# Solving the Schrödinger equation for hydrogen atom

• raul_l
In summary, the Schrödinger equation for the hydrogen atom is a complex mathematical equation that describes the energy states and wave functions of an electron in the hydrogen atom. Solving this equation is important for understanding and predicting the behavior of electrons in the hydrogen atom, which has practical applications in various fields such as chemistry, material science, and engineering. The equation is also fundamental in quantum mechanics, incorporating principles such as particle-wave duality and probabilistic behavior. While it may be difficult to solve, advanced computational methods make it possible to obtain accurate results.
raul_l

## Homework Statement

Hello. I'd like to solve this: $$-\frac{\hbar^2}{2m}\nabla^2 \Psi(r,\theta,\phi) -U(r) \Psi(r,\theta,\phi) = E\Psi(r,\theta,\phi)$$

## The Attempt at a Solution

I can separate the variables, but that's about it.

$$\frac{1}{R(r)} \frac{d}{dr}(r^2 \frac{d}{dr}R(r))+\frac{2mEr^2}{\hbar^2} + \frac{2m\gamma r}{\hbar^2} = C_r$$

$$\frac{1}{F(\phi)} \frac{d^2}{d\phi^2} F(\phi) = C_\phi$$

$$-C_r-\frac{sin\theta}{P(\theta)} \frac{d}{d\theta} (sin\theta \frac{d}{d\theta} P(\theta))=C_\phi$$

The answer should be something like this but I don't know how to get there.
$$\psi_{nlm}(r,\vartheta,\varphi)=\sqrt{{\left(\frac{2}{n a_0} \right)}^3\frac{(n-l-1)!}{2n(n+1)!}}e^{-\rho /2} \rho^{1} L_{n-l-1}^{2l+1}(\rho)\cdot Y_{lm}(\vartheta, \varphi)$$

If somebody could offer me any idea on how to proceed that would be great. I'm not even sure of how to choose the righ boundary conditions.

raul_l said:

## Homework Statement

Hello. I'd like to solve this: $$-\frac{\hbar^2}{2m}\nabla^2 \Psi(r,\theta,\phi) -U(r) \Psi(r,\theta,\phi) = E\Psi(r,\theta,\phi)$$

## The Attempt at a Solution

I can separate the variables, but that's about it.

$$\frac{1}{R(r)} \frac{d}{dr}(r^2 \frac{d}{dr}R(r))+\frac{2mEr^2}{\hbar^2} + \frac{2m\gamma r}{\hbar^2} = C_r$$

$$\frac{1}{F(\phi)} \frac{d^2}{d\phi^2} F(\phi) = C_\phi$$

$$-C_r-\frac{sin\theta}{P(\theta)} \frac{d}{d\theta} (sin\theta \frac{d}{d\theta} P(\theta))=C_\phi$$

The answer should be something like this but I don't know how to get there.
$$\psi_{nlm}(r,\vartheta,\varphi)=\sqrt{{\left(\frac{2}{n a_0} \right)}^3\frac{(n-l-1)!}{2n(n+1)!}}e^{-\rho /2} \rho^{1} L_{n-l-1}^{2l+1}(\rho)\cdot Y_{lm}(\vartheta, \varphi)$$

If somebody could offer me any idea on how to proceed that would be great. I'm not even sure of how to choose the righ boundary conditions.

the derivation is done in almost all quantum mechanics textbooks so you should look on up.
It's a long derivation.

There are two apporaches to this problem: the DE's are standard ones and one can simply look up the solutions. However, if you have to prove the solutions from scratch, you must start with a series expansion. Again, all the details are in QM textbooks

Ok, this looks pretty complicated (not suprisingly).

$$\frac{d^2}{d\phi^2} F(\phi) = C_\phi F(\phi)$$

The solution is in the form $$F(\phi)=Ae^{B\phi}$$

which gives me
$$\frac{d^2}{d\phi^2} Ae^{B\phi} = C_\phi Ae^{B\phi} \Rightarrow -AB^2 e^{B\phi} = C_\phi Ae^{B\phi}$$ and therefore $$C_\phi=-B^2$$

Due to constraints on the wavefunction $$F(\phi)=F(\phi+2\pi)$$

which means that $$Ae^{B\phi}=Ae^{B(\phi+2\pi)}=Ae^{B\phi} e^{2B\pi} \Rightarrow e^{2B\pi}=1 \Rightarrow B=im$$ where m is an integer.

Because of normalization $$\int_{-\infty}^{\infty} F(\phi) d\phi=1$$ and therefore A=1.

I have arrived at $$F(\phi)=e^{im\phi}$$
and $$C_\phi=-m^2$$

which means that $$-C_r-\frac{sin\theta}{P(\theta)} \frac{d}{d\theta} (sin\theta \frac{d}{d\theta} P(\theta))=-m^2$$

Thanks for your help so far. I'll keep on working on this.

I'm half way solving the colatitude equation (I know how to solve differential equations using series expansion).

However, this is a lot of work I think it would probably be a better idea to just look it up in a QM textbook.

I have one question though.
Why is $$C_r=l(l+1)$$ ? I can't find the answer anywhere.

I have a problem with solving the general Legendre equation.

As we know the series solution is in the form
$$\sum a_s x^s$$

I have found the formula for the coefficients
$$a_{s+2}=-a_s \frac{(l-s)(l+s+1)}{(s+1)(s+2)}$$

Now, if $$l=s$$ then $$a_{l+2}$$ and all subsequent coefficients become zero.

In the book it says that the highest coefficient $$a_l$$ is given by the formula $$a_l=\frac{(2l)!}{2^l (l!)^2}$$ but I don't see how.

If I set $$a_0=1$$ I get
$$a_s=\frac{(l-s+2)(l-s+4)...l(l+1)(l+3)...(l+s-1)}{s!}$$

If $$l=s$$ then
$$a_l=\frac{2*4*...*l(l+1)(l+3)...(2l-1)}{l!}$$

I don't know how to proceed. The best I can do is write it like this:
$$a_l=\frac{(2l)!}{1*3*...(l-1)(l+2)(l+4)...2l*l!}$$

which is pretty close to $$\frac{(2l)!}{2^l (l!)^2}$$ but not exactly.

which book?

Mathematical Methods For Physicists by Tai L. Chow

Of course, there is always the possibiliy of a mistake in the book.

At the bottom of page 298.

Yes, as far as I know.

Actually, they use the series expansion method to solve the Laguerre equation. But the Legendre equation is solved differently. I think it's quite clever.

Hello again.

I'm having trouble evaluating $$\frac{(-1)^l}{2^{2l} (l!)^2} \frac{(2l)!(l+m)!}{(l-m)!} \int_{-1}^{1} (x^2-1)^l dx$$

which should be $$\frac{2}{2l+1} \frac{(l+m)!}{(l-m)!}$$

My first thought was to use the Newton's binomial formula to evaluate the integral.
$$\int_{-1}^{1} (x^2-1)^l dx = \int_{-1}^{1} \sum_{k=0}^l \frac{l!}{k!(l-k)!} (x^2)^{l-k} (-1)^k = (\sum_{k=0}^l \frac{l!}{k!(l-k)!(2l-2k+1)} x^{2l-2k+1} (-1)^k) _{x=-1}^{x=1}$$

Since $$(2l-2k+1)$$ is always odd $$(x^{2l-2k+1})_{x=-1}^{x=1}$$ always yields 2.

Therefore
$$\frac{(-1)^l}{2^{2l} (l!)^2} \frac{(2l)!(l+m)!}{(l-m)!} \int_{-1}^{1} (x^2-1)^l dx = \frac{(-1)^l}{2^{2l} (l!)^2} \frac{(2l)!(l+m)!}{(l-m)!} \sum_{k=0}^l \frac{l!}{k!(l-k)!(2l-2k+1)} (-1)^k = \frac{(-1)^l}{2^{2l} l!} \frac{(2l)!(l+m)!}{(l-m)!} \sum_{k=0}^l \frac{1}{k!(l-k)!(2l-2k+1)} (-1)^k$$

I don't know how to proceed (or if I've been correct so far).
Btw, here x substitutes $$cos \theta$$ but I don't think that changes anything.

## 1. What is the Schrödinger equation for the hydrogen atom?

The Schrödinger equation for the hydrogen atom is a mathematical equation that describes the energy states and wave functions of an electron in the hydrogen atom. It takes into account the electron's mass, charge, and the attractive force between the electron and the nucleus.

## 2. Why is solving the Schrödinger equation for the hydrogen atom important?

Solving the Schrödinger equation for the hydrogen atom is important because it allows us to understand and predict the behavior of electrons in the hydrogen atom. This is crucial for understanding the properties of atoms, molecules, and materials, as well as for developing technologies such as semiconductors and lasers.

## 3. Is the Schrödinger equation for the hydrogen atom difficult to solve?

The Schrödinger equation for the hydrogen atom is a complex mathematical equation that requires advanced mathematical techniques to solve. However, with the use of computers and advanced computational methods, it is possible to solve the equation and obtain accurate results.

## 4. How does the Schrödinger equation for the hydrogen atom relate to quantum mechanics?

The Schrödinger equation for the hydrogen atom is a fundamental equation in quantum mechanics, which is the branch of physics that describes the behavior of particles at the atomic and subatomic level. The equation incorporates the principles of quantum mechanics, such as particle-wave duality and probabilistic behavior, to describe the behavior of electrons in the hydrogen atom.

## 5. Are there any practical applications of the Schrödinger equation for the hydrogen atom?

Yes, the Schrödinger equation for the hydrogen atom has many practical applications. It is used in fields such as chemistry, material science, and engineering to understand and predict the properties of atoms and molecules. It is also the basis for technologies such as MRI machines, transistors, and solar cells.

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