# Atom Stability and harmonic oscillator

1. Jun 1, 2009

### feynmann

Why does in QM the electron does not fall toward the nucleus? After all, the only force between nucleus and electron is attractive. It seems that the electron can and does indeed fall toward the center in <simple harmonic oscillator>?
My question is what's so different in these two systems that cause two different results?

It seems that the argument in the FAQ would fail in the case of s.h.o.
If that's the case, does it mean the argument is wrong in the first place?

Last edited: Jun 1, 2009
2. Jun 1, 2009

### Staff: Mentor

Have you ever seen the wave functions and probability distributions for the quantum harmonic oscillator? Note in particular the ground state wave function and probability distribution. The particle has maximum probability of being at the center, but it also has a significant probability of being away from the center.

The same is true of the electron in the hydrogen atom! Note in particular the ground state wave function n = 1, l = 0, m_l = 0. This page doesn't show graphs of the hydrogen wave functions, but if you plot $|\psi|^2$ for the ground state, you'll see that the shape is rather similar to the ground state of the SHO, except for being "pointed" at the center instead of "rounded" as in the SHO.

3. Jun 1, 2009

### alxm

First, and god knows how many times this has been pointed out already, but the the electron does enter the nucleus in a quantum-mechanical atom. Specifically, every s-type (i.e. l=0) orbital (solution to the Schrödinger equation) has r=0 as not only a point of non-zero location-probability, but the most probable point.

Second, the classical-mechanical problem of "why the electron doesn't fall into the nucleus" isn't strictly about why the electron doesn't pass near to or within the nucleus, it's about why it doesn't do so and stay there - since that is obviously the point of lowest potential energy.

The answer is that energy and momentum operators don't commute, in short: The uncertainty principle, which is implicit in the Schrödinger equation. You cannot get a solution which is completely located at any single point, and the energy of the system will be higher than the lowest possible potential energy.

Which is precisely the same result as you get if you compare the quantum harmonic oscillator versus a classical harmonic oscillator. The latter's lowest energy state is obviously to be completely stationary (and with precisely defined locations), with an energy of zero. The quantum harmonic oscillator, on the other hand, can never be completely stationary, its ground-state energy level is above zero (aka zero-point energy), and its location is no longer exactly defined.

An atomic electron in its lowest state is likely to be nearest the nucleus and a quantum harmonic oscillator is likely to be near the equilibrium point. Now, the atom is a three-dimensional problem and r=0 is not the most probable radius - if one sums the probabilities over the surface of a sphere with the radius r - but understanding that is a simple matter of geometry.

4. Jun 1, 2009

### feynmann

That's the point of the question. Why the particle of <simple harmonic oscillator> has maximum probability of being at the center, while the electron of the atom has maximum probability of being at the Bohr's radius, but not at the center?

Last edited: Jun 1, 2009
5. Jun 1, 2009

### alxm

The probability is maximal at the center. An s-type orbital is (ignoring normalization constants) $$\psi = e^{-\frac{r}{a_0}}.$$. - the probability is $$|\psi|^2 = e^{-\frac{2r}{a_0}}.$$. Sum that over the surface of a sphere of radius r by multiplying by $$4 \pi r^2$$ and integrate over r and you get the radial probability distribution. Which has a maximum at $$r=a_0$$.

This isn't even a failure to understand QM, this is failing to understand geometry.

6. Jun 1, 2009

### feynmann

>>> The probability is maximal at the center.
Do you mean the <probability density> is maximal at the center, not the probability?

7. Jun 2, 2009

### alxm

Very well then, the probability density of an infinitesimal volume element has a maximum at r=0, whereas the probability density of an infinitesimal line segment at a given radius that is the sum of the probability densities at that radius, has a maximum at r=a0.

That is, if one assumes that you're using the same quantum mechanical formalism as everyone else, where $$|\psi|^2$$ is the probability distribution. You https://www.physicsforums.com/showpost.php?p=2189766&postcount=63" last time I said as much.

Last edited by a moderator: Apr 24, 2017
8. Jun 2, 2009

### ytuab

How about Helium?

In Helium atom, the orbital angular momentum of both the two electrons are zero.
The two electrons do enter the nucleus. OK.

But is it probable that the two electrons crush into each other?
When two electrons are close to each other, the potential energy will be beyond the ground state energy of Helium.
If two electrons avoid each other, this means they turn into a side not entering the nucleus.

So, this is contrary to the fact that the orbital angular momentum is zero?

9. Jun 2, 2009

### alxm

Helium has a doubly-occupied s-orbital.

Correct.

The probability of both electrons being at the same spot simultaneously is exactly zero, for two reasons. First, the Pauli exclusion principle prohibits it - they're fermions. Second, the potential energy would be infinite. That's termed the 'Fermi hole' and 'Coulomb hole', respectively.

True, but not relevant, since in QM, the electrons don't have exactly defined positions. Their actual energy is gotten from solving the Schrödinger equation, which essentially means taking into account all possible locations.

No, just that if one is in the nucleus the other cannot be at that point, at that moment. It's true that if one electron is (theoretically) held stationary, the other will most often be located directly opposite it on the other side of the nucleus. But the overall effect of this (correlation energy) is small, 0.042107 Hartrees, as opposed to the total electronic energy of -2.903724.

And the most probable location to find either electron is still at the nucleus. In fact, the helium wave function looks very much like the hydrogen one, only with the radius coordinate scaled by 0.5 (due to twice the nuclear charge).

Nope.

10. Jun 2, 2009

### ytuab

The probability of both electrons being at the same spot simultaneously is exactly zero. OK.

The hydrogen atom has only one electron, so it has magnetic moment.
The helium atom has spin-up and spin-down electrons, so it does not have magnetic moment.

If two electrons of Helium are apart from each other,
at the point which is closer to the spin-up electron than the spin-down electron,
upward magnetic moment will be occurred?

11. Jun 2, 2009

### alxm

It cancels out. They don't occupy any particular point in space at any particular moment (unless measured). Again, you must take into account all positions, in which case it's all spherically symmetrical and cancels out.

Also, the electrons don't have definite "spin-up or spin-down" 'identities' - unless they're measured.

12. Jun 2, 2009

### ytuab

I didn't say about "at any particular space or moment".
But it is fact that two electrons are apart as you said.

The average value was zero.
But as two electrons are apart, the magnetic (electronic) field are changing.
(If the field is not changing, it will not cancel out because the two electrons are apart.)
So the electromagnetic waves are emitted?

13. Jun 2, 2009

### alxm

No, the magnetic field is static and zero. Because the location-probability density is static. Even though the electrons 'move', it's not in the classical sense. Their location probabilities doesn't change with time. (as long as the system remains in the same energetic state) Another way of looking at it is to consider the electrons to be a static 'charge density cloud'. But that picture tends to obscure the fact they are actually 'moving', albeit not in the classical sense. (the energetic component of their correlated motion - a purely dynamical effect - is evidence of that)

That's just how the wave-particle 'duality' works.

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