Showing wavefunctions are orthonormal.

In summary, I'm having trouble showing that the wavefunctions are orthonormal (i.e. Orthogonal and normal). I tried setting the limits from 0 to infinity for the radial part and 0 to π for cos²θ , so: 1/(16πa^5) . ∫ (r² . exp(-r/a)) dr . ∫ cos²θ dθ = 1. However, when I integrate the radial part I get 2a³ and when I integrate cos²θ I get π/2 . I don't know what I'm doing wrong. Thanks very much.
  • #1
hhhmortal
176
0

Homework Statement



I've attached the question as a jpeg.

I'm having trouble showing that the wavefunctions are orthonormal (i.e. Orthogonal and normal)

When I try to show U_210 is normal I don't get 1:

My working:


∫ |U_210|² = 1

I set the limits from 0 to infinity for the radial part and 0 to π for cos²θ , so:

1/(16πa^5) . ∫ (r² . exp(-r/a)) dr . ∫ cos²θ dθ = 1


I used the identity given in the question to integrate the radial part and I get 2a³

Then I integrate the cos²θ and I get π/2

Therefore,

1/(16πa^5) . 2a³ . π/2 = 1 (But of course this does not equal to 1, and I don't know what I'm doing wrong)

Thanks very much.
 

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  • #2
You forgot the Jacobian for spherical coordinates.
 
  • #3
Cyosis said:
You forgot the Jacobian for spherical coordinates.

Aha! So I'm integrating cos²θ w.r.t to dr and not dθ?

If so how can I convert the integral from dr to dθ..
 
  • #4
No you will need to integrate cos²θ with respect to θ of course. You are integrating over a volume, in Cartesian coordinates the volume element would be dxdydz. However we're not using Cartesian coordinates, but spherical coordinates. Therefore you need to integrate over the volume element in spherical coordinates.
 
  • #5
To put what Cyosis said (in post #2) in other words, the normalization integral is a volume integral:

[tex]\int {\int {\int {|\psi_{210}|^2 dV}}} = 1[/tex]

where dV is a volume element. If you were doing the integral in rectangular coordinates, dV would be dx dy dz. But you're actually using spherical polar coordinates. What's dV in spherical polar coordinates?

(Cyosis beat me to it while I was proofreading my LaTeX!)
 
  • #6
Oh right yeap, I completely forgot I was integrating w.r.t to the volume. Thanks!

There is something else I am not exactly sure on. The very first part of the question asks me to prove that the degeneracy is n² .

I know that the number of values of m for a particular value of l is (2l + 1) and that there are n values of l. When you add all the combinations together it is actually n² but how can I show this.
 
  • #7
Just add them up for a given n. For each value of l, you have 2l+1 states, so the total number of states is

[tex]\sum_l (2l+1)[/tex]

You just need to fill in the correct limits for the summation and evaluate the sum.
 
  • #8
vela said:
Just add them up for a given n. For each value of l, you have 2l+1 states, so the total number of states is

[tex]\sum_l (2l+1)[/tex]

You just need to fill in the correct limits for the summation and evaluate the sum.

So there is no neat way of proving it is equal to n² . You just have to show that the summation of all the states is the same as n²?
 
  • #9
hhhmortal said:
So there is no neat way of proving it is equal to n² . You just have to show that the summation of all the states is the same as n²?
Yeah, there's no clever way I'm aware of, but you don't need one. Showing the sum is equal to n2 is straightforward.
 
  • #10
One last thing that is bothering me, is when I try to normalise the wavefunction U_211

I get part of the integral:

∫ exp(2iφ) dφ

limits between 2Pi to 0. How can I evaluate this, to show that the wavefunction is normalised.
 
  • #11
I am sure you can find the primitive of a simple exponential function. Once you do this you will notice that something is very wrong!
 
  • #12
hhhmortal said:
One last thing that is bothering me, is when I try to normalise the wavefunction U_211

I get part of the integral:

∫ exp(2iφ) dφ

limits between 2Pi to 0. How can I evaluate this, to show that the wavefunction is normalised.
Is it bothering you because what you tried gave you 0 as the answer? If so, you integrated correctly. That factor shouldn't be there.
 
  • #13
vela said:
Is it bothering you because what you tried gave you 0 as the answer? If so, you integrated correctly. That factor shouldn't be there.

Of course I forgot about the complex conjugate when normalising the wavefunctions, so essentially this will get rid of the exponential of the wavefunction I am taking the complex conjugate of, but there will still be an exponential term in the integral. I am not sure where this is going..
 
  • #14
U_210 had an exponential in the integral as well. What's the problem? The complex exponential disappears, but the real exponential, just as for u210, stays.
 
  • #15
Cyosis said:
U_210 had an exponential in the integral as well. What's the problem? The complex exponential disappears, but the real exponential, just as for u210, stays.

What I'm trying to do is integrate the following:

∫exp(φ).exp(iφ) dφ

Before, I could easily integrate the exponential because I had the identity, but when I try integrating this I get a factor of 'i'. My maths is terrible, this is partly the reason why I can't see the problem very well.
 
  • #16
You need to show us what you're doing from the beginning because you shouldn't have a term like that.
 
  • #17
What I'm trying to do is integrate the following:

∫exp(φ).exp(iφ) dφ

As Vela said that term shouldn't be there. Remember [itex]|\psi|^2=\psi^* \psi[/itex].

That said I find it a little odd that you aren't able to compute a simple exponential function and I suggest you brush up your calculus immediately or you will run into a lot of difficulties during your QM course.

Before, I could easily integrate the exponential because I had the identity, but when I try integrating this I get a factor of 'i'. My maths is terrible, this is partly the reason why I can't see the problem very well.

If you integrate that expression you will indeed get a factor 'i'. However to not confuse you I will say it again, those exponentials shouldn't be there in the first place!
 
  • #18
Cyosis said:
As Vela said that term shouldn't be there. Remember [itex]|\psi|^2=\psi^* \psi[/itex].

That said I find it a little odd that you aren't able to compute a simple exponential function and I suggest you brush up your calculus immediately or you will run into a lot of difficulties during your QM course.



If you integrate that expression you will indeed get a factor 'i'. However to not confuse you I will say it again, those exponentials shouldn't be there in the first place!

Wow! that was very stupid of me! The mix up was with the complex conjugate, very often examples in QM from my course that involve normalisation of a wavefunction don't have complex numbers. After fixing that I got a factor of Pi, hence normalising the wavefunction. I assume the limits are from 0 to Pi
 

FAQ: Showing wavefunctions are orthonormal.

What does it mean for wavefunctions to be orthonormal?

Orthonormality refers to a mathematical property of a set of functions where they are mutually orthogonal and normalized. This means that the functions are perpendicular to each other and have a magnitude of 1.

How is the orthonormality of wavefunctions shown?

To show that wavefunctions are orthonormal, we must demonstrate that their inner product (integral of the product of two functions) is equal to 0 if the functions are not the same and equal to 1 if the functions are the same.

Why is it important for wavefunctions to be orthonormal?

Orthonormality is important in quantum mechanics because it allows us to easily calculate the probability of finding a particle in a certain state. It also simplifies calculations and allows us to easily manipulate wavefunctions in mathematical equations.

Can wavefunctions be both orthogonal and normalized, but not orthonormal?

Yes, it is possible for wavefunctions to be orthogonal and normalized, but not orthonormal. This means that the functions are perpendicular to each other, but do not have a magnitude of 1.

Are orthonormal wavefunctions unique?

Yes, orthonormal wavefunctions are unique. This means that no two orthonormal wavefunctions can be the same, as they must be completely perpendicular to each other and have a magnitude of 1.

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