Solve Hydrogen Atom: 1/(r^3) 2p State Constant Problem by Jan 6

In summary, the problem with the hydrogen atom is that the r^(-2) expectation value with quantum numbers does not match the integral calculators. This divergence is caused by a mistake in the r^(-2) expectation value formula.
  • #1
alariv8
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Hydrogen atom: 1/(r^2 )2p state constant problem //1/(r^3) exp. value

Homework Statement


Hello, I am new to this forum. I hope it is all right I uploaded problems as image files (in handwriting).

I calculated all (r^(-1), r^(-2) and r^(-3) expectation values for Hydrogen atom for 1s and 2p. Almost all calculations gave the required result - the same as the theoretical formulas with quantum numbers, but r^(-3) for 1s and 2s state give an infinity when calculating using quantum numbers and should give the same when integrating. My question: Why does this integral diverge?
The first problem:
1sdivergeproblemlarger.jpg


I found out that there is a similar formula on integral-table.com

integraltableformula.jpg


but I guess there n is a natural number, am I right?



The Attempt at a Solution



Homework Statement




r^(-2) for 2p - there I have a constant problem:
2pconstantproblem.jpg





Homework Equations





The Attempt at a Solution





 
Last edited:
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  • #2


from the simple fact that [tex] \Gamma(z+1) = z\Gamma(z)[/tex] where [tex] \Gamma(z) = \int_{0}^{+ \infty} t^{z-1} e^{-t} dt [/tex]
 
  • #3


Yeah, I knew it would settle down for Gamma function. If I got it then:

When to reparametrize r=r0t and calculate, we get the definition of Gamma function. Because in my problem, n was -1, I would have to calculate Gamma(-1). It's modulus is +infinity, because Gamma(n)= [Gamma(n+2)]/[n*(n+1)] and if n=-1, then we get division by zero and it's limit is infinity.

I hope that is more or less it? Can anybody validate this?


The constant problem still stands.
 
  • #4


I think understand the divergence problem, quite simple actually. Integrating by parts gives Gamma(0), because one can show, that u*v limits to 0, so no recurrence relation for Gamma function needed at all, because modulus of Gamma(0) tends to infinity

GammaFunction_1000.gif


PS! I was also right at the beginning. It can be also done by the upper incomplete Gamma function. Consider the formula:

integraltableformula.jpg


What matters is on the right - I get the special case for upper incomplete gamma function (0,0) that in fact equals to Gamma(0) and this limit tends to infinity. At the same time this is also the special case for the lower incomplete gamma function, so they coincide for 0.
As you can read from http://mathworld.wolfram.com/IncompleteGammaFunction.html
and http://en.wikipedia.org/wiki/Incomplete_gamma_function the incomplete gamma function is the generalized gamma function.



The second problem is more fundamental now - actually I had differences (*1/2) in 2 cases. In one case, Alpha got the same answer as me (integrating), but theoretical formula differs by *1/2.
The other one is the one of which I have the second picture here - there I had a mistype, now Alpha gives the same answer there too, but the result with theoretical formula still differs *1/2.

So now the problem is - the theoretical formula with quantum numbers doesn't give the same answer as my integral and Wolfram Alpha!


Can anybody give me an insight what is going on? I don't know if I should find any similar formulas from a textbook, these formulas I took from professor's notes.

Thanks!
 
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  • #5


I figured even integration by parts wasn't necessary, only a change of variable.

I passed the exam and the professor did not find any mistakes in my integrals.

There is a chance that at least r^(-2) expectation value formula with quantum numbers may be slightly faulty. I did not have time to check them before the exam but the professor promised to check and maybe some day I will know where the rub was.
 

1. What is the significance of the 2p state constant in the hydrogen atom?

The 2p state constant in the hydrogen atom represents the energy of the electron in the first excited state. This state is characterized by the electron being in the second energy level, or shell, and having a specific angular momentum and magnetic moment.

2. How does the 1/(r^3) function relate to the hydrogen atom?

The 1/(r^3) function, also known as the inverse-cube potential, is used to describe the Coulomb interaction between the positively charged nucleus and the negatively charged electron in the hydrogen atom. It is a fundamental part of the Schrödinger equation used to solve for the energy levels and wavefunction of the atom.

3. Why is the hydrogen atom often used as a model system in quantum mechanics?

The hydrogen atom is a simple system with only one electron and one proton, making it easier to solve mathematically compared to more complex atoms. It also exhibits important quantum mechanical phenomena, such as discrete energy levels and wave-particle duality, making it a good model for studying these concepts.

4. What is the Jan 6 deadline for solving the 2p state constant problem in the hydrogen atom?

The Jan 6 deadline is a hypothetical date that may have been given in a specific context, such as a class assignment or research project. It is not a real deadline for solving the problem and does not have any scientific significance.

5. How is the 2p state constant problem in the hydrogen atom solved?

The 2p state constant problem is typically solved using the Schrödinger equation, a mathematical equation that describes the behavior of quantum systems. The equation is solved for the energy levels of the hydrogen atom, which can then be used to calculate the 2p state constant. This process involves using mathematical techniques such as integration and solving differential equations.

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