To check if these wave functions are normalized to 1

In summary: R(1,0) is the radius of the particle at the origin, R(2,0) is the radius of the particle at the point (2,1). You should integrate with respect to r to find the most probable radius.
  • #1
quantum_prince
29
0
[SOLVED] to check if these wave functions are normalized to 1

I need to check if the following radial functions are properly normalized to unit probability

R(1,0) (r) =


2(1/ao)^3/2 e^(-r/ao)

R(2,1) (r) =


(1/2*ao)^3/2 *[ r/ sqrt(3)*a0] e^(-r/2ao)

We do know that


∫ u ^n e^(-u) du = factorial(n)
0

To normalize the wave function in the following way



∫ [tex] \phi ^2[/tex] = 1
-∞

Now applying the same for R(1,0)


∫ [2(1/ao)^3/2 e^(-r/ao)]^2 dr
-∞

=


∫ 4(1/ao)^3 e^(-2r/ao) dr
-∞


How do I proceed further.

Regards.
QP
 
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  • #2
Hint: Radial wave functions are only defined from r = 0 to +infinty.
 
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  • #3
Yes.Thats useful thanks.How do I still perform the integration.I don't know how to integrate it.
 
  • #4
but have the general formula for it, just do a substitution.

let 2r/ao = u

what is then n? Well n = 3. And what is 1/ao if 2r/ao = u ?

Have you done calculus classes?

Now try do this, and show me/us what you get.
 
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  • #5
As I understand if we integrate it with respect to r then we don't need to apply in that general formulae at all since everrything else is a constant except e raised to -r/ao. Suppose we integrate with respect to ao then this formulae can be applied.What should we integrate with respect to ao or r.

Regards
QP.
 
  • #6
ao is just a constant, infact it is the bohr radius. r is the guy that you should integrate over.

And in the R(2,0) case, you must use it. And now also, i don't know if your professor had said this, but the radial part of the wave function, there are "two ways" to have a radial wave function.

When you solve the radial Schrödinger eq, you do this ansatz:
[tex] R(r) = u(r)/r [/tex]

And then you solve the radial part with u(r) as your radial wave function; and the total wave function is normalized according to:

[tex] \int ^{\infty} _0 |R(r)|^2 r^2 dr = 1 [/tex]

due to the [tex] r^2 [/tex] that comes from the volume integration. [tex]d\vec{r} = r^2 sin \theta drd\theta d\phi [/tex]

And this then implies that:
[tex] \int ^{\infty} _0 |u(r)|^2 dr = 1 [/tex]

So what is your R(1,0) , R(2,0)? is it the R or u that I denoted here?
 
  • #7
It is R which you denoted here so an r(square) term must be introduced when I integrate with respect to r is it not.

So it this should be finally integrated is it not

∫ 4(1/ao)^3 e^(-2r/ao) r^2 dr
0
 
  • #8
Now I have hard to understand what you are saying, but if the R(1,0) is the R(r) i wrote, then yes, you should do the integration that you wrote. Now try to do the substitution i wrote eariler, and see if this becomes unity. If it doesn't, try figure out why =)
 
  • #9
I solved it no problems.It comes to one.All that needs to be done is substitution. The most crucial part of this problem is that additional factor r^2.Without that we cannot proceed anywhere.Thanks a lot for the hints.

How do I find the most probable value of radius r in the same question?
 
  • #10
where the wave-function has maximum. Just as you do with an "ordinary" function. The wave function represents probablity right? =)
 
  • #11
I still don't understand how to proceed.

This is what I need to compute.The radius r needs to be found for both radial wave functions.
 
  • #12
You don't know how to find maximum for a function?

First let me first show you the definitions again:

[tex] | \psi (x') |^2 dx' [/tex] is the probabilty to find the particle between x' and x' + dx', if the wave function [tex] \psi (x) [/tex] is normalized to unity.

Now for a 3-d wave function:

[tex] | \Phi (\vec{r'}) |^2 d\vec{r'} [/tex] is the probability to find the particle inside the infinitesimal volume element. (if it is normalized to unity of course).

Now the radial wave function comes from the ansatz of separation of variables when we solve the 3D shrödinger Equation.
[tex] \Phi (\vec{r}) = R(r) \cdot h(\theta , \phi ) [/tex]

Now what do you think you must do to find the most probable radius for the R(1,0) and R(2,0) ? Remember the volume element.

This is a good link: http://www.physics.gatech.edu/gcuo/lectures/ModernPhysicsLectures/MP16HydrogenAtom.ppt
OBS it is a power point file, so it can take some time to load. Slide 31 shows you the wave function vs. probability.

Now good luck =)
 
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  • #13
Its solved now.

Thanks a lot.

Regards.

QP.
 

Related to To check if these wave functions are normalized to 1

1. What does it mean for a wave function to be normalized to 1?

A wave function that is normalized to 1 means that the total probability of finding a particle in any possible state is equal to 1. This is a fundamental requirement for any valid wave function in quantum mechanics.

2. How do you check if a wave function is normalized to 1?

To check if a wave function is normalized to 1, you need to integrate the absolute square of the wave function over all possible states. If the result is equal to 1, then the wave function is normalized. This can also be checked using the normalization condition, which states that the integral of the wave function multiplied by its complex conjugate must equal 1.

3. Why is it important for wave functions to be normalized to 1?

Normalized wave functions are important because they represent the probability amplitudes for finding a particle in a particular state. If a wave function is not normalized, the probabilities will not be accurate and the results of any calculations will be incorrect.

4. Can a wave function be normalized to a value other than 1?

Yes, a wave function can be normalized to a value other than 1. This is known as partial normalization and is commonly used in quantum mechanics for mathematical convenience. However, in most cases, wave functions are normalized to 1 for simplicity and to accurately represent the probabilities of finding a particle in a particular state.

5. What happens if a wave function is not normalized to 1?

If a wave function is not normalized to 1, the probabilities of finding a particle in different states will not add up to 1. This means that the results of any calculations using that wave function will not be accurate. In some cases, it may also violate the laws of quantum mechanics and lead to physically impossible results.

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