Executive summary: Energy isn’t necessarily quantized. Momentum isn’t either. Many things that are commonly considered “always quantized” are not.
There are extremely common misconceptions about this. Almost all introductory-level textbooks get it wrong. Encyclopedias and even dictionaries get it wrong. Einstein got it wrong. That doesn’t make it any less wrong.
The misconception can be traced back to the earliest days of quantum mechanics. Prof. Planck found his famous equation mostly by equation-hunting. Obviously he didn’t derive it, because at that time there weren’t any established principles from which it could have been derived. The equation agrees with experiment, and that should have been enough. As Galileo famously pointed out, physics needs to say what happens. The fundamental laws rarely say how it happens, and virtually never why it happens. Newton went to school on Galileo, and summed this up as Hypotheses non fingo.
On the other hand, it is nicer if you can explain the mechanism of how it happens. This has explanatory and pedagogical value. Planck pointed out that “quantized energy levels” was a possible mechanism that could explain the results. On the third hand, he wisely warned people that this was just an interpretation, and other interpretations were possible.
Alas, almost everybody fetishized the simple interpretation and ignored the warning. The right answer wasn’t clearly understood until 50 years later; see reference 1. Doing quantum mechanics requires you to count the basis states. You can enumerate them using the energy eigenstate basis or any other basis!
Here’s a well-known situation where non-quantized energy shows up experimentally: Fourier transform spectroscopy. A π pulse will flip the system from one quantized energy state to another, while a π/2 will flip it exactly half way. The halfway state is a perfectly valid state.
For more about state-counting and half-way states, see section 4.2.
Old-school optical spectroscopy can produce some very strong sharp lines, which might tempt you to believe in stable energy states, but I insist that’s an artifact of the way the experiment is set up.
There are at least three different basis sets up for discussion:
Those could easily be three different sets. In some cases they might happen to be the same, but they don’t have to be.
In other words: Energy states are not the only states. They’re not the only basis states. They’re not the only stable states.
Sometimes what passes for quote “the” quote “ground state” isn’t the lowest-energy state.
Y’all know about resonance stabilization, where an electron flips back and forth, so the lowest energy state is the gerade superposition of the two classical ball-and-stick model states. Benzene is the poster child for this. Each of the ball-and-stick structures has threefold rotational symmetry. The actual observed structure is the superposition of the two, and has sixfold rotational symmetry.
The carboxylate group is another poster child.
A particularly interesting example is ammonia. In this example it is not merely electrons that flip back and forth, but protons. Those have 2000 times more mass, so the tunneling rate is orders of magnitude smaller compared to elections. The energy gained by forming the superposition is corresponding less. This small energy is exploited in the ammonia maser, which is famously and brilliantly discussed in reference 2
So far, so good. So far, conventional quantum ideas get the right answer. In all the cases discussed in this subsection, the ground state is a state of definite parity.
We are in for a surprise when we consider molecules heavier than ammonia. The superposition that occurs in ammonia occurs in almost no other molecules.
As a specific example, consider lactic acid. The carbon highlighted in magenta in figure 3 is chiral; it has four different ligands. By the same logic as before, if the ligands could exchange locations by tunneling, the lowest energy state would be the gerade superposition of the two ball-and-stick structures ... but this is not observed. You get either the stable L form or the stable R form or a classical mixture of the two, even though the quantum superposition provably has lower energy.
The conventional rule of thumb is: electrons resonate but nuclei do not. This rule works for everything except a few things like ammonia, where the nuclei have not much mass and not far to travel.
The explanation goes like this: The classical ball-and-stick models are stabilized by the environment that the molecule finds itself in, namely the electromagnetic field. The electromagetic reaction is weak, but it enough to quench the tunneling process. The EM field couples to dipole moment, so what we are observe are states of definite dipole moment (rather than states of definite parity).
The moral of the story is, sometimes seemingly-minor attributes of the environment determine which quantum states will be stable.
When you buy electricity from the power company, you do not get states of definite photon number. That is, you do not get energy eigenstates. Instead you get something else entirely, something closer to coherent states of definite voltage. For the next level of detail, see reference 3.
You get this because of how the power-plant machinery is designed.
You can measure a sinusoidal voltage using an AC voltmeter, and it will tell you that what you have is indeed an AC voltage eigenstate, very nearly. This is determined by the physics of the measuring device.
As always, you can re-express the state in any basis you like, but if you choose a basis that is mismatched to the physical situation, the state (as provided to you by the power company) will be a superposition of a huge number of basis states, which will not be very convenient. In particular, if you measure the power line using a photon counter, you will get some wildly variable number of photons. This stands in contrast to the much more predictable AC voltage reading.
As shown in figure 4 and discussed in reference 3, there exist coherent states with almost definite voltage. They are perfectly fine states, but they are not definite energy states. Really not.
Recall that we said there are at least three different basis sets up for discussion:
Those could easily be three different sets. In some cases they might happen to be the same, but they don’t have to be.
The physical situation determines what’s quantized and what’s not. This is more physical than mathematical. Typically the quantized states correspond to the eigenstates of the measurement setup. The eigenstates of any operator are mutually orthogonal (or can be arranged to be), so they form a basis set, but that is of dubious relevance, because you don’t have to choose that basis. The quantized states (a) are quantized by the physical situation, no matter what basis (b) you choose to describe the situation. One or both of those might happen to be energy states (c), but they don’t have to be.
You need to be able to enumerate the states in order to calculate the partition function. Since this is a sum over all states, i.e. a trace, it doesn’t matter which basis you use. This is mathematically guaranteed. Some choices will be more convenient than others, but any choice will work. More generally this is true of any vector space: The choice of basis doesn’t affect the results of the calculation.
You can formalize this as follows: The trace is independent of basis (which is an easy corollary of the cyclic property of the trace). The trace of the density matrix gives you the partition function, which gives you all of stat mech.
Remember, Z stands for Zustandssumme, i.e. the sum over states. They don’t have to be energy states.
Here’s a simple example. In atomic physics, sometimes we use one of the following bases:
There are five different states mentioned in equation 1, and they are all perfectly fine states, but there are only three states in each basis. The Zustandssumme will consist of three terms, no matter which basis you choose.
In the presence of a magnetic field, in simple cases, the states of definite angular momentum (equation 1b) will be the quantized-energy states. The x state is halfway in between two of the energy states, and similarly for the y state:
| (2) |
That factor of −i is ugly, but it is needed for consistency with the inverse mapping:
| (3) |
In the presence of a magnetic field, the |p_{x}⟩ and |p_{y}⟩ states are not stationary states, but they are still perfectly fine states. It is routine to observe them experimentally. They can be created using a π/2 tipping pulse.
By way of analogy: In high-school physics, vertical motion is physically different from horizontal motion, because of gravity. That’s physics, not mathematics. You can describe the motion using any vector basis you like, aligned or not with the horizontal and vertical directions. This is mathematically guaranteed. For motion on an inclined ramp, an inclined basis may in fact be convenient. You may even choose to switch back and forth between the horizontal/vertical basis and the inclined basis.
The same goes for quantum mechanics: You can calculate using any basis (or bases) of your choosing, and the results of the calculation will be the same. Sometimes this-or-that basis will be more convenient, but that’s not an issue of principle.
Bottom line: Do not imagine that energy has to be quantized. The reality is that QM requires being able to count the basis states. They could be energy states, but they don’t have to be.