# Finding Hydrogen eigenfunction u(2,0)

• unscientific
In summary, the conversation is about solving the equation ##\left(\frac{\partial}{\partial r} + \frac{2-n}{r} + \frac{Z}{(n-1)a_0} \right) u_{n}^{l=n-2} = 0## using the integrating factor method. The solution obtained is similar to ##u_n^{l=n-1}## but with n replaced by n-1. It is later mentioned that applying A† to ##u_n^{l=n-1}## will give ##u_n^{l=n-2}##, and this method is used to obtain the correct result.
unscientific
Taken from Physics of Quantum Mechanics, by James Binney.

I try to calculate ##u_{n}^{l=n-2}##, something goes wrong:

Starting, we define operator A by:

$$A_{n-2} = \frac{a_0}{\sqrt 2}\left(\frac{i}{\hbar}p_r + \frac{1-n}{r} + \frac{Z}{(n-1)a_0}\right)$$

Substituting ##p_r = -i\hbar (\frac{\partial}{\partial r} + \frac{1}{r})##:

$$A_{n-2} = \frac{a_0}{\sqrt 2}\left( \frac{\partial}{\partial r} + \frac{2-n}{r} + \frac{Z}{(n-1)a_0}\right)$$

Thus, we want to solve:

$$\left(\frac{\partial}{\partial r} + \frac{2-n}{r} + \frac{Z}{(n-1)a_0} \right) u_{n}^{l=n-2} = 0$$

Solving by integrating factor method, we obtain:

$$u_n^{l=n-2} = A r^{n-2} e^{\frac{Z}{(n-1)a_0}r}$$

Normalizing,

$$u_n^{l=n-2} = \frac{1}{\sqrt{[2(n-1)]!}} \left(\frac{2Z}{(n-1)a_0}\right)^{\frac{3}{2}} \left(\frac{2Z}{(n-1)a_0}\right)^{n-2} e^{-\frac{Z}{(n-1)a_0}r}$$

This is similar to ##u_n^{l=n-1}##, simply replace n by n-1:

But when I substitute n = 2, so l = 0, I get ##u_2^0 = \frac{1}{\sqrt 2} \left(\frac{2Z}{a_0}\right)^{\frac{3}{2}} e^{-\frac{Z}{a_0}r}##

I get a completely different result from the book:

I'm not sure what's wrong with my derivation?

The copy of Binney I'm looking at is slightly different from yours, so bear with me, but the equation you're trying to use,

An-1|n, n-1> = 0

is only for (as he says) "the radial wavefunction of the circular orbit with angular momentum ℓ = n - 1." That is, it gives the wavefunctions along the diagonal in Table 8.1. To get the other wavefunctions you need to start with one of these wavefunctions and one or more times apply A. This will step you up in the table, along one of the columns.

"Applications of A should generate the wavefunctions for ℓ < n - 1."

Bill_K said:
The copy of Binney I'm looking at is slightly different from yours, so bear with me, but the equation you're trying to use,

An-1|n, n-1> = 0

is only for (as he says) "the radial wavefunction of the circular orbit with angular momentum ℓ = n - 1." That is, it gives the wavefunctions along the diagonal in Table 8.1. To get the other wavefunctions you need to start with one of these wavefunctions and one or more times apply A. This will step you up in the table, along one of the columns.

"Applications of A should generate the wavefunctions for ℓ < n - 1."

Ah, so by applying A to ##u_n^{l=n-1}## it will give ##u_n^{l=n-2}##?

unscientific said:
Ah, so by applying A to ##u_n^{l=n-1}## it will give ##u_n^{l=n-2}##?
That's my understanding. Try it and see!

Bill_K said:
That's my understanding. Try it and see!

It's mentioned right below the screenshot, I can't believe I missed it..

I tried it, and normalized and it came out right!

## 1. What is the significance of finding the hydrogen eigenfunction u(2,0)?

The hydrogen eigenfunction u(2,0) represents the probability amplitude of finding the electron in a hydrogen atom at a distance of 2 units from the nucleus and with zero angular momentum. This is an important calculation in understanding the energy states and behavior of the electron in a hydrogen atom.

## 2. How is the hydrogen eigenfunction u(2,0) calculated?

The hydrogen eigenfunction u(2,0) is calculated using the Schrödinger equation, which is a mathematical equation that describes the behavior of quantum systems. The specific formula for calculating u(2,0) involves solving a differential equation and applying boundary conditions.

## 3. What information can be obtained from the hydrogen eigenfunction u(2,0)?

The hydrogen eigenfunction u(2,0) provides information about the energy state and position of the electron in a hydrogen atom. Additionally, it can be used to calculate other important properties such as the probability density and radial distribution function.

## 4. How does the hydrogen eigenfunction u(2,0) relate to the concept of quantum numbers?

The hydrogen eigenfunction u(2,0) is associated with the principal quantum number, n=2. This number represents the energy level of the electron in a hydrogen atom. Other quantum numbers, such as the angular momentum quantum number and the magnetic quantum number, also play a role in determining the specific eigenfunction for a given state.

## 5. Is the hydrogen eigenfunction u(2,0) the same for all hydrogen atoms?

No, the hydrogen eigenfunction u(2,0) will vary for different atoms depending on their energy states and quantum numbers. However, for a given atom, the eigenfunction will remain the same and can be used to make predictions about the behavior of the electron in that atom.

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