you will find like this is the normalization constant for the wavefunction. What is the normalization constant for the probability density?

16. Aug 5, 2012

### gabbagabbahey

n Z and a are real and positive, and moreover n is and integer. You can tell mathematica this explicitly by adding assumptions to your integrate statement, and thus avoid the conditional statement (and the gamma function) in the answer

17. Aug 5, 2012

### zhillyz

I thought I would leave it as -cos pi + cos 0 just to show I had understood the process of integration haha.

Is the the normalization constant of a pdf just integrating the pdf from 0 to infinity?

18. Aug 5, 2012

### gabbagabbahey

Sort of. Assume that your pdf is normalized by multiplying by a constant $\alpha$ so that $f_n(r)_{ \text{normalized} } = \alpha f_n(r)$. Integrating a normalized pdf over all space should give 1 (the probability of finding the electron somewhere in the universe), so you have

$$\int_{ \text{all space} } f_n(r)_{ \text{normalized} } d^3 \mathbf{r} = \int_{ \text{all space} } \alpha f_n(r) d^3 \mathbf{r} = 1$$

How does $\alpha$ relate to $N$?

19. Aug 5, 2012

### zhillyz

Integrate[(((r Z)/(a))^(2 n))*(Exp[(-2 r Z)/a n])*(r^2), {r, 0, Infinity}, Assumptions -> {a \[Element] Reals, Z \[Element] Reals,
r \[Element] Reals, a > 0, Z > 0, r > 0, n > 0,
n \[Element] Integers}]

Out[25]= (2^(-3 - 2 n) a^3 n^(-3 - 2 n) Gamma[3 + 2 n])/Z^3

20. Aug 5, 2012

### zhillyz

$N^2 = \int_0^{\infty} f_n d^3r = \int_0^{\infty} \alpha f_n d^3r$

So

$α=\frac{d^3N^2}{d^3r f_n}$

Is that right?