TISE solution for a hydrogen atom

• Kynsuo
In summary: Can you solve this equation for ##a_0##?$$\frac{a_0}{r} - \frac 2 r = 0$$In summary, the conversation discusses finding the value of a constant, a_0, in the radial component of the TISE for the electron in the Hydrogen Atom. The equation is solved and it is shown that the expression in the brackets on the right-hand side of the equation must be constant and not depend on the variable r. This leads to the conclusion that a_0 can be written in terms of fundamental constants. The value of a_0 can be found by solving an equation and using the fact that the energy should not depend on r.
Kynsuo
Homework Statement
Show that psi is a solution to the radial component of the TISE for the electron in the Hydrogen Atom, find the value of ##a_0## in terms of fundamental constants and find the energy for this state.
Relevant Equations
$$- \frac{\hbar^2}{2mr^2} \frac d {dr} \left(r^2 \frac {d \psi}{dr} \right) + V(r) \psi = E \psi$$
$$V(r) = -\dfrac 1 {4 \pi \varepsilon_0} \dfrac{e^2}{r}$$
$$\psi(r) = \dfrac 1 {\sqrt{\pi a_0^3}} \exp\left(-\dfrac r {a_0}\right)$$
I am unable to complete the first part of the question. After I plug in the function for psi into the differential equation I am stuck:

$$\frac {d \psi (r)}{dr} = -\frac 1 a_0 \psi (r), \frac d{dr} \biggl(r^2 \frac {d\psi (r)}{dr} \biggr) = -\frac 1 {a_0}\frac d {dr} \bigl[r^2 \psi(r) \bigr] = - \frac 1 {a_0}\biggl[ 2r\psi(r) - \frac{r^2}{a_0} \psi(r) \biggr]$$

$$\frac{\hbar }{2mr^2} \frac 1 {a_0} \biggl[ 2r \psi(r) - \frac {r^2}{a_0} \psi(r) \biggr] + V(r) \psi(r) = E \psi(r)$$
$$\Leftrightarrow \psi (r) \biggl[ \frac {\hbar}{2mr^2} \frac {2ra_0-r^2}{a_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r} \biggr] = E \psi(r)$$

I don't the how the expression I get in the brackets in the RHS at the end is equal to the total energy.

The solution requires a specific value of ##a_0##. The parameter ##a_0## can't just be any value.

PeroK said:
The solution requires a specific value of ##a_0##. The parameter ##a_0## can't just be any value.
Hi, thank you so. much for your reply, but I'm not sure I understand your point. Does this mean that ##a_0## has to be such that the expression in the brackets is equal to the sum of the potential and kinetic energies?

Kynsuo said:
Hi, thank you so. much for your reply, but I'm not sure I understand your point. Does this mean that ##a_0## has to be such that the expression in the brackets is equal to the sum of the potential and kinetic energies?
It means that ##a_0## has to be such that the expression in the brackets is constant! I.e. not a function of ##r##.

PeroK said:
It means that ##a_0## has to be such that the expression in the brackets is constant! I.e. not a function of ##r##.
Hi, I think that this is my problem: I don't understand why the total energy/expression the brackets has no dependency on the variable ##r##.

Kynsuo said:
Hi, I think that this is my problem: I don't understand why the total energy/expression the brackets has no dependency on the variable ##r##.
You need to try to find ##a_0## to achieve that. Hint: expand the expression explicitly in powers of ##r##.

Kynsuo said:
Homework Statement:: Show that psi is a solution to the radial component of the TISE for the electron in the Hydrogen Atom, find the value of ##a_0## in terms of fundamental constants and find the energy for this state.
Relevant Equations:: $$- \frac{\hbar^2}{2mr^2} \frac d {dr} \left(r^2 \frac {d \psi}{dr} \right) + V(r) \psi = E \psi$$
$$V(r) = -\dfrac 1 {4 \pi \varepsilon_0} \dfrac{e^2}{r}$$
$$\psi(r) = \dfrac 1 {\sqrt{\pi a_0^3}} \exp\left(-\dfrac r {a_0}\right)$$

I am unable to complete the first part of the question. After I plug in the function for psi into the differential equation I am stuck:

$$\frac {d \psi (r)}{dr} = -\frac 1 a_0 \psi (r), \frac d{dr} \biggl(r^2 \frac {d\psi (r)}{dr} \biggr) = -\frac 1 {a_0}\frac d {dr} \bigl[r^2 \psi(r) \bigr] = - \frac 1 {a_0}\biggl[ 2r\psi(r) - \frac{r^2}{a_0} \psi(r) \biggr]$$

$$\frac{\hbar }{2mr^2} \frac 1 {a_0} \biggl[ 2r \psi(r) - \frac {r^2}{a_0} \psi(r) \biggr] + V(r) \psi(r) = E \psi(r)$$
$$\Leftrightarrow \psi (r) \biggl[ \frac {\hbar}{2mr^2} \frac {2ra_0-r^2}{a_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r} \biggr] = E \psi(r)$$

I don't the how the expression I get in the brackets in the RHS at the end is equal to the total energy.
Your calculations are all correct.
What Perok is pointing out is that ##a_0## is not an independent constant. It can actually be written in terms of ##e,\epsilon_0, \hbar, m_e##. Use this and then you will see that the energy will be a constant and will give the correct value.

nrqed said:
Your calculations are all correct.
What Perok is pointing out is that ##a_0## is not an independent constant. It can actually be written in terms of ##e,\epsilon_0, \hbar, m_e##. You can either 1) look it up and you will see that the energy will be constant or 2) impose that the energy does not depend on ##r##, in other words, force the term proportional to ##1/r## to be zero. That will give the value of ##a_0## in terms of these other parameters.

Then you will see that the energy will be a constant and will give the correct value.

Thanks nqred, but I think I'm still missing something. Can this expression for ##a_0## be found purely mathematically or do I need to make assumptions/ rewrite other terms?

Kynsuo said:
Thanks nqred, but I think I'm still missing something. Can this expression for ##a_0## be found purely mathematically or do I need to make assumptions/ rewrite other terms?

Can you solve this equation for ##a_0##?
$$\frac{a_0}{r} - \frac 2 r = 0$$

Hi PeroK, thanks for replying. The solution would be ##a_0 =2## but I'm not sure I understand why I need this equation.

Kynsuo said:
Hi PeroK, thanks for replying. The solution would be ##a_0 =2## but I'm not sure I understand why I need this equation.
That equation was simpler, but the same as this one. Which you seem to be so frightened of that you're afraid to go anywhere near it!

Kynsuo said:
$$\Leftrightarrow \biggl[ \frac {\hbar}{2mr^2} \frac {2ra_0-r^2}{a_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r} \biggr] = E$$

Hi, I still don't get it (I'm really sorry :(! ): when I replace ##a_0## by 2, the expression inside the brackets is still dependent on ##r##

Kynsuo said:
Hi, I still don't get it (I'm really sorry :(! ): when I replace ##a_0## by 2, the expression inside the brackets is still dependent on ##r##
$$\frac {\hbar}{2mr^2} \frac {2ra_0-r^2}{a_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r} = \frac {\hbar}{ma_0r} - \frac {\hbar}{2ma_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r}$$
So, if we have:
$$\frac {\hbar}{ma_0r} - \frac{e^2}{4 \pi \varepsilon_0 r} = 0$$
Then we get rid of all terms in ##r##. To do this we need:
$$\frac {\hbar}{ma_0} = \frac{e^2}{4 \pi \varepsilon_0}$$
Hopefully you can solve for ##a_0## from there.

PeroK said:
$$\frac {\hbar}{2mr^2} \frac {2ra_0-r^2}{a_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r} = \frac {\hbar}{ma_0r} - \frac {\hbar}{2ma_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r}$$
So, if we have:
$$\frac {\hbar}{ma_0r} - \frac{e^2}{4 \pi \varepsilon_0 r} = 0$$
Then we get rid of all terms in ##r##. To do this we need:
$$\frac {\hbar}{ma_0} = \frac{e^2}{4 \pi \varepsilon_0}$$
Hopefully you can solve for ##a_0## from there.
Hi peroK, this is really helpful, and I am confident I can solve this equation for ##a_0##. I think that my problem is that I don't understand why the kinetic energy of the electron in the hydrogen atom is: $$\frac {\hbar}{ma_0r} - \frac {\hbar}{2ma_0^2}$$

Kynsuo said:
Hi peroK, this is really helpful, and I am confident I can solve this equation for ##a_0##. I think that my problem is that I don't understand why the kinetic energy of the electron in the hydrogen atom is: $$\frac {\hbar}{ma_0r} - \frac {\hbar}{2ma_0^2}$$

It depends what you mean by understand. It all came out of your calculation in the original post. Which in turn were based on expressing the Hamiltonian in spherical coordinates.

PeroK said:
It depends what you mean by understand. It all came out of your calculation in the original post. Which in turn were based on expressing the Hamiltonian in spherical coordinates.
I haven't studied the Hamiltonian in much detail as this came from an introductory foundation course for quantum physics, which I will study in detail next year (all I know is that it is an operator that returns the total energy of a system). What I mean by understand is that I am not able to see how this expression for the kinetic energy stems directly from my calculation.

Kynsuo said:
$$\psi (r) \biggl[ \frac {\hbar}{2mr^2} \frac {2ra_0-r^2}{a_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r} \biggr] = E \psi(r)$$
I don't the how the expression I get in the brackets in the RHS at the end is equal to the total energy.
I'll try explaining this. You arrived at the expression shown above all by yourself therefore you understand it. You have one equation and apparently two unknowns, ##a_0## and ##E## which the problem is asking you to find. Furthermore, you are also asked to show that the given wavefunction is a solution. There are only specific values for ##a_0## and ##E## that make the wavefunction a solution. The underlying logic then is to assume that the given wavefunction is indeed a solution and then proceed to find ##a_0## and ##E## that will make this a correct assumption.

Under the assumption you can cancel ##\psi(r)## that appears on both sides, noting that it is zero only at infinity. Then you have the energy, $$E= \biggl[ \frac {\hbar}{2mr^2} \frac {2ra_0-r^2}{a_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r} \biggr]$$ Again under the assumption, this equality must hold for any value of ##r## otherwise we don't have a solution.

In post #14, @PeroK showed how this can be achieved and obtained the equality$$\frac {\hbar}{ma_0} = \frac{e^2}{4 \pi \varepsilon_0}.$$At this point, all you have to do is solve this last equation for ##a_0##, then go back to the energy equation and substitute ##a_0##. The ##r## terms in the expression should drop out because ##a_0## was chosen specifically to make this happen. This then will give you a constant for the energy ##E##. Not only that, but also you have justified that ##\psi(x)## is a solution. If you don't believe me that this is the case, you can always take the given ##\psi(x)##, the value for ##a_0## and ##E## that you found and put them all in the original TISE. If you make no algebraic mistakes, you should end up with ##0=0##.

Last edited:
PeroK
Hi! Sorry for taking so long to get back. I understand PeroK's logic better now. I alsorealised that there is a typo in my equation (all the ##\hbar##'s are actually ##\hbar^2##) . Thank you for that. The only thing I'm missing is how PeroK went from

##\frac{\hbar^2}{2mr} \frac{2a_0-r}{a_0^2} - V(r) = \frac{\hbar^2}{ma_0r}- \frac{\hbar^2} {2ma_0^2} - V(r)##To: ##\frac{\hbar^2}{ma_0r} - V(r) =0 ##

Kynsuo said:
Hi! Sorry for taking so long to get back. I understand PeroK's logic better now. I alsorealised that there is a typo in my equation (all the ##\hbar##'s are actually ##\hbar^2##) . Thank you for that. The only thing I'm missing is how PeroK went from

##\frac{\hbar^2}{2mr} \frac{2a_0-r}{a_0^2} - V(r) = \frac{\hbar^2}{ma_0r}- \frac{\hbar^2} {2ma_0^2} - V(r)##To: ##\frac{\hbar^2}{ma_0r} - V(r) =0 ##

I didn't really do anything there. I just noted that if we choose ##a_0## such that:
$$\frac{\hbar^2}{ma_0r} - V(r) =0$$
Then:
$$\frac{\hbar^2}{ma_0r}- \frac{\hbar^2} {2ma_0^2} - V(r) = - \frac{\hbar^2} {2ma_0^2}$$
is constant, which is what we want.

PeroK said:
I didn't really do anything there. I just noted that if we choose ##a_0## such that:
$$\frac{\hbar^2}{ma_0r} - V(r) =0$$
Then:
$$\frac{\hbar^2}{ma_0r}- \frac{\hbar^2} {2ma_0^2} - V(r) = - \frac{\hbar^2} {2ma_0^2}$$
is constant, which is what we want.
Hi PeroK, thank you so much for getting back so fast. This might sound stupid, but are we allowed to do this? Can we 'chose a value for ##a_0## such that the expression is equal to zero' only because this equation is true for all values of ##r##, IE this is not an equation, but an identity?

Kynsuo said:
Hi PeroK, thank you so much for getting back so fast. This might sound stupid, but are we allowed to do this? Can we 'chose a value for ##a_0## such that the expression is equal to zero' only because this equation is true for all values of ##r##, IE this is not an equation, but an identity?
We can if ##V(r) = \frac k r##. That's the point.

High Perok: I think I understand. Basically, the energy must be a constant that doesn't depend on ##r##. Since we have 2 terms of the form ## \frac cr, \; c \in \mathbb C ##, their sum must equate to 0. This also makes sense because the expression gives the first energy state for the H atom when I plug in the value of ##a_0##.
Thanks!

PeroK

1. What is TISE solution for a hydrogen atom?

The Time Independent Schrödinger Equation (TISE) is a mathematical equation used to describe the behavior of a quantum system, such as a hydrogen atom, in a stationary state. The TISE solution for a hydrogen atom involves finding the energy levels and corresponding wavefunctions that describe the electron's motion around the nucleus.

2. How is the TISE solution for a hydrogen atom derived?

The TISE solution for a hydrogen atom is derived by solving the Schrödinger equation for a spherically symmetric potential, which is the case for the Coulomb potential of the hydrogen atom. This involves applying separation of variables and solving for the radial and angular parts of the wavefunction.

3. What does the TISE solution tell us about the hydrogen atom?

The TISE solution provides information about the energy levels and wavefunctions of the hydrogen atom. This allows us to understand the quantized nature of the electron's energy and its probability distribution in space. It also helps us predict the behavior of the hydrogen atom under different conditions.

4. Can the TISE solution be applied to other atoms?

Yes, the TISE solution can be applied to any quantum system with a spherically symmetric potential, such as other atoms or even molecules. However, the specific form of the potential will vary depending on the system, and the solution will be different for each case.

5. What are the limitations of the TISE solution for a hydrogen atom?

The TISE solution for a hydrogen atom assumes that the electron and nucleus are point particles and neglects the effects of relativity. It also does not take into account the interactions between multiple electrons in an atom. These limitations can be addressed by using more advanced models, such as the Schrödinger equation with relativistic corrections or the Hartree-Fock method.

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