# Schrodinger Equation and Energy Quantization

1. Aug 30, 2008

### quantumlaser

I'm a bit embarrased to ask this (thats why I'm asking here and not asking one of my professors), as a grad student in Physics I've had a good deal of quantum mechanics, but one thing I haven't fully understood yet is the mechanism in the Schrodinger Equation that forces eigenvalue quantization (energy, angular momentum, etc...) in bound state solutions. For quantum well potentials, energy quantization is forced by the boundary conditions, but in the harmonic oscillator and hydrogen atom potentials, the eigenvalue quantization seems to just pop out as a consequence of the math. It seems there should be some deeper explaination as to why the bound state eigenvalues are quantized. Could someone enlighten me on this?

2. Aug 31, 2008

### malawi_glenn

what is wrong with math? A particles eq of motion is described by the Schrödinger eq, and as a consequence of that, eigenvalues for bound states are discrete.

And what is a "quantum well", and why is a "quantum well" different from the "the harmonic oscillator and hydrogen atom potentials" ???

The discrete states comes from, as you say, the boundary conditions with you impose on your solution, this you do for ALL potentials no matter how they look like. (well = potential)

And this B.C has to do with the definition of a bound state, that it's wavefunction goes to zero for large radii, and in spherical systems, that the wavefunction is periodic, etc. Of course, the VERY definition for bound state is E < 0, but you also impose the B.C, otherwise the prob. to find particle outside well is very large (infinity) and that is not so nice.

Last edited: Aug 31, 2008
3. Aug 31, 2008

### quantumlaser

By "quantum well" I mean a well potential, i.e square wells, spherical wells, etc... I just thought that there might be something deeper in eigenvalue quantization than just boundary conditions. I guess I'm just looking for something deeper where there is nothing. My original impetus for posting this was me trying to explain quantization to my friends outside of physics and I didn't really have a suitable answer for why energy is quantized other than "its in the math."

4. Aug 31, 2008

### malawi_glenn

The discrete values for square wells are also due to mathematics of the solution to the boundary conditions... same thing for all wells remember.

Explaining quantum physics and special relativity etc without math is quite impossible since those branches are non-intutive with our ordinary life language and reasoning, which has developed during thousands of years without encountering neither high velocites nor very very tiny objects. Look at math as the language of physics, and also how math and physics has developed hand in hand during the centuries, maybe you'll stop thinking "it's just math" and similar.

5. Aug 31, 2008

### Marty

If you think about it, the quantization of energy really does not come from the Schroedinger equation. It's just a differential equation, which we solve by separation of variables, automatically leading to an orthogonal set of basis states. How is this different from the Bessel equation, which we might use to solve for the vibration modes of a circular membrane streched across a drum. Do we then say that the energy levels of the drum are quantized because of the Bessel equation?

6. Aug 31, 2008

### ZapperZ

Staff Emeritus
Humm.. I'm not sure I understand this.

The quantization in both the harmonic oscillator and the hydrogen atom ARE due to the boundary conditions. The harmonic oscillator potential imposes such boundary condition for the harmonic oscillator, and the central force potential imposes the boundary condition for the hydrogen atom. So there are boundary conditions for both cases.

When you remove any form of bound states (i.e. your boundary conditions are at infinity), then you get the continuous states, as what you get out of a free-particle scenario, for example.

Zz.

7. Aug 31, 2008

### Staff: Mentor

In both cases (a vibrating circular drumhead and a circular quantum well), we get solutions that are standing-wave modes with discrete frequencies, determined by the boundary conditions. In the circular drumhead, the energy is determined by the amplitude of the wave, which can have a continuous range of values. In the circular quantum well, the energy is determined by the frequency of the wave, so it has discrete values.

8. Aug 31, 2008

### Marty

But this doesn't explain why the hydrogen atom can only have certain fixed energy levels, for two different reasons:

1. The circular drumhead can vibrate in mixed modes with cominations of different frequencies. So why can't the hydrogen atom vibrate in mixed modes with different combinations of energy?

2. The modes of the circular drum can be excited to any desired amplitude. Why not the modes of the hydrogen atom?

I'm not asking these questions because I expect answers. I'm asking them because I want to point out that you simply can't explain things away by saying "it comes out of the equations".

9. Aug 31, 2008

### Dr Transport

The discretization of the energy eigen-values is a result of the conditions placed on the wave-functions. When you solve the SE for the radial component, the general form of the wave function is a hypergeometric function or sometimes a confluent hypergeometric equation. The coefficients are required to be integer values leading to quantization of the energy levels. See Schiff, Mertzbacher, Messiah or any host of QM texts.

10. Aug 31, 2008

### malawi_glenn

Marty, what if I or or someone else can show you that discretization of energy levels in an atom follows the Quantum mechanics formalism? The Language of physics is math, Dirac predicted existence of antiparticles just from the solution from the Dirac equation and so on.

And if you know your atomic physics, hydrogen atom don't vibrate...

And as Jtbell said, the energy of an electron in hydrogen atom is determined by its frequency of the wavefunction, and the freq's are discrete, hence discrete energy levels.

So what you really want to ask is why an electron in hydrogen atom can't take any wave function that is a superposition of the eigenfunctions, the other stuff you mentioned is not applicable to the hydrogen atom.

11. Aug 31, 2008

### Marty

Malawi Glen already said that there's no essential difference between the discrete boundary conditions of the square well versus the more or less extended boundary conditions of the hydrogen atom. Yes, we know the eigenvalues are discrete. But so are the eigenvalues of the circular drum. You haven't explained why the energy is quantized in one case and not the other.

12. Aug 31, 2008

### malawi_glenn

Marty, it is a trivial exercise in introQM to solve the Schrodinger eq and find bound states and show that the energies are quantized. Do you want us to show you all this? Otherwise see the sources Dr Transport posted, those are really good QM-textbooks.

This was the first hit on google; http://musr.physics.ubc.ca/~jess/p200/sq_well/sq_well.html ("Equation (11) implies restrictions on the allowed values of E").

Tell me/us if you want to have some middle steps or whatever you need to understand how the shrodinger eq + boundary conditions leads to discrete energy levels of the bound states.

Also, the fundemental reason why there is a difference is that the physical systems are different... quite trivial.

Last edited: Aug 31, 2008
13. Aug 31, 2008

### Staff: Mentor

It can, under the right circumstances. For example, during a transition from one energy eigenstate to another, the atom is briefly in a superposition (linear combination) of the two states:

$$\psi = a(t) \psi_1 + b(t) \psi_2$$

As the transition proceeds, a(t) decreases from 1 to 0 and b(t) increases from 0 to 1.

Also, some people study "Rydberg atoms" which are hydrogen (or other) atoms in highly-excited states, with large values of n, and energies just below zero (i.e. almost ionized). If I remember correctly, they actually produce states that are superpositions of energy eigenstates with close-together values of n, to form a localized wave-packet that travels around the nucleus in a classical-like orbit. In other words, they're working in the boundary zone between "quantum-like" behavior and "classical-like" behavior.

The amplitude of the QM wave function is fixed by the requirement that the total probability of the electron being found somewhere equals 1.

The absolute amplitude of the quantum wave function (taken as a whole) actually doesn't have any physical significance. It can be anything, or can be left unspecified. It's merely a convenience to "normalize" it so the integral of $\psi^*\psi$ over all space equals 1. What really matters are the relative values of $\psi^*\psi$ at different locations, because that determines the relative probabilities of the particle being at those locations (e.g. twice as likely to be at point 1 as at point 2).

Last edited: Aug 31, 2008
14. Aug 31, 2008

### nrqed

It's still the boundary conditions that forces quantization upon us.
Consider the harmonic oscillator. This time, the boundary conditions are that the wavefunction must go to zero fast enought at spatial inifnity to make the wavefunction square integrable.

Let's say you would solve the equation grpahically and you would adjust the value of the parameter E to whatever value you please and observe the behavior of the wavefunction. What you would observe is that for most values of E the wavefunction is not normalizable. It might approach zero for a while but then start to divereg and shoots off to an infinite value. As you woudl approach one the quantized values of the enery of the harmonic oscillator you would see the tail of the wavefunction at spatial infinities get closer and closer to zero but would still would eventually shoot off to infinity. If you reach exactly one of the special values of the enery, the wavefunction would die off and not ever shoot to infinity. Then the wavefunction is normalizable.

Hope this helps

15. Aug 31, 2008

### Marty

Yes, this is a good point. But you haven't quite explained why the superposition of two states cannot persist indefinitely. In which case the energy would have an intermediate value.

Then your seem to be arguing that ultimatley the quantization of energy is a consequence of the quantization of charge. Because if you had half an electron orbitin a hydrogen nucleus, then the energy would be different than the case of a whole electron.

I'm not sure this argument is conclusive. Consider two hydrogen atoms, or rather two nuclei, located one centimeter apart. If you write the Schroedinger equation for this arrangement, there should be a ground state solution that looks very much like the expected ground states for two independent atoms. (Actually there would be two degenerate solutions - one odd, and one even. But that's not my point.) If you "fill" this ground state with a single electron, wouldn't you have half an electron in the vicinity of each nucleus? And then what becomes of energy quantization?

16. Aug 31, 2008

### Avodyne

It would persist indefinitely if the atom was not coupled to the quantized radiation field. In general, if you have a completely isolated system with a time-independent hamiltonian, ANY state can be written as a superposition of energy eigenstates. For such a superposition, the energy does not have a definite value.

17. Aug 31, 2008

### Dr Transport

Let's think about the hydrogen atom. From my well worn copy of Schiff, pg 92, when you determine the series solution for the radial portion of the wave function, "you must terminate the series, thus the highest power of $\rho$ in $L$ is $\rho^{n'} (n' \ge 0)$, we must chose the parameter $\lambda$ to be a positive integer $n$....."

In other words for the series to terminate, you must pick an integer, thus leading to quantization of the energy.

18. Aug 31, 2008

### Staff: Mentor

A superposition state does not have an intermediate value of the energy. It has an indeterminate value of the energy until it is measured, at which point the energy becomes one of the two values corresponding to the states that make up the superposition. The "choice" is random, with probabilities that depend on the relative weights of the two states in the superposition.

19. Sep 1, 2008

### Marty

I don't know how you distinguish between what I call an intermediate value and what you call an indeterminate value. Experimentally that is.

20. Sep 1, 2008

### malawi_glenn

Recall some postulates of quantum mechanics:

A measurment of an observable makes the wavefunction collapse into one eigenstates of the observable and the outcome of a measurement gives you ONE of the eigenvalues to the observable.

i.e suppose:

$$\Psi _E = a\psi _{E1} + b\psi _{E2} + d\psi _{E4}$$

where $a,b,d$ are complex numbers.

A measurment can give you $E_1$ with probability $|a|^2$, $E_2$ with prob. $|b|^2 etc.$ You never get a mixture and hence you'll only get discrete energy values.

Another postulate of QM is that is meaningless to ask what properties (such as energy etc.) a state has before a measurment, so talking about the energy of the wavefunction $\Psi _E$ is meaningless.

Last edited: Sep 1, 2008