How can energy be discrete when momentum and position are not?

Question

In quantum mechanics, it's said that the total energy of a system can only take certain discrete values. That is, the set of all possible energies of a system can be indexed by the natural numbers and is countably infinite. However, $E=T+U$, where $T$ is the kinetic energy and is a function of momentum, and $U$ is the potential energy and is a function of position. The problem is that the domain of $T$ and $U$ is the set of all real numbers (or arguably only positive reals, but the cardinality is equivalent).

More formally, think of E as the cartesian product of T and U, where each ordered pair (t,u) in E represents the quantity t+u (the multiplicity of different ordered pairs representing the same sum shouldn't make a difference). The cartesian product of two countable sets is countable, but the cartesian product of one uncountable set and one countable set, or of two uncountable sets, is uncountable. So for E to be countable, and therefore discrete, both T and U most be countable. Analyzing U can be complicated since it can take several different forms, but T has only one form, that is: T=(p^2)/2m. Notice that T is strictly monotonic (strictly increasing for increasing input p). This implies that there is a 1-1 correspondence between the set of allowed p and the set of allowed t, and therefore the cardinalities of these sets is equal. However, p can take any real value and is uncountable. Therefore, T must also be uncountable which implies that E is uncountable and therefore not discrete.

I'm rather confident that the mathematical logic I use here is correct, so my mistake must be a misinterpretation of the underlying physics. Can someone please explain what exactly this misinterpretation is and where I went wrong?

Answer 1

The actual mathematical way of showing that the spectrum of Hamiltonians of the form $H = p^2 + V(x)$ can be discrete under certain conditions is discussed elsewhere, e.g. in this question (h/t Qmechanic).

But the flaw in your argument is an elementary flaw we should address separately because it is fundamental to understanding how the nature of observables as operators means our classical intuition fails.

More formally, think of E as the cartesian product of T and U, where each ordered pair (t,u) in E represents the quantity t+u

This is not how it works. Since $x$ and $p$ do not commute - the canonical commutation relation is $[x,p] = \mathrm{i}$, after all - $T$, as a function of $p$, and $V$, as a function of $x$, also do not commute. This means there is no joint eigenbasis where the idea that the eigenvalues of $H = T+U$ are sums of eigenvalues of $T$ and $U$ would hold. Emilio Pisanty gives an elementary two-dimensional counterexample to this idea here.

Answer 2

In quantum mechanics, physical quantities (called observables) are not functions or set, so the reasoning used here is incorrect. Instead, observables are represented by hermitian operators, while the allowed values of those observables are encoded in the spectrum of the operator (roughly, their eigenvalues). The key point is that operators need not commute with each other : this is what invalidates the classical intuition underlying OP's reasoning. The eigenvalues of the sum of two operators which commute are the sum of eigenvalues of these operators, but this fails in general.

In the case of a particle living in a $1$-dimensional space, the canonical commutation relation imposes that the position and momentum observable do not commute : $$xp-px=i\hbar$$ In turn, this implies that : $$TU -UT = \frac{-i\hbar}{2m}(pU'(x)+U'(x)p) \neq 0$$

To see why some operators have a discrete spectrum while others have a continuous spectrum, one needs to dig deeper into the math (namely the functional analysis). This has been discussed in other posts like :

• Discreteness of set of energy eigenvalues,
• Must bounded operators have normalisable eigenfunctions and discrete eigenvalues?
• What does it mean for an operator to have both a continuous and discrete spectrum?
• How to prove bound state's spectrum must be discrete and scattering state's spectrum must be continuous?

Answer 3

Other answers say that the kinetic and potential energy operators don't commute in general, which is true, but I think it's not really the issue here. In the case of an infinite square well, they do commute and you can write $E=T+U$ if you like, but the energy levels are still discrete.

I think your argument fails already at the classical level, if you allow for the existence of systems with a fixed total energy. E.g., look at a particle bouncing elastically between the walls of a box. It was set in motion with a kinetic energy $E$ and has that energy at all times. If you measure the momentum you'll get either $\sqrt{2mE}$ or $-\sqrt{2mE}$, violating your assumption that $p$ can take any real value. Just like energy, $p$ can have any value in general, but it may not in specific systems. For another example, look at a classical particle in a harmonic oscillator potential, again with a fixed total energy $E$. If you measure $T$ or $U$ you may get any value in the interval $[0,E]$ of cardinality $2^{\aleph_0}$, but the number of possible values of $T+U$ is not $(2^{\aleph_0})^2$ but just $1$.

Answer 4

Your argument is a bit of a tangle. Let me present something on the conceptual side that might help:

Coordinates and momenta come in conjugate pairs. Position and momentum are a pair, time and energy are a pair, and angle and angular momentum are a pair. There are others pairs. (The Lorentz Transformation and something often called the 'boosts' are a pair).

If there is symmetry under the coordinate, then the momentum is conserved (Noether's theorem). There is also an uncertainty relationship between these conjugate pairs.

If the angle is completely uncertain (like in an orbit) then it's conjugate momentum, the angular momentum, is completely certain.

Also, when there are periodic boundary conditions on a coordinate (meaning that the state returns to it's original state after a fixed distance along that coordinate) then the conjugate momentum is quantized. We also see this effect when an electron is moving through a crystal lattice. The periodicity of what the electron 'feel' as it travels through the crystal gives it discrete momentum states.

It's a little tricky to imagine what happens with time and energy. When there are periodic boundary conditions on the time, we get quantized energy states.