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Given that the atomic orbitals are fuzzy, why are the energy levels and energy transitions sharp?

Qmechanic
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Brian B
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1 Answers1

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They are "fuzzy" in position space because they have well defined energies (and therefore give rise to sharp distributions of photon energy when they change states).

Foundation

Quantum states (the class to which orbitals belong) are elements in a Hilbert Space (which physicist sometime call "infinite dimensional vector spaces" just to watch the mathematicians cringe ...), and like a position in physical space can be expressed in terms of differing sets of coordinates depending on how you define your coordinate system, quantum states have different "coordinates" depending on the coordinate system you chose.

The big difference between quantum mechanics and analytic geometry, however, is that physical observables (things you can measure about the system) are represented by different coordinate systems in the Hilbert space, and for each such coordinate system, each coordinate direction is associated with a particular measured value of that observable (this is why the space must be infinite dimensional: some measurements have an unbounded number of outcomes).

Furthermore, the odds of getting that value of the measurement are equal to the square of the component1 in the direction of that axis for the current condition of the states. Consequently, a system whose vector lies along one of the coordinate direction for an observable will only ever give you one possible value for the measurements. This state of the system is called an eigenstate of that observable, and the value of the measurement is the associated eigenvalue.

If the system is not an eigenstate for an observable, then it may return more than one possible value of the measurement.

Lastly, a measurement leaves the system in the eigenstate corresponding to the eigenvalue that was measured.

Application to atomic orbitals

The orbitals are the eigenstates of the Hamiltonian operator (which corresponds to measurements of energy. That means that each and every orbital corresponds to a single well defined energy for the system. (Well, almost. See Emilio Pisanty's answer to a related question.)

But measurement of the electron position corresponds to a different set of axis in the Hilbert space for the state of the atom. It can be shown that in that set of coordinates, the energy eigenstates (i.e. the orbitals) are not along a single coordinate axis; In fact they point in a direction that has a non-zero component is the direction of almost every possible outcome of measuring the electrons position.

What the pictures of orbitals show is some visualization of the mostly likely possible outcomes of a position measurement starting from the assumption that the system was in an energy-eigenstate.

That means that the well-definedness of the energy is an assumption that goes into the calculation of what positions to expect, and the fuzziness of position is a consequence of those two sets of coordinate-axis (the one for energies and the one for positions) not being lined up with one-another.

Follow-up

We really should go one step farther. The evolution of quantum systems in time is governed by the Hamiltonian (the operator for energy measurements), and so the kinds of states that don't change in time are eigenstates of that operator. Excepting the ground state the orbitals are not exactly changeless (they decay to lower states releasing light, right?) so they are not exactly energy eigenstates. That's what Emilio is talking about in that link. But even excited orbitals last "a long time" on the proper quantum scale of energy (given by $(\hbar c)/\text{energy of atomic transitions}$), so they are very close to being eigenstates of the energy operator.

1 A wrinkle here is that the coordinates are complex and "square" means "the component times the-complex-conjugate-of-the-component". But I'm trying to avoid actually writing down the math.

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dmckee --- ex-moderator kitten
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    Wow, that is a remarkably sharp answer to a particularly fuzzy question. – Emilio Pisanty Aug 10 '16 at 22:42
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    Great answer, but something confuses me: as a layman I thought we had a time-energy uncertainty principle, not a position-energy uncertainty principle? – user541686 Aug 11 '16 at 07:43
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    It's position - momentum uncertainty principle; in fact, for all canonically conjugated variables you have an uncertainty principle. The operators corresponding to these variables commute, and from that relationship, one can obtain the uncertainty principle. Energy - time pair is more interesting because time is not an operator. – user3653831 Aug 11 '16 at 10:15
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    @Mehrdad The Hamiltonian (energy) operator is $p^2/2m + V$, so it's not a long proof to start from the canonical commuter of positions and momentum and arrive at the conclusion that energy also fails to commute with position. – dmckee --- ex-moderator kitten Aug 11 '16 at 13:02