In classical mechanics, physical quantities, such as, e.g. the coordinates of position, velocity, momentum, energy, etc, are real numbers, but in quantum mechanics they become operators. Why is this so?

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    $\begingroup$ Related: physics.stackexchange.com/q/46015/2451 $\endgroup$
    – Qmechanic
    Commented Sep 14, 2013 at 16:52
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    $\begingroup$ A major point is that observables in quantum mechanics do not always commute. So you can at least determine relatively quickly that real numbers (and complex numbers) won't be sufficient. $\endgroup$ Commented Sep 15, 2013 at 5:33

3 Answers 3


The short answer is that you can treat classical observables as operators if you want, but the algebra of observables that actually described the world turns out to be non-commutative. Or in other words, to the question:

  • If the state of a physical system is completely determined in terms of definite outcomes of some observable(s), are there still measurements that can be made that are not definitively determined?

Quantum mechanics answers 'yes', and that's just what we find in the world, e.g., for a simple spin-$0$ particle a definite position completely determines the state (position eigenstate), but does not result in a definite outcome for a momentum measurement. Etc.

In classical mechanics in the Hamiltonian formulation, observables are not numbers, but functions (or distributions) over the phase space. In terms of the canonical coordinates of position and momentum $(q;p)$, some of the usual suspects have particularly simple forms, e.g., kinetic energy of a particle $T(q;p) = p^2/2m$. The time-evolution of a classical observable is given by Hamilton's equation $$\frac{\mathrm{d}f}{\mathrm{d}t} = \frac{\partial f}{\partial t} + \{f,\mathcal{H}\}\text{,}$$ where $\{\cdot,\cdot\}$ is the Poisson bracket and $\mathcal{H}$ is the Hamiltonian.

The product or sum of any two observables is an observable, and mathematically, the Poisson bracket is both a Lie bracket and a derivation. The details of this you can look up on your own, but the immediately relevant effect is that the classical observables form a commutative Poisson algebra, a subtype of a Lie algebra.

Once one gets used to thinking of observables as forming an algebra, some questions naturally appear:

  • Does our algebra actually get all the physical observables, does everything we can measure correspond to something in the algebra of classical observables? If not, do physical observables require loosening the rules of this algebra somewhat, e.g., making it non-commutative?

It turns out that the answer is that the algebra required to describe the world is indeed non-commutative, although of course it wasn't historically discovered by this analogy.

As for why linear operators in particular, it's because it works--though the reason we could have expected it to work is that a fairly general class of algebras can be represented as a linear operators on a complex Hilbert space. That's the essence of the Gel'fand-Naĭmark theorem regarding $C^\star$-algebras.

As an interesting curiosity, classical mechanics can be formulated using wavefunctions/kets on a complex Hilbert space, with physical observables represented by operators and measurement being probabilistic, just as in quantum mechanics. It's not even all that strange, turning into a complexified reformulation of the classical Liouville equation (which handles probability distributions over the phase space).

The common formalism of QM is quite general and is completely capable of handling classical mechanics; where they disagree is on which operators represent things we actually measure in the world.


In classical mechanics, physical quantities, such as, e.g. the coordinates of position, velocity, momentum, energy, etc, are real numbers, but in quantum mechanics they become operators. Why is this so?

Discrepancies from classical theories with experimental results necessitated the formulation of a new theory. The black body radiation required a model of harmonic oscillators which could emit in units of h*nu in order to give finite results and not go into the ultraviolet catastrophe mode. The spectra of the newly discovered atoms showed discrete energy levels necessitating the Bohr model . The data could not fit into the classical mold. The energy levels suggested that they obeyed equations which would give standing waves, similar to the classical string with its standing waves. Read up on the historical background leading to equations that described the quantum nature observed.

The theory was slowly built up, it was found that complex numbers were necessary for the solutions of the equations and the assumption of the wave function squared being a probability distribution was found to explain the data. Then the analogy of the forms of the differential equations to the Hamiltonian led to the identification of operators instead of simple variables for the expected average values of the variables. It was a process by trial and error and then it took off.


That is because in quantum mechanics the mathematical language is linear algebra. Operators used in quantum mechanics can decide the eigenvalues, eigenstates and also make transformation on the quantum states and so on.

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    $\begingroup$ This doesn't answer the question. The question amounts to why the language is linear algebra. $\endgroup$
    – user4552
    Commented Oct 15, 2013 at 19:17

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