5

Imagine an electron in the hourglass-shaped $2p_x$-orbital. It has two lobes. The probability of it being in the centre of the orbital is zero. This is the point that connects the two lobes of the orbital.

Now if I make an observation I find the electron in the right lobe of the orbital. I make another observation and still it's in the right lobe. Now when I make the third observation the electron is in the left lobe. How did it go into the left lobe without passing through the centre of the orbital. The probability of it being at the centre is zero!

Qmechanic
  • 201,751
Black Dagger
  • 1,283

2 Answers2

3

The orbital is a probability density distribution, the square of the wavefunction characterizing the electron bound in an atom.

The "density" part of the probabilistic definition of quantum mechanical solutions is important, because it means a dx is multiplying the probability density to get a probability around the point x+dx. Thus not only zero, but any point in the function of the orbital will give zero probability if not multiplied by a finite dx.

The electron is not on a path, as the moon around the earth. Only the probability density can describe its location, not a motion within the atom.It is in a quantum mechanical state.

The probability density distribution is what can be measured, and that can be done having a large number of atoms with electrons in the same orbital. To see the lobe structure of the p orbital density distribution one would in addition have to orient all the atoms in the sample.

So at measurement of the probability distribution, an electron is found in the left lobe for one atom, and the right lobe for another atom etc. Intervening twice in the same atom is not possible because measurements change the boundary conditions and hence the solution of the equations.

anna v
  • 233,453
  • I believe the question was about measuring the same atom trice. I mean it is probably not possible in practice, but as certainly possible as an useful thought experiment. It is a nice way to understand wave packet dynamics for example. If the electron is measured to be at right, so it must be, and unitary quantun dynamics will follow from that boundary condition. – Mikael Kuisma Jul 31 '16 at 20:21
1

If you make a measurement, the electron will no longer be in the px orbital. Alternatively, if you build a localized wave function as a linear combination of ground state orbitals which is at the right, it will consist of more than just that px orbital. This will correspond to the situation after the measurement. After that, due to different oscillation phases of these orbitals, the quantum evolution will move it to the left hand side (and back and forth, untill the phases completely become incoherent).

As the electron undergoes ordinary quantum dynamics, as dictated by the Scrödinger equation, there will be currents and change of the density, but these will fullfill the continuity equation (since Scrödinger equation fullfills it). Hence, no sudden jumping.

So about going through center of the atom: In order to have a localized wave packet there will be higher excited states involved and evolving, and these (along with the original px orbital) carry the current through the center.

Mikael Kuisma
  • 1,652