# Some basic derivations need improvement

1. Dec 15, 2013

### devang2

The traditional Schrodinger equation is written as
Laplacian ψ+constant(E-V)ψ=0 the above equation derivation is based on the formation of hydrogen after the combination of proton and electron. E is said to be total energy and V potential. energy of hydrogen atom. On solving the value of E is found to be negative which indicates that hydrogen atom is expected to stable but on the contrary it is highly unstable. Actually E is the energy evolved when when proton and electron combine to form hydrogen atom. The above form of equation needs modification
The equation given below is based on the stages of formation of hydrogen atom by the combination of electron and proton . Suppose an electron starts from infinity towards proton to form hydrogen atom,if V is the electrostatic potential energy evolved at any stage the whole of it is not given out as electromagnetic radiation , otherwise Schrodinger equation will be meaningless E is just energy as the energy of wound up spring. E is converted to kinetic energy of electron and electromagnetic radiation
v=kineticenergy+E here E the electromagnetic energy radiated at any stage
Kinetic energy =V-E v is just energy with positive sign .if this value of kinetic energy is usedto derive schrodinger equation the equation gets the form Laplacian operatorψ +constant(V-E)ψ=O
if above equation is solved the value of E is positive .E is the electromagnetic radiation emotted at any stage . the rest of the solution of the modified equation is the same as that of traditional equation. Please tell me if i am right or where i have gone wrong.

Click the sigm

2. Dec 15, 2013

### tom.stoer

The Schrödinger equation

$(H-E)\psi = 0$

$H=-\frac{1}{2m}\nabla ^2 + V$

cannot be used to discuss electromagnetic radiation or energy of the electromagnetic field b/c the field is not a dynamical entity in this equation. You would have to use quantum electrodynamics instead.

3. Dec 16, 2013

### tom.stoer

it is rather simple; w/o any external el.-mag. field energy is conserved and the Schrödinger equation applies; the "E" is nothing else but the energy of the energy eigenstate;

if you want to study el.-mag. radiation in the non.-rel. Schrödunger equation you add an external, time-dependent el.-mag. field; via this field you get non-zero matrix elements between different states, i.e. non-vanishing transition probabilities; but due to the time-dependent el.-mag. field the Hamiltonian is time-dependent and therefore the energy "E" which is the energy of the quantum state of teh hydrogen atom is not conserved;

so via time-dependent perturbation theory you can calcluate matrix elements and transition probabilities, but the method seems to violate energy conservation; in order to correct this you must not add the el.-mag. field by hand but you have to introduce it as independent dynamical d.o.f.; this is done in QED, and then the energy of the atom + the radiation is field is conserved

4. Dec 18, 2013

### devang2

Thank you for the detailed reply but still it is not clear but is the real source of radiation associated with hydrogen atom , i think Schrodinger equation is silent about it . The fundamental ladder of role played by electron to study properties of hydrogen has been discarded by negating the motion of electron which plays very important role. Properties of bare proton and electron are not same . No motion no wave function and hence kinetic energy of electron has also be considered while deriving schrodinger equATION

5. Dec 18, 2013

### ChrisVer

The E just gives you the possible energy states of the Hydrogen atom (that's why when you solve the Schrodinger equation you don't just get 1 energy value that would correspond to your particle's energy, but you get the energy spectrum)

6. Dec 18, 2013

### ChrisVer

Could you also please explain what you mean by solving quantum problems without Schrodinger equation?
In what means did you do that? and for which kind of problems?

7. Dec 18, 2013

### tom.stoer

Yes, THIS non-rel. Schrödinger equation does not contain any dynamical el.- mag. field.

8. Dec 19, 2013

### devang2

i mean the source of energy that is radiated when electron and proton initially separated by large distance combine to form ground state of hydrogen atom. The nature of source of energy remains the same when excited state forms spectrum.Schrodinger equation gives only an equation withe different values of energies depending on the value n but is silent about the real source of energy

9. Dec 19, 2013

### ChrisVer

Is there a source of energy?

10. Dec 19, 2013

### tom.stoer

I can repeat this a couple of times:

11. Dec 23, 2013

### devang2

i am baffled . Please remove my doubts . it is taught to the students that Lyman Balmer etc.series based on energy equation derived from Schrodinger equation forum the electromagnetic spectrum of hydrogen atom . I am seeking the real source of that electromagnetic energy but your learned opinion is that Schrodinger equation can not be used to discuss electromagnetic radiation .I am puzzled

12. Dec 23, 2013

### Staff: Mentor

SchrÃ¶dinger's equation gives us the discrete energy levels for hydrogen (and other systems). When a hydrogen atom makes a transition between two energy levels, we observe that a photon is radiated, and we use conservation of energy to predict the energy of the photon.

However, SchrÃ¶dinger's equation by itself does not describe the actual radiation process. For that we need quantum electrodynamics.

13. Dec 24, 2013

### devang2

Thank you for taking trouble to give satisfactory answers to my queries . It is not still clear to me what is the real source of energy when electron and proton initially separated by large distance combine to form hydrogen.

14. Dec 24, 2013

### Staff: Mentor

That comes from electrostatic attraction between the positive-charged proton and the negative-charged electron. They attract one another, so there's a force between them and potential energy when they're separated; as they move closer the potential energy is released just as with a falling object or a stretched spring relaxing.

15. Dec 25, 2013

### devang2

thank you very much for the answer.please allow me to elaborate your answer .Suppose r is the distance of minimum approach of electron and proton which is equal to the radius of K orbit as calculated by schrodinger equation.By putting the value r as calculated by schrodinger equation in the the relation for elctrostatric potential as the electron moves from very large distance to r from the proton the value of potential energy is U which is actually found to be twice than the energy emitted when ground state of hydrogen atom is formed. Can you please tell me where the half of the emitted potential goes

16. Dec 25, 2013

### Staff: Mentor

Before I answer that question, I'm going to ask you to think about a related but completely classical question: if I drop an object to the earth's surface from infinity, where does the gravitational potential energy go? In that case, the answer is obvious: it ends up as kinetic energy of the dropped object, which gains speed as it falls.

An electron at a distance from a proton has potential energy from the electrostatic force between them. Allow the electron to fall towards the proton under the influence of that force, and some of that potential energy will be radiated away and some of it will end up as kinetic energy of the electron, which is accelerating under the influence of the electrical force.

Schrodinger's equation and the methods of quantum mechanics only come into play when you want to know how much energy is radiated away and how much stays with the electron. Solving the equation tells us how much energy a bound electron will have; if while it was unbound the electron had more energy (electrostatic potential plus kinetic) than that, the excess is what's radiated away when the electron is captured to form a hydrogen atom.

17. Dec 26, 2013

### devang2

your logical answer in keeping with the law of conservation of energy imparts half of the potential energy to electron as kinetic energy hence when associated with proton electron must be moving which leads to my confusion because now it is taught that according to quantum mechanics electron in hydrogen atom is not moving in discrete orbits but is present as sort of electronic cloud.Please take some trouble to reconcile the two view points i shall be thankful to you

18. Dec 26, 2013

### Staff: Mentor

Quantum mechanics tells us that we cannot assign a definite position to a bound electron, but we can still assign a definite energy to it.

19. Dec 26, 2013

### Naty1

Wikipedia paints a concise picture:

..

Last edited by a moderator: May 6, 2017
20. Dec 28, 2013

### devang2

21. Dec 28, 2013

### tom.stoer

Let's make this more precise (we do not have to use quantum mechanics).

We have a charged particle at radius r with kinetic energy Ekin and potential energy Epot. For the total energy we have E = Ekin + Epot. This is just Newtonian mechanics with a particle e.g. in a potential ~ 1/r.

Now we compare diffent stationary orbits with different radii r, r', r'', ... We find different energies E, E', E'', ... In Newtonian mechanics each orbit is stable, there are no transitions between orbits. And of course for each orbit the energy E is conserved.

Now quantize this classical system, using the well-known Schrödinger equation (H-E)ψ = 0. All what happens is that the classical orbits with arbitrary E are replaced by stationary states ψ with discrete energy E0, E1, E2, ...

Again the states are stationary i.e. there are no transitions between different states, and the energy is conserved.

Now we want to be more realistic and describe i) transitions between states and ii) radiation.

In non-rel. QM only i) is achieved - via a trick!! One adds a time-dependent perturbation, a kind of external electromagnetic field, which does itself not carry energy (the total energy of a plane-wave would be infinite!) Therefore the Schrödinger equation now looks like

$i\frac{d}{dt}\psi(x,t) = \left[ -\frac{1}{2m}\Delta - \frac{\alpha}{r} + h(x,t \right]\psi(x,t)$

b/c of the time dependent term h(x,t) which represents the external electromagnetic field total energy is no longer conserved . You can no longer use the Ansatz

$\psi(x,t) = e^{-iEt}\,u(x)$

to solve the Schrödinger equation. But what you can do is to calculate transition probabilities between different states using time-dependent perturbation theory. These transitions are induced by the external field. Via another trick one can relate these probabilities to probabilities for spontaneous emission, i.e. transitions that would happen w/o any external perturbation h(x,t).

So non-rel. QM provides a means to calculate transition probabilities between different energy levels E and E'. But in these transitions the total energy E is not conserved. Unfortunately non-rel. QM does not provide a dynamical entity describing the photon carrying away this energy. It's not h(x,t) which is an artificial field. This is obvious b/c the Schrödinger equation acts on the wave function ψ which represents the electron, not the photon.
----------

In order to achieve ii) you have to use relativistic quantum electrodynamics. The 1st step is to look at Maxwell's equations which do explicitly contain a dynamical electromagnetic field A. Then you replace the current j by an expression containing the electron field ψ. The result is a theory where the Schrödinger equation acts on states representing both the electron and the photon.

If you do that there is no need for an artificial term h(x,t), and energy will be conserved, simply b/c you added an entity (the photon) which is able to carry away energy.

Last edited: Dec 28, 2013
22. Dec 28, 2013

### tom.stoer

Unfortunately I did not find any paper presenting the full QED Hamiltonian in spacetime formalism plus a calculation of electron transition plus spontaneous photon emission.

23. Jan 3, 2014

### Simon Bridge

No it isn't. The correct equation has already been given though ... moving along...

The equation you wrote was general - the hydrogen one would be:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html
... you can derive them from the SE using an appropriate choice of V though.

This would actually be a model of a Universe with only one electron and one proton in it, and we ask how the electron moves with respect to the proton.
It makes a useful approximation only.

The negative value for total energy means that the electrons are in bound states.
These states need not be "stable" - that's a whole other kettle of fish.
If you want to know if a state is stable, you have to modify V to include a small perturbation
(due to the definition of "stable" in this context)

"energy evolved"??
Are you talking about binding energy?

The energy eigenstate would be the negative of the binding energy - in this model.

Recap: in the formulation you started with, you have a Universe with only an electron and a proton in it - now we want the initial state to have the electron at infinity.

Time passes - nothing can happen. Total energy is zero.

However, if there were some additional mechanism that changes this, so that the electron is later found in the ground state. This represents a total energy change for the electron ... which we notice is accounted for by the presence of a third particle - a photon.

That would be how EM radiation is normally taught.

The model does not include this situation because you didn't include it.
But - the fact we can do that, shows how useful such an incomplete model can be.

But we don't put specific values of kinetic energy into the schodinger equation - there is no place for it to go.

Instead the kinetic energy is represented by the H operator.