Undergrad Solve Schrodinger Equation for Electrons in Atom: Potential Used?

[ZapperZ]

I am very familiar with this problem. The discrete solutions come from the boundary condition because a general solution of the Schrodinger equation is in the form Asin(kx)+Bcos(kx) (i.e. is periodic). I guess my quantum number question all along was what are the boundary conditions associated the electron in an atom.

On a side note, if you apply boundary conditions to a particle, would the possible values of linear momentum be discrete? Is that how we get the discrete values of angular momentum?

No, I don't think you are "familiar" with this problem, at least, not from the way you've asked your questions in this thread.

Again, as has been mentioned in this thread, why aren't you looking at the various solutions to this exact problem that you can find all over the 'net? Your question on how to know if the solution is "discrete" is very puzzling. And to say that the "general solution" is "periodic" is even more puzzling, especially considering that the radial solution is more complicated than that! And the angular part of the solution has "single-valued" requirement as its "boundary" condition.

This link provides one of the typical work that one has to do to solve the Schrodinger equation for a hydrogenic atom.

http://www.harding.edu/lmurray/modern_files/hydro.pdf

The "n,l,m" are the quantum numbers, i.e. this is where they are "discrete" and not continuous.

Now, read that, and then describe what you did not understand. Otherwise, this is going nowhere.

Zz.

 

[Nugatory]

That's a bit strange coordinate system. The usual physicist's spherical coordinates (r,ϑ,φ)(r,\vartheta,\varphi) have the domains r>0r>0, 0<ϑ<π00≤φ<2π0\leq \varphi

D'oh - yes, of course. I'm not sure where my factor of two came from.

 

[ftr]

@Isaac0427

Sorry about the little disagreement about the quantum numbers. We seem to a different view of what is important to say to you given that you may be new to this subject and may not as yet developed the tools to work through the problem. The quantum numbers n,l,m come out naturally in the solution of the Schrodinger eq. what l and m mean at this point is not appreciated by the novice. The energy of the shells (orbit is a misleading term) is determined only by one of the quantum number, n with l and m just numbers associated with n.and n being the specification of the shells outward from the nucleus. The quantum numbers l and m have no influence on the energy of the electron. We should not at this time quibble about the significance of l or m.. As it turns out l is related to the angular momentum of the electron and m are related to the values of the projection that l can take along a preferred axis once it is establish as for example the direction of an applied magnetic field to the hydrogen atom.

I would recommend that you get a pre quantum mechanics modern physics book to get an historical overview of experimets and theoretical attempt to deal with quantum phenomena and atomic models. This will give you a background and help you navigate the quantum mechanical approach to the solution of pre quantum mechanics issues.

from wiki
https://en.wikipedia.org/wiki/Hydrogen_atom

http://blob:https://www.physicsforums.com/fc709ddf-c6b7-452b-9d31-d36725f5bedb

 

[gleem]

from wiki
https://en.wikipedia.org/wiki/Hydrogen_atom

I apologize in advance to those who might be offended but so many times the thread becomes a debate among responders and miss the chance to provide a helpful learning experience.

The OP author I believe is a high school student. The question was straight forward. The explanation does not need, nor is IMO of much value, to provide a state of the art dissertation or erudite dialog on elements of the OP.

The incidental remark that I made regarding the independence of the eigenstate energies of the Hydrogen atom is based on a simple application of Schrodinger eg. for a coulomb potential. The spin of the electron is not included an as you may recall the Schrodinger eq. must be corrected for this. As the article stated the correction is small, in fact about 0.005%. This adds nothing to the explanation requested by the OP.

 

[vanhees71]

I apologize in advance to those who might be offended but so many times the thread becomes a debate among responders and miss the chance to provide a helpful learning experience.

The OP author I believe is a high school student. The question was straight forward. The explanation does not need, nor is IMO of much value, to provide a state of the art dissertation or erudite dialog on elements of the OP.

This is utter nonsense. The debate among physicists is the highest value a forum like this can provide to a student. To just get an answer, it's most efficient to read a good textbook on the subject (although particularly in QT you have the problem that there are many textbooks, and it's not easy to find the good ones out of the noise). The discussion in the forum provides additional insight, how problems are attacked. You also learn that the arguments are not so simple even for experts, which should be encouraging to learn more and to build an own opinion on the problem and its possible solutions.

The tendency in modern physics didactics is very dangerous for the education: The tendency is to marginalize proper mathematical and scientific methods in favor of pseudoscientific qualitative narratives. The impact on the German high school system in the STEM curricula is already now very worrisome. Now they even discuss to give up the distinction between the classical fundamental subjects in the natural sciences (physics, chemistry, biology) and mix everything together in a kind of "natural history" class. This is a step back behind the achievements of the 19th century where a high scientific standard has been reached in the German high school curriculum. I fear the real reason is to cut even more on the educational budgets of the states and finally to reduce the overall hours in the STEM part of high school education.

The consequences are well visible here at the university: At the university entrance level you have to provide more and more auxilliary lectures and courses to close the gap between what was standard in the high school curriculum, particularly in mathematics (which used to include a solid basis in calculus, linear algebra, and probability theory which are all very useful for studying any subject with a quantitative empirical basis, and these are not only the obvious "classical" natural sciences but also part of the humanities, medicine, psychology, and of course economy).

So, as part of the scientific community, we should fight against these tendencies, which are unfortunately not restricted to Germany, as good as we can and note give in against superficial "studies" in questionable didactics!

Sorry for getting off topic.

 

[gleem]

I apologize in advance to those who might be offended but so many times the thread becomes a debate among responders and miss the chance to provide a helpful learning experience.

The OP author I believe is a high school student. The question was straight forward. The explanation does not need, nor is IMO of much value, to provide a state of the art dissertation or erudite dialog on elements of the OP

This is utter nonsense. The debate among physicists is the highest value a forum like this can provide to a student. To just get an answer, it's most efficient to read a good textbook on the subject (although particularly in QT you have the problem that there are many textbooks, and it's not easy to find the good ones out of the noise). The discussion in the forum provides additional insight, how problems are attacked. You also learn that the arguments are not so simple even for experts, which should be encouraging to learn more and to build an own opinion on the problem and its possible solutions.

The tendency in modern physics didactics is very dangerous for the education: The tendency is to marginalize proper mathematical and scientific methods in favor of pseudoscientific qualitative narratives. The impact on the German high school system in the STEM curricula is already now very worrisome. Now they even discuss to give up the distinction between the classical fundamental subjects in the natural sciences (physics, chemistry, biology) and mix everything together in a kind of "natural history" class. This is a step back behind the achievements of the 19th century where a high scientific standard has been reached in the German high school curriculum. I fear the real reason is to cut even more on the educational budgets of the states and finally to reduce the overall hours in the STEM part of high school education.

The consequences are well visible here at the university: At the university entrance level you have to provide more and more auxilliary lectures and courses to close the gap between what was standard in the high school curriculum, particularly in mathematics (which used to include a solid basis in calculus, linear algebra, and probability theory which are all very useful for studying any subject with a quantitative empirical basis, and these are not only the obvious "classical" natural sciences but also part of the humanities, medicine, psychology, and of course economy).

So, as part of the scientific community, we should fight against these tendencies, which are unfortunately not restricted to Germany, as good as we can and note give in against superficial "studies" in questionable didactics!

Sorry for getting off topic.

I respectively disagree wrt this thread. Some of the discussion in this thread are of little value to the question asked.. IMO some of the responses just distract from the presenting of the best answers to the question. I might add that there where about 8 response to the OP when the poster repeated his question having had no satisfactory response to that point.

Well, that's a pity since if you are only after the explicitly analytically solvable models you need to solve only for the (3D symmetric) harmonic oscillator. With this you get the solution for the angular-momentum algebra su(2), and this can be used for so(4)=su(2)⊕su(2)so(4)=su(2)⊕su(2)\mathrm{so}(4)=\mathrm{su}(2) \oplus \mathrm{su}(2) and their deformations needed for the Kepler problem/hydrogen atom in its most simple form :-).

How does this clarify the OP, when the question obviously indicates a lack a familiarity with partial differential equation solutions.?

Sorry about staying off the topic.

 

[Nugatory]

Both of you (thats @gleem and @vanhees71) - enough with the off-topic digression!

There is a real issue here about how to manage B-level threads, but you should take it to the Feedback forum if you want to discuss further.

 

[Isaac0427]

Ok, so I watched a video and I think I can put this together a little. So, for this purpose, can we just deal with the basic 2-D model of the hydrogen atom, and just the z component of angular momentum. Our particle is just going around a circle, so we only need the variable $\theta$. The angular momentum eigenvalue equation is $-i\hbar \partial_{\theta}\psi=L\psi$. I get the solution $\psi=ce^{il\theta}$. Thus, $L=l\hbar$. But, if l is a solution (solution is probably not the right term but you know what I mean), so is l+2$\pi$. Then, my solution is $\psi=ce^{i\theta(l+2\pi n)}$ where n is an integer. Assuming that l can be zero, a possible value for l is zero and thus 2$\pi$, 4$\pi$, etc. This shows that l is a multiple of 2$\pi$. If the equation were $L=lh$, l would be an integer for obvious reasons. I know I am on the right track, but I am doing something wrong, because I believe l should be an integer with $L=l\hbar$. Please don't go into three dimensions, let's just stick with this case.

I then move on to the Schrodinger equation, where $\frac{-\hbar^2}{2m} \partial^2_{\theta}\psi+V\psi=E\psi$. Rearranging,$\partial^2_{\theta}\psi=\frac{-2m}{\hbar^2}(E-V)\psi$. This differential equation also has a general solution of $\psi=ce^{i\theta2\pi(l+n)}$,(I just took out the 2$\pi$ and made l an integer) right? Doing some differentiation and more rearranging, I get $(E-V)=\frac{4\pi^2n^2l^2\hbar^2}{2m}$. I'm arriving at the correct (or in the case of angular momentum very close to correct) equations, so I must be doing something right. Can I have some feedback about my work? Again, I'd appreciate staying out of three dimensions.

Thanks!

EDIT: I have no Idea why I previously said that I think the l term should be ignored. I have changed it.

 

[gleem]

Ok, so I watched a video and I think I can put this together a little. So, for this purpose, can we just deal with the basic 2-D model of the hydrogen atom, and just the z component of angular momentum.

Unfortunately in two dimensions there is no z component. Even in the three dimensional model nothing can be said about angular momentum. from just the Schrodinger eq. I think the answer to the question you are seeking is a little more complex than you wish.
To find its value you should find the expectation value of the angular momentum operator squared. Why squared because you are looking for the magnitude of the angular moment since it is a vector quantity. <R_nl Θ _lmΦ_m | L²_op | R_nl Θ _lmΦ_m >. for the two dimensional case I suspect the eigenfunction will be of the form R_nlΘ_l .

I think Schrodinger's eq. two dimensions in polar coordinates is

$-h^2/2m\left \{ \frac{\partial^2 }{\partial\rho^2 }+ \frac{1}{\rho }\left ( \frac{\partial }{\partial \rho } \right ) +\frac{1}{\rho ^2}\frac{\partial ^2}{\partial \theta^2 }\right \}\psi + V(\rho )\psi = E\psi$

I think if you take time to study the three dimensional case by following the logic and accepting the math you can appreciate the origin of the quantum numbers.

 

[Isaac0427]

Unfortunately in two dimensions there is no z component. Even in the three dimensional model nothing can be said about angular momentum. from just the Schrodinger eq. I think the answer to the question you are seeking is a little more complex than you wish.
To find its value you should find the expectation value of the angular momentum operator squared. Why squared because you are looking for the magnitude of the angular moment since it is a vector quantity. <R_nl Θ _lmΦ_m | L²_op | R_nl Θ _lmΦ_m >. for the two dimensional case I suspect the eigenfunction will be of the form R_nlΘ_l .

I think Schrodinger's eq. two dimensions in polar coordinates is

$-h^2/2m\left \{ \frac{\partial^2 }{\partial\rho^2 }+ \frac{1}{\rho }\left ( \frac{\partial }{\partial \rho } \right ) +\frac{1}{\rho ^2}\frac{\partial ^2}{\partial \theta^2 }\right \}\psi + V(\rho )\psi = E\psi$

I think if you take time to study the three dimensional case by following the logic and accepting the math you can appreciate the origin of the quantum numbers.

Again, I'd like comments on the work I have shown.

 

[ZapperZ]

Again, I'd like comments on the work I have shown.

You seem to not be getting it. Your Laplacian in 2D plane polar coordinates is wrong!

But why would you even do this? The topic of this thread that you created is on electron in an atom. What atom is 2D? If you are hoping that the 2D solution will look like a 3D solution, or that you can simply "add" another 1D to the 2D solution, you are sadly mistaken. So why even waste your time (and our time) on something useless like this? What is wrong with (i) working directly in 3D spherical coordinates and (ii) all the various links, solutions, and "lessons" that you have been given?

What exactly is the point to all this if you are not willing to do as you're asked or instructed?

Zz.

 

[Isaac0427]

Ok, so let's say this is called the particle in a circular fixed orbit. The particle can only move around the orbit, and thus we only need one variable/coordinate, $\theta$. Given this situation, how does the work I showed in post 38 look?

 

[ZapperZ]

Ok, so let's say this is called the particle in a circular fixed orbit. The particle can only move around the orbit, and thus we only need one variable/coordinate, $\theta$. Given this situation, how does the work I showed in post 38 look?

Back up a bit. Explain! Is this now a CLASSICAL PHYSICS question? You have imposed a constraint to the motion ("circular fixed orbit") which isn't there in a quantum mechanical problem. Do you REALLY want to do this? Because if you do, why not just simply solve this classically and be done with?

You never did answer ANY of my queries to you. This appears to be a one way, "gimme, gimme, gimme" street.

Zz.

 

[Isaac0427]

Back up a bit. Explain! Is this now a CLASSICAL PHYSICS question? You have imposed a constraint to the motion ("circular fixed orbit") which isn't there in a quantum mechanical problem. Do you REALLY want to do this? Because if you do, why not just simply solve this classically and be done with?

You never did answer ANY of my queries to you. This appears to be a one way, "gimme, gimme, gimme" street.

Zz.

I currently have just one question that is in post 38 that I have not been able to get a straight answer to. Can we just ignore the technical things (this is a quantum mechanics problem. If I make it sound classical, I don't mean to) and actually look and respond to what I have done in post 38. If I am doing things completely incorrectly, how is my final solution to the problem similar if not the same as what quantum mechanics gives me.

 

[Nugatory]

Ok, so let's say this is called the particle in a circular fixed orbit. The particle can only move around the orbit, and thus we only need one variable/coordinate, $\theta$. Given this situation, how does the work I showed in post 38 look?

What you're doing in #38 is not a solution for a fixed circular orbit (not surprising, because there is no such thing in QM). It's a solution for a particle in an ideal circular tube; $r$ and $\phi$ are both fixed and the potential $V$ is constant (so you might as well take it to be zero). Your solution may or may not be right, but it has nothing to do with electrons in an atom.

To solve the problem of the hydrogen atom, you have to find a set of functions that are all of:
1) Eigenfunctions of $H$ when the potential is the three-dimensional Coulomb potential.
2) Eigenfunctions of $L^2$, so the angular momentum has definite value. Note that that is $L^2$, not $\vec{L}$.
3) Eigenfunctions of $L_z$, where we've chosen some arbitrary axis to be the $z$ axis.

 

[Isaac0427]

What you're doing in #38 is not a solution for a fixed circular orbit (not surprising, because there is no such thing in QM). It's a solution for a particle in an ideal circular tube; $r$ and $\phi$ are both fixed and the potential $V$ is constant (so you might as well take it to be zero). Your solution may or may not be right, but it has nothing to do with electrons in an atom.

To solve the problem of the hydrogen atom, you have to find a set of functions that are all of:
1) Eigenfunctions of $H$ when the potential is the three-dimensional Coulomb potential.
2) Eigenfunctions of $L^2$, so the angular momentum has definite value. Note that that is $L^2$, not $\vec{L}$.
3) Eigenfunctions of $L_z$, where we've chosen some arbitrary axis to be the $z$ axis.

Ok, thank you very much.

 

[ftr]

Ok, so let's say this is called the particle in a circular fixed orbit. The particle can only move around the orbit, and thus we only need one variable/coordinate, $\theta$. Given this situation, how does the work I showed in post 38 look?

In 2D you need to specify two variables r, theta. The hole point of QM is that the particle is not restricted to a an "orbit". Your post #38 is all some jumble like Zapper said. I think you are way ahead of yourself( your enthusiasm appreciated), you cannot do this stuff without doing all the math exercises for calculus I, II, III as in university mathematics courses.

 

[my2cts]

Any atom with only one electron can be solved exactly, this is just the hydrogen problem with Z>1.
I believe that H2+ can be solved exactly as well, if the nuclei or nuclear wavefunction is kept fixed.
If you do not insist on analytical solution then Quantum Chemistry is what you are looking for.
QC solves the atomic Schrodinger or Dirac equation for you probably in an app by now.
It can deal with quite large molecules and with local properties in solids as well.
You may also want to take a look at local density approximations.

 

[Charles Link]

In studying some of the more advanced models, it still is very worthwhile to have the Bohr atom derivation of the hydrogen atom at your fingertips because it does get the correct energies for the principal quantum number and it does quantize the orbital angular momentum, although perhaps not in the same way that the Schrodinger equation does it. The derivation is a simple one, and it really offers a major return for the amount of effort that is required. The derivation involves 3 equations and 3 unknowns: $v, r,$ and $E_n$. $\\$ (Working in c.g.s. units.) $\\$
1) $L=mvr=n \hbar$ ( $n =$ non-zero integer; $m=$ electron mass). $\\$ 2) $mv^2/r=Ze^2/r^2$. $\\$ 3) $E_n=(1/2)mv^2-Ze^2/r$. $\\$ Solving equation 3) (eliminating $v$) using 2) gives $E_n=-Ze^2/(2r)$. Solving the first two equations for $1/r$ (again eliminating $v$) gives $1/r=mZe^2/(n^2 \hbar^2)$ so that $\\$ $E_n=-mZ^2e^4/(2 n^2 \hbar^2)$. $\\$ It really is worthwhile to memorize the derivation. I have actually had exams where I needed the Bohr atom result (to memorize the formula for the energy levels is putting too much into memory), and I was able to get the correct result. For more precise computations of the hydrogen atom energy levels, (which gives the correct answer for wavelengths of the observed spectral lines), the mass of the electron $m$ can be replaced by the reduced mass $\mu=m_p m_e/(m_p+m_e)$ of the proton-electron system. Studying the more detailed quantum mechanical derivations is also important, but the Bohr atom really offers quite a good return for minimal effort without even requiring any calculus.

 

[ZapperZ]

In studying some of the more advanced models, it still is very worthwhile to have the Bohr atom derivation of the hydrogen atom at your fingertips because it does get the correct energies for the principal quantum number and it does quantize the orbital angular momentum, although perhaps not in the same way that the Schrodinger equation does it. The derivation is a simple one, and it really offers a major return for the amount of effort that is required. The derivation involves 3 equations and 3 unknowns: $v, r,$ and $E_n$. $\\$ (Working in c.g.s. units.) $\\$
1) $L=mvr=n \hbar$ ( $n =$ non-zero integer; $m=$ electron mass). $\\$ 2) $mv^2/r=Ze^2/r^2$. $\\$ 3) $E_n=(1/2)mv^2-Ze^2/r$. $\\$ Solving equation 3) (eliminating $v$) using 2) gives $E_n=-Ze^2/(2r)$. Solving the first two equations for $1/r$ (again eliminating $v$) gives $1/r=mZe^2/(n^2 \hbar^2)$ so that $\\$ $E_n=-mZ^2e^4/(2 n^2 \hbar^2)$. $\\$ It really is worthwhile to memorize the derivation. I have actually had exams where I needed the Bohr atom result (to memorize the formula for the energy levels is putting too much into memory), and I was able to get the correct result. For more precise computations of the hydrogen atom energy levels, (which gives the correct answer for wavelengths of the observed spectral lines), the mass of the electron $m$ can be replaced by the reduced mass $\mu=m_p m_e/(m_p+m_e)$ of the proton-electron system. Studying the more detailed quantum mechanical derivations is also important, but the Bohr atom really offers quite a good return for minimal effort without even requiring any calculus.

Unfortunately, this is not what was asked by the OP. If you look back at the very first post, the OP explicitly indicated that the starting point of this derivation is the Schrodinger equation.

If the OP had wanted the Bohr-Sommerfield model and the energy quantization from it, this would have been dealt with within the first few post. But noooooo... that is not what he wanted.

This is definitely the case where the eyes are bigger than the stomach.

Zz.

 

[Charles Link]

Unfortunately, this is not what was asked by the OP. If you look back at the very first post, the OP explicitly indicated that the starting point of this derivation is the Schrodinger equation.

If the OP had wanted the Bohr-Sommerfield model and the energy quantization from it, this would have been dealt with within the first few post. But noooooo... that is not what he wanted.

This is definitely the case where the eyes are bigger than the stomach.

Zz.

I realize this is not exactly what the OP asked for, but I'm also trying to make sure the OP doesn't lose sight of the basic physics when he finds himself swamped in difficult mathematics. In this instance, I wouldn't even call the Bohr atom a watered-down approach. It is a very important stepping stone that provides a very important part of the foundation. One thing I think the OP is looking for is some useful mathematics, and the Bohr model of the atom provides it. There is no simple way that I know of to learn the more complex models, including the one using the Dirac equation which I never completely understood. (presented in Bjorken and Drell's textbooks).

 

[ZapperZ]

I realize this is not exactly what the OP asked for, but I'm also trying to make sure the OP doesn't lose sight of the basic physics when he finds himself swamped in difficult mathematics. In this instance, I wouldn't even call the Bohr atom a watered-down approach. It is a very important stepping stone that provides a very important part of the foundation. One thing I think the OP is looking for is some useful mathematics, and the Bohr model of the atom provides it. There is no simple way that I know of to learn the more complex models, including the one using the Dirac equation which I never completely understood. (presented in Bjorken and Drell's textbooks).

You are more than welcomed to try and convince the OP that he should learn the simpler case first before attempting what he had been asking.

Zz.

 

[gleem]

@Isaac0427 Here is a link to a similar problem that you have be asking about . http://www.physics.csbsju.edu/QM/square.08.html The round square well. Not a hydrogen atom but is instructive.