Solve Schrodinger Equation for Electrons in Atom: Potential Used?

[Isaac0427]

How would you solve the Schrödinger equation for an electron in an atom. What potential do you use?

Thanks!

 

[drvrm]

How would you solve the Schrödinger equation for an electron in an atom. What potential do you use?

The potential felt by an electron has to be defined as it is under the field of nuclear charge as well as other electrons if the atom is a multi-electron set up...
In case of simple atom having single electron under the influence/field of a proton in the nucleus the potential is electrostatic potential of the charges.
I guess that if you have your electron in the outermost shell than an effective potential had to be defined as the electron's perception of the nuclear charge will be screened by the electrons present in the shells in between. such screening effects have been observed or screened potential with effective z value have been used.

 

[dextercioby]

How would you solve the Schrödinger equation for an electron in an atom. What potential do you use?

Thanks!

You can only solve the H-atom in the Schrödinger (no spin, no SR) / Klein-Gordon (no spin, SR) / Pauli (spin 1/2, no SR) / Dirac (spin 1/2, SR) equations, as this is a 2-particle system. Once you have at least 3 particles (2 electrons + nucleus = He atom), you can only find approximate solutions. The potential is a sum of Coulomb electrostatic terms for the multielectron atom in the Schrödinger approach. To these you can also add spin interactions. Any book on atomic physics (try the one by Bransden & Joachain) should provide you with the exact details.

 

[Isaac0427]

You can only solve the H-atom in the Schrödinger (no spin, no SR) / Klein-Gordon (no spin, SR) / Pauli (spin 1/2, no SR) / Dirac (spin 1/2, SR) equations, as this is a 2-particle system. Once you have at least 3 particles (2 electrons + nucleus = He atom), you can only find approximate solutions. The potential is a sum of Coulomb electrostatic terms for the multielectron atom in the Schrödinger approach. To these you can also add spin interactions. Any book on atomic physics (try the one by Bransden & Joachain) should provide you with the exact details.

So how do we get from the Schrödinger equation to quantum numbers, and how do we get from quantum numbers to atomic orbitals?

 

[dextercioby]

Answers to these questions could be the subject of a book. :) I am not really sure on your preparation for understanding even a super-shortened answer. Anyway, to such generic questions, we urge members to read the (fairly decent,) relevant articles on wikipedia, then come here if they still have unclear aspects.

 

[PeroK]

So how do we get from the Schrödinger equation to quantum numbers, and how do we get from quantum numbers to atomic orbitals?

That should be covered in any textbook on QM.

 

[DrClaude]

How would you solve the Schrödinger equation for an electron in an atom. What potential do you use?

Coulomb potential.

I guess that if you have your electron in the outermost shell than an effective potential had to be defined as the electron's perception of the nuclear charge will be screened by the electrons present in the shells in between. such screening effects have been observed or screened potential with effective z value have been used.

That's an approximation. The full Hamiltonian simply contains the electron-nucleus and electron-electron Coulomb interactions.

 

[bhobba]

So how do we get from the Schrödinger equation to quantum numbers, and how do we get from quantum numbers to atomic orbitals?

You need to wait until uni for that one.

Its from the solution to Schrödinger's equation. You will either learn how to do that as you do your physics degree (it uses the method of separation of variables) or if you do an applied math degree as part of a standard course on partial differential equations where it's one the the PDE's usually studied.

I did a degree in math and studied it in my PDE subject which is why math students were allowed to skip some introductory QM courses should they wish to study QM.

Thanks
Bill

 

[Nugatory]

So how do we get from the Schrödinger equation to quantum numbers, and how do we get from quantum numbers to atomic orbitals?

If you google for "Schrödinger equation hydrogen" you'll find many explanations for how Schrödinger's equation is solved for a single electron around a positive-charged nucleus. You'll need to be comfortable with the separation of variables technique used to solve some multi-variable differential equations, so you'll usually encounter this towards the end of your first semester of QM after you've been through a fair amount of college-level math.

Multi-electron problems are much harder because you can't use the simple $1/r$ potential associated with $1/r^2$ forces because the repulsion between the electrons has to be included as well.

 

[Isaac0427]

If you google for "Schrödinger equation hydrogen" you'll find many explanations for how Schrödinger's equation is solved for a single electron around a positive-charged nucleus. You'll need to be comfortable with the separation of variables technique used to solve some multi-variable differential equations, so you'll usually encounter this towards the end of your first semester of QM after you've been through a fair amount of college-level math.

Multi-electron problems are much harder because you can't use the simple $1/r$ potential associated with $1/r^2$ forces because the repulsion between the electrons has to be included as well.

But how do we know that their solutions are discrete?

 

[Khashishi]

The hydrogen atom is part of any standard quantum mechanics curriculum. If you are interested, you should read some standard textbooks on the subject, and come back with smaller questions.
The energy solutions are discrete because it has to do with the number of nodes in the wavefunction, which should be an integer. The nodes can be either in the radial or angular direction. Of course, you can take linear combinations of solutions with different numbers of nodes -- this means the energy is in superposition.

 

[gleem]

But how do we know that their solutions are discrete?

The hydrogen atom can be assumed to be spherically symmetric so Schrödinger's equation is put into spherical coordinates. Using the method of separation of variable as suggested above results in three independent differential equations one for each of the spherical coordinates. The solution of these equations plus certain criteria as requiring the wave function to be single valued, finite and lead to physically possible results ends up with functions that are indexed integrally resulting in a principle (Orbit) quantum number n ( index in the radial component of the wave function) plus two other indices l and m which do not effect the energy of the electron orbits in this simple model. The energy of an orbit in this model only depends on n.

Study the solution of the single electron atom as described any elementary QM books

 

[Khashishi]

quantum number n plus two other indices l and m which do not effect the energy of the electron orbits in this simple model.

Actually, it's a little more correct to say that $\ell$ does affect the energy. The principal quantum number $n = n_r + \ell + 1$, where $n_r$ is the number of nodes in the radial wavefunction, and $\ell$ is the number of planar nodes in the spherical harmonic. The energy increases with the number of nodes. By defining the principal quantum number, we hide the energy dependence of $\ell$, but it is an important detail if we apply Hellmann-Feynman theorem to the hydrogen atom.

 

[Nugatory]

But how do we know that their solutions are discrete?

I don't know of any better way than to work through the solution and see how it turns out.

However, you can get a general feel for how solutions to Schrödinger's equation may end up with discrete eigenvalues by working with simpler problems. For example... Are you familiar with the problem of the one-dimensional infinite potential well, which models a particle trapped in a one-dimensional box? Although this problem is much simpler, it also has discrete solutions and they appear for the same general reason: only for certain discrete values of $E$ will a function $\psi$ that satisfies $H\psi=E\psi$ also satisfy the boundary conditions. Similar thinking (but appreciably more complicated) applies to the bound electron problem; this is what @Khashishi is getting at in #11 of this thread.

 

[gleem]

Actually, it's a little more correct to say that ℓℓ\ell does affect the energy. The principal quantum number n=nr+ℓ+1n=nr+ℓ+1n = n_r + \ell + 1, where nrnrn_r is the number of nodes in the radial wavefunction, and ℓℓ\ell is the number of planar nodes in the spherical harmonic. The energy increases with the number of nodes. By defining the principal quantum number, we hide the energy dependence of ℓℓ\ell, but it is an important detail if we apply Hellmann-Feynman theorem to the hydrogen atom.

I think it is misleading to say that the eigenvalue depends on l . Dependence implies an influence which in fact l does not have. The energy of the various eigenstates depends only on n from the solution of Schrödinger's equation. The energy of an eigenstate does not change when the angular momentum is different.for a given n. The relation n= n_r + l +1 is an interesting association of the characteristics of the components of the eigenfunctions .

 

[vanhees71]

The hydrogen energy levels are highly degenerate (in the most simple non-relativistic model without spin the degeneracy of $E_n$ is $n^2$-fold). The reason is the large symmetry of the quantized Kepler problem, which is the reason that you get easily a completely representation free algebraic solution (as has been done first by Pauli to solve the hydrogen energy-level problem in modern quantum theory in its formulation as matrix mechanics by Heisenberg, Born, and Jordan in 1925).

 

[Khashishi]

The principal quantum number $n$ is historically important, because it is a parameter in the Balmer series, and in related series. But the simple models lead to the idea of Bohr orbits, which are incorrect. In Bohr orbits, the only degree of freedom is $n$, which labels a circular orbit around the nucleus. There is some arbitrariness in saying what influences what, but for a deeper understanding, it would have been better to label the states with $n_r$ and $\ell$.
See https://en.wikipedia.org/wiki/Hellmann–Feynman_theorem#Expectation_values for an application of Hellmann-Feynman theorem on (promoted) $\ell$. If there was no $\ell$ dependence, then $\frac{\partial H}{\partial \ell} = 0$.

 

[dextercioby]

If we're bringing into discussion obsolete physics, then why mention the Bohr's model at all? Sommerfeld's model had elliptic i/o circular orbits and as a topping the correct flavor of special relativity. Of course n and l are related and not completely independent of another, that's why we got SO(4) group coming in. The discovery of the SO(4) symmetry for the Kepler problem was a major breakthrough and not many textbooks put it into the spotlight it diserves.

 

[gleem]

@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 Schrödinger 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 experiments 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.

 

[jtbell]

a pre quantum mechanics modern physics book

There are "introductory modern physics" textbooks that are intended to fit in between a standard calculus-based intro physics course (classical mechanics, E&M, thermo, optics) and a full upper-division undergraduate quantum mechanics course based on e.g. Griffiths or Sakurai. I taught such a course for many years using first Beiser (which now seems to be out of print, latest edition was 14 years ago) and then https://www.amazon.com/dp/1938787757/?tag=pfamazon01-20. There are similar books by Krane and by Serway/Moses, and probably some others.

These books usually (at least Beiser and Taylor/Zafiratos) require only basic calculus (derivatives and integrals) as mathematical prerequisite, and introduce more advanced stuff as necessary. They introduce the Schrödinger equation and solve it for simple systems (e.g. one-dimensional particle in a box). For the hydrogen atom they don't give all the gory details (e.g. the complete derivation of Legendre polynomials or associated Laguerre polynomials), but they show enough to see where the quantization comes from. And they give some of the historical and experimental background.

And they cover other areas such as relativity, radioactive decay, nuclear physics, elementary particles and solid-state physics at a basic overview-type level.

 

[gleem]

If we're bringing into discussion obsolete physics, then why mention the Bohr's model at all? Sommerfeld's model had elliptic i/o circular orbits and as a topping the correct flavor of special relativity. Of course n and l are related and not completely independent of another, that's why we got SO(4) group coming in. The discovery of the SO(4) symmetry for the Kepler problem was a major breakthrough and not many textbooks put it into the spotlight it diserves.

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.

 

[Isaac0427]

However, you can get a general feel for how solutions to Schrödinger's equation may end up with discrete eigenvalues by working with simpler problems. For example... Are you familiar with the problem of the one-dimensional infinite potential well, which models a particle trapped in a one-dimensional box? Although this problem is much simpler, it also has discrete solutions and they appear for the same general reason: only for certain discrete values of EEE will a function ψψ\psi that satisfies Hψ=EψHψ=EψH\psi=E\psi also satisfy the boundary conditions. Similar thinking (but appreciably more complicated) applies to the bound electron problem; this is what @Khashishi is getting at in #11 of this thread.

I am very familiar with this problem. The discrete solutions come from the boundary condition because a general solution of the Schrödinger 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?

 

[Khashishi]

Do you know what spherical harmonics are? It is essentially a 2D spherical analog to the Fourier series. The Asin(kx)+Bcos(kx) are just terms of a Fourier series, which can be used to construct any smooth periodic function, which is the same (isomorphic) as a function defined on a circle. Spherical harmonics are functions defined on a sphere which can be used to construct any smooth function on a sphere. (They aren't the only functions that can be used.). The periodic nature is due to the fact that going around 360 degrees brings you back to the same place. This gives you angular momentum quantization.

 

[bhobba]

I don't know of any better way than to work through the solution and see how it turns out.

Exactamundo - the Fonz.

Its simply the solution.

There is a very deep reason but that involves some really advanced deep math that's a little scary even for someone like me that has a degree in math:


If you really really want to do it the following will give you the math:
http://www.physics.miami.edu/~nearing/mathmethods/mathematical_methods-one.pdf

The perquisite would be calculus BC in the US systen, Malth HL in the IB program or Specialist math in the Australian system for which many textbooks can be found eg:
https://www.haesemathematics.com.au/books/mathematics-hl-core-3rd-edition

Thanks
Bill

 

[bhobba]

I guess my quantum number question all along was what are the boundary conditions associated the electron in an atom.

The answer is the in video lectures I linked to.

Thanks
Bill

 

[vanhees71]

If we're bringing into discussion obsolete physics, then why mention the Bohr's model at all? Sommerfeld's model had elliptic i/o circular orbits and as a topping the correct flavor of special relativity. Of course n and l are related and not completely independent of another, that's why we got SO(4) group coming in. The discovery of the SO(4) symmetry for the Kepler problem was a major breakthrough and not many textbooks put it into the spotlight it diserves.

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 $\mathrm{so}(4)=\mathrm{su}(2) \oplus \mathrm{su}(2)$ and their deformations needed for the Kepler problem/hydrogen atom in its most simple form :-).

 

[Nugatory]

I am very familiar with this problem. The discrete solutions come from the boundary condition because a general solution of the Schrödinger 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.

[Edit: Pre-coffee silliness corrected]
Think about what the wave function of a bound electron must look like. Because of the spherical symmetry of the problem, we'll write it in polar coordinates as $\psi(r,\theta,\phi)$ instead of the Cartesian x, y, and z coordinates we use for particles in a box. (I'll be using $\theta$ as the "longitude", the coordinate that runs from zero to $2pi$, and $\phi$ as the "latitude" running from zero to $\pi$).

Choose an arbitrary radius $R$, and consider $\psi(R,\theta,\pi/2)$; this is just the value of $\psi$ along a ring of radius $R$ in the equatorial plane.

Obviously $\psi$ must have the property that $\psi(R,\theta,\pi/2)=\psi(R,\theta+2\pi,\pi/2)$; otherwise the wave function would have multiple values at the same point on the ring. So there's a boundary condition right there, one that basically says that we have to be able to fit an exact integer number of nodes along the circumference of that ring of radius $R$. This leads to discrete solutions for pretty much the same reason that a similar constraint leads to discrete solutions for the particle in a box.

Of course this isn't the only boundary condition (bound electron is a way more complex problem than the particle in a box), but it's an example of how the boundary conditions can arise from the symmetry of the physical problem.

 

[Nugatory]

On a side note, if you apply boundary conditions to a particle, would the possible values of linear momentum be discrete?

Not necessarily. A counterexample would be the reflection/transmission of an unbound particle; the boundary conditions are such that both the energy and the linear momentum take on a continuous spectrum instead of discrete values.

 

[vanhees71]

That's a bit strange coordinate system. The usual physicist's spherical coordinates $(r,\vartheta,\varphi)$ have the domains $r>0$, $0<\vartheta<\pi$, and $0\leq \varphi<2 \pi$. They are defined by the transformation of the position vector from spherical to Cartesian components
$$(x,y,z)=r (\cos \varphi \sin \vartheta,\sin \varphi \sin \vartheta,\cos \vartheta).$$
The conventional way to address a spherically symmetric problem is to write the Laplacian in spherical coordinates and then make an separation ansatz
$$\psi(r,\vartheta,\varphi)=\frac{1}{r} R(r) \Theta(\vartheta) \Phi(\varphi).$$
To take out the factor $1/r$ is for convenience. The boundary conditions are not so trivial to determine as it seems on the first glance. They boil down to the fact that without loss of generality you can assume that the wave function is a scalar under rotations and that the radial Schrödinger equation should be governed by an effective Hamiltonian that is self-adjoint on $\mathrm{L}^2(0,\infty)$ (where $R(r)$ lives). Particularly this yields $\psi(r,\vartheta,\varphi+2 \pi)=\psi(r,\vartheta,\varphi)$. From the self-adjointness of the Hamiltonian you also get that $R(0)=0$ as a boundary condition. Further the solutions for $\Theta$ should be square integrable with the weight $\sin \vartheta$. This leads to the set $\Theta(\vartheta) \Phi(\varphi)=Y_{lm}(\vartheta,\varphi) \propto \exp(\mathrm{i} m \varphi)$, where for each $l \in \{0,1,2,\ldots \}$ the $m \in \{-l,-l+1,\ldots,l-1,l \}$. The radial solution should also stay finite (go to 0) with $r \rightarrow \infty$ for the scattering (bound states). With these conditions you also get the radial solutions for the case of the hydrogen atom. They are given by the Laguerre polynomials times an exponential function (for the bound states). A very good treatment, simplifying a lot by making use of the angular-momentum algebra to determine the spherical harmonics, can be found in the textbook by Messiah, vol. I.

 

[gleem]

n 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?

You can determine the angular moment by taking the expectation value of the square of the total angular momentum operator which reults in the well know relations L² = l(l+1)ħ² and L_z = mħ