vmelkon
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Schrodinger wave equation for the 1 electron atom
I'm reading my chemistry textbook about quantum physics.
There are a list of equations for the "Schrodinger wave equation for the 1 electron atom".
For the 1s orbital, the equation is
1/sqrt(Pi)*(1/5.29e-11)^(3/2)*E^(-x/5.29e-11)
I know, it is rather hard to read that.
Anyway, it says squaring the wave equation is interpreted as giving the probability of finding the electron.
So I used Maple V to integrate
int((1/sqrt(Pi)*(1/5.29e-11)^(3/2)*E^(-x/5.29e-11))^2, x=0..0.00000000002);
and it gives a very high number, which is impossible since the RESULT MUST BE LESS THAN 1.
I got 0.3017282358 * 10^20.
So, what did I do wrong?
I defined E as 2.7182818284590452353602874713527.
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smaerd
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I take it your E is, e?
I solved the problem and also came to an unusually large number. Edit - Oh uhm, do it in spherical polar coordinates. You're X should be an r. That's
why. It's a 3-D problem.
edit - If you have any issues doing that post back and I'll help. Unless it's homework... If it's homework, do your own goddamn home-work and don't
use maple for integrations you can do in your head!
[Edited on 20-9-2014 by smaerd]
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phlogiston
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I have mediocre math skills and it has been a long time since I used them , but for the sake of exercise I had a go:
First, radial wave function for 1s in spherical coordinates (from PW Atkins):
psi = sqrt(1/(pi*a^3)) * e^(-r/a)
where a = bohr radius = 52.9E-12 m
pi = 3.1415926535...
psi = radial wavefunction
square it:
psi*psi = 1/(pi*a^3) * e ^ (-2r/a)
find integral (imagine S is integral sign):
S psi*psi dr = 1/(pi*a^3)*a/-2 * e^(-2r/a)
which rearranges to
= -1/(2*pi*a^2) * e ^ (-2r/a)
With r = 2E-12 m
I get -1.096E20
Also a large number (same order of magnitude), but different and negative.
I'd be interested to learn where I went wrong, I went over it a couple of times and can't find any mistake.
I am not sure this wave function is normalised.
[Edited on 20-9-2014 by phlogiston]
[Edited on 20-9-2014 by phlogiston]
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smaerd
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The wave function is normalized. Nice attempt! The normalization coefficient is: sqrt(1/(π*a^3)) But you must square it as it is now part of Psy
(without fancy terminology).
It's a triple integral when you pop it into spherical polar.
The volume integral goes as follows
Integral φ=0 to 2π Integral θ = 0 to π Integral r = 0 to 0.00000000002 sqrt(1/(pi*a^3)) ^2 * (e^(-r/a))^2 * r^2 sin(θ) dr dθ dφ
Notice that when we take this to spherical polar we multiply by : r^2 sin(θ) dr dθ dφ and the two other integrands appear(not arbitrary review
spherical polar probably also in atkins).
So let's handle the square of the normalization coefficient...
sqrt(1/(π*a^3)) * sqrt(1/(π*a^3)) = 1/(π*a^3)
Nice...
So I will break up the integrations as that can be the easiest way to approach these problems...
Integral θ = 0 to π sin(θ) dθ
= -cos(θ) from 0 to 2π
= -(cos(2π) - cos(0))
= -(-1 -1)
= 2
Now,
Integral φ=0 to 2π dφ
simply equal to 2π
Okay now the integral you probably do not want to solve by hand but is actually quite easy... Can approach this using integration by parts or using a
table of integrals... Basically it comes out to be a very simple series. I won't show my work but I typically do tanzalin integration by parts. In
this case, in my head. If I make an error here it's for you to figure out because I don't want to actually do this with pencil and paper...
Integral r = 0 to 0.00000000002 r^2 * (e^(-2r/a))
= e^(-2r/a) *[ (-a * r^2/2) + (-2*a^2*r/4) + (-2*a^3/8)] from r=0 to 0.00000000002
simplify
= e^(-2r/a) *[ (-a * r^2/2) + (-a^2*r/2) + (-a^3/4)] from r=0 to 0.00000000002
evaluate...
= e^(-4E-11/a) *[ (-a * 2E-11^2/2) + (-a^2 * 2E-11/2) + (-a^3/4)] - e^(0) *[ (0 + 0 + (-a^3/4)]
simplify...
= e^(-4E-11/a) *[ (-a * 2E-22) + (a^2 * 1E-11) + (-a^3/4)] + [a^3/4]
where a = 52.9E-12, s.t...
= 0.4694733 * [-1.058E-32 - 2.79841E-32 - 3.7008E-32] + [3.7008E-32]
= 1.528916E-33
so the answer I get is....
1.528916E-33 * 4 * π * (1/(π*a^3))
= 6.115664E-32 * 1/(a^3)
=0.4131
Feel like I made a small algebraic mistake but I think you get the jist... Also I forget whether or not you have to use the bohr radius constant in
meters or not. I think you keep it as a whole number.
[Edited on 20-9-2014 by smaerd]
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blogfast25
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From: Laguerres&hydrogenatom.pdf
The first six normalised radial wave functions of the hydrogen atom (a<sub>0</sub> = 0.529 Angstrom):
How to normalise a general expression of the Radial:
(Where L(n,l) stands for the Laguerre polynomial and A is the normalisation constant)
smaerd: your triple integration isn't necessary. These wavefunctions are radially symmetrical.
So, if normalised then:
∫<sub>0</sub><sup>∞</sup> [R<sub>1,0</sub>(r)]<sup>2</sup> r<sup>2</sup> dr = 1
If I have time I’ll check that.
[Edited on 20-9-2014 by blogfast25]
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smaerd
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Actually the bohr radius should be in meters not pm not sure why I even said that. haven't had my morning espresso yet hah.
Edit again - Blogfast it's already normalized. Thats what the sqrt(1/(π*a^3)) term comes from. Also where my error came from.
[Edited on 20-9-2014 by smaerd]
[Edited on 20-9-2014 by smaerd]
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blogfast25
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∫<sub>0</sub><sup>∞</sup> [R<sub>1,0</sub>(r)] <sup>2</sup> r<sup>2</sup> dr = 1
... is likely to be correct as the e<sup>-r/a0</sup> on integration to infinity becomes 0.
The general expression of the Normalisation Constant in function of n and l:
[Edited on 20-9-2014 by blogfast25]
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smaerd
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It actually does not become zero. it becomes [(a^3)/4]. Well if you square the wave function and use e^(-2r/a0)
[Edited on 20-9-2014 by smaerd]
The terms evaluated at infinity become zero but one term evaluated at zero is a constant. If you see what I did above you'll quickly see the wave
function is normalized.
[Edited on 20-9-2014 by smaerd]
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blogfast25
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It fits.
Substitute -2r/a<sub>0</sub> = x
Then r<sup>2</sup> = a<sub>0</sub><sup>2</sup>/4
And dr = - a<sub>0</sub>/2 dx
Plug in: the constant before the integral becomes -1/2 and the integral becomes:
∫ x<sup>2</sup> e<sup>x</sup> dx = (x<sup>2</sup> – 2 x + 2) e<sup>x</sup> + C
Integrated between 0 and - ∞ [because of -2r/a<sub>0</sub> = x], that yields – 2
-2 . – ½ = 1
[Edited on 20-9-2014 by blogfast25]
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blogfast25
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Quote: Originally posted by smaerd  | If you see what I did above you'll quickly see the wave function is normalized.
[Edited on 20-9-2014 by smaerd] |
Where did I say it wasn't??? It's very clear it is.
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smaerd
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I was confused about what you were doing to be honest. Now I see that we were pretty much on the same page. Basically you changed the normalization
coefficient in the original problem to eliminate the two other really easy integrations and only deal with the radial direction. I see now.
There's a way to solve these using only the volume of a sphere and the normalized eigenfunction but I don't really have time to crunch the numbers
right now. Working on a lab report, a pre-lab, some homework, a research presentation, graduate proposal for funding, and scheduling for the GRE's
this weekend...
[Edited on 20-9-2014 by smaerd]
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blogfast25
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| Quote: Originally posted by smaerd | Basically you changed the normalization coefficient in the original problem to eliminate the two other really easy integrations and only deal with the
radial direction.
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Hmmm. Not sure about that.
“1/sqrt(Pi)*(1/5.29e-11)^(3/2)*E^(-x/5.29e-11)”
…is essentially unintelligible.
Instead I used:
R(1,0) = 2a<sub>0</sub><sup>-3/2</sup> e<sup>-r/a0</sup>
If this is normalised (which of course it is) then:
∫<sub>0</sub><sup>∞</sup> 2a<sub>0</sub><sup>-3/2</sup> e<sup>-r/a0</sup>
r<sup>2</sup> dr = 1
… applies.
To prove this identity I carried out the simple substitution described above, then integrated.
Substitute -2r/a<sub>0</sub>= x
Then r<sup>2</sup> = a<sub>0</sub><sup>2</sup>/4
And dr = - a<sub>0</sub>/2 dx
Plug in: the constant before the integral becomes -1/2 and the integral itself becomes:
∫ x<sup>2</sup> e<sup>x</sup> dx = (x<sup>2</sup> – 2 x + 2) e<sup>x</sup> + C
Integrated between 0 and - ∞, that yields – 2
-2 . – ½ = 1
[Edited on 20-9-2014 by blogfast25]
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smaerd
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Yea I got that, and I am sure about it... The problem is originally written as a 3-D spherical harmonic. Notice the square root of the recipricol Pi
and how the normalization coefficient you used does not contain the pi which comes from the Phi integrand? So that's what ya did, it works(obviously),
but it is a different approach to the problem(a pretty cool one).
But all you really eliminated from the problem are the easy theta and phi integrals. For more interesting wave functions that are radially symmetric
I'll have to keep this in mind though. Not sure how many of those I'll be seeing hah, but you never know.
Definitely agree the original problem was not presented very well, but what can ya do. Definitely would recommend using scientific notation and proper
constant denotion (e for eulers constant not E... etc). Clearly they are still learning the basics though.
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vmelkon
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| Quote: Originally posted by smaerd | The wave function is normalized. Nice attempt! The normalization coefficient is: sqrt(1/(π*a^3)) But you must square it as it is now part of Psy
(without fancy terminology).
It's a triple integral when you pop it into spherical polar.
The volume integral goes as follows
Integral φ=0 to 2π Integral θ = 0 to π Integral r = 0 to 0.00000000002 sqrt(1/(pi*a^3)) ^2 * (e^(-r/a))^2 * r^2 sin(θ) dr dθ dφ
Notice that when we take this to spherical polar we multiply by : r^2 sin(θ) dr dθ dφ and the two other integrands appear(not arbitrary review
spherical polar probably also in atkins).
So let's handle the square of the normalization coefficient...
sqrt(1/(π*a^3)) * sqrt(1/(π*a^3)) = 1/(π*a^3)
Nice...
So I will break up the integrations as that can be the easiest way to approach these problems...
Integral θ = 0 to π sin(θ) dθ
= -cos(θ) from 0 to 2π
= -(cos(2π) - cos(0))
= -(-1 -1)
= 2
Now,
Integral φ=0 to 2π dφ
simply equal to 2π
Okay now the integral you probably do not want to solve by hand but is actually quite easy... Can approach this using integration by parts or using a
table of integrals... Basically it comes out to be a very simple series. I won't show my work but I typically do tanzalin integration by parts. In
this case, in my head. If I make an error here it's for you to figure out because I don't want to actually do this with pencil and paper...
Integral r = 0 to 0.00000000002 r^2 * (e^(-2r/a))
= e^(-2r/a) *[ (-a * r^2/2) + (-2*a^2*r/4) + (-2*a^3/8)] from r=0 to 0.00000000002
simplify
= e^(-2r/a) *[ (-a * r^2/2) + (-a^2*r/2) + (-a^3/4)] from r=0 to 0.00000000002
evaluate...
= e^(-4E-11/a) *[ (-a * 2E-11^2/2) + (-a^2 * 2E-11/2) + (-a^3/4)] - e^(0) *[ (0 + 0 + (-a^3/4)]
simplify...
= e^(-4E-11/a) *[ (-a * 2E-22) + (a^2 * 1E-11) + (-a^3/4)] + [a^3/4]
where a = 52.9E-12, s.t...
= 0.4694733 * [-1.058E-32 - 2.79841E-32 - 3.7008E-32] + [3.7008E-32]
= 1.528916E-33
so the answer I get is....
1.528916E-33 * 4 * π * (1/(π*a^3))
= 6.115664E-32 * 1/(a^3)
=0.4131
Feel like I made a small algebraic mistake but I think you get the jist... Also I forget whether or not you have to use the bohr radius constant in
meters or not. I think you keep it as a whole number.
[Edited on 20-9-2014 by smaerd] |
Thanks.
I checked your work. Basically, you are injecting r^2 sin(θ) into the equation. I didn't realize I had to do that. What is the logic behind this?
I get 0.0415298304.
Perhaps I made a mistake somewhere.
If I do r = 0 to 5.29e-10, I get 1.000199584. This can't be right because it exceeds a probability of 1.
No, this isn't homework. I have taken physics over 10 years ago at university but never studied quantum physics. It is a rather nutty notion but I am
getting a feel to why approaching the problem from the point of view of waves makes sense.
As for the E, I decided to switch from e to E to make sure Maple V doesn't get confused. It had no effect. I'm not a Maple V expert or calculus
expert.
[Edited on 25-9-2014 by vmelkon]
[Edited on 25-9-2014 by vmelkon]
Signature ==== Is this my youtube page? https://www.youtube.com/watch?v=tA5PYtul5aU
We must attach the electrodes of knowledge to the nipples of ignorance and give a few good jolts.
Yes my evolutionary friends. We are all homos here. |
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