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After looking around in the internet, I could not find a sufficient proof how every operator in QM has to be linear. Many sources claim that the linearity of the Schrödinger equation implies that, however I was not able to find a proof for this. I even asked ChatGTP for help but he only provided me with a Zirkelschluss, which wasn't very helpful.

I initially assumed that it is related to Sturm-Liouville Differential equations, but then I got even more confused, as to why the scalar product in the Hilbert-space of state-vectors (always?) has a weight-function of $\rho = 1$. I am curious to know, whether that is always the case.

I would be very thankful for any guidance in proving this.

Qmechanic
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Susp1cious
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    "Proving this" starting from which assumptions? – ACuriousMind May 11 '23 at 19:30
  • @ACuriousMind Proving it from the axioms of Quantummechanics. Perhaps I overlooked the fact that the linearity of operators is an axiom by itself. However, I am only aware of the linearity of the Schrödinger equation. – Susp1cious May 11 '23 at 19:32
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    It is indeed part of the axioms of the theory: https://en.wikipedia.org/wiki/Dirac%E2%80%93von_Neumann_axioms – d_b May 11 '23 at 19:35
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    What axioms do you mean? There are various different axiomatizations, but I don't see any common ones where the linearity is not more or less immediate. – ACuriousMind May 11 '23 at 19:37
  • @ACuriousMind My bad. I overlooked the fact that self-adjointness already implies linearity. I thought that self-adjointness is more general. – Susp1cious May 11 '23 at 19:41
  • Related: https://physics.stackexchange.com/q/134503/2451 , https://physics.stackexchange.com/q/33344/2451 , https://physics.stackexchange.com/q/1201/2451 and links therein. – Qmechanic May 11 '23 at 22:54
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    "I even asked ChatGTP for help..." Why? Please stop. – hft May 12 '23 at 02:06
  • Because all the experiments in a quantum mechanical ensemble are completely independent. The linearity of QM is an artifact of the structure of the theory. It has nothing to do with actual linearity/non-linearity in nature. – FlatterMann May 12 '23 at 02:36
  • See references in Why quantum dynamics is linear and Are operators in quantum mechanics linear transformations? – Quillo May 12 '23 at 07:12
  • @hft I know that the usage of ChatGTP for answering physics related questions is generally frowned upon and I have observed several mistakes it did myself. However, I still believe that it can be very useful, especially in finding links between different branches of physics and mathematics. I also make sure to read through its proofs and derivations thouroughly and to always take his "wisdom" with a grain of salt. – Susp1cious May 16 '23 at 13:57

3 Answers3

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Quote from wiki on QM operators:

The operators must yield real eigenvalues, since they are values which may come up as the result of the experiment. Mathematically this means the operators must be Hermitian.

This means that for operator $A$ and it's Hermitian adjoint operator $A^\star$ must be satisfied :

$$\langle Ax,y\rangle =\langle x,A^{*}y\rangle , \tag 1$$

where $\langle \cdot ,\cdot \rangle$ is inner product on vector space. (1) condition is satisfied if $A$ operator is linear.

Agnius Vasiliauskas
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  • More on "Hermitian" in QM: https://physics.stackexchange.com/q/54154/226902 – Quillo May 12 '23 at 07:09
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    Thank you very much for your help! I was able to prove the linearity of $A$ if it is hermitian: $$\langle x|A(y+z) \rangle = \langle Ax|y+z \rangle = \langle Ax|y \rangle + \langle Ax|z \rangle = \langle x|Ay \rangle + \langle x|Az \rangle = \langle x|Ay + Az \rangle \qquad \square$$ – Susp1cious May 16 '23 at 14:17
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You need to be more precise about which operators you're referring to.

The fact that observables correspond to Hermitian (and therefore linear) operators is generally taken as a fundamental postulate of QM, which cannot be derived from everything else.

The fact that the time-evolution operator for a closed quantum system is unitary (and therefore linear) follows directly from the linearity of the Schrodinger equation.

tparker
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  • The "postulate" derives trivially from the fact that all copies of an experiment in a quantum mechanical ensemble are completely independent. – FlatterMann May 12 '23 at 02:37
  • @FlatterMann could you flesh that out a bit? It is not an argument I have encountered. Why do we use linear operators in quantum mechanics and not when describing classical systems of independent events? – By Symmetry May 12 '23 at 10:27
  • @BySymmetry The rule of addition for probabilities is P(A∪B) = P(A) + P(B) – P(A∩B). If the events are completely independent, then the third term is zero and the probabilities are behaving in a linear fashion. The construction of a quantum mechanical ensemble assumes that all the repetitions of the experiment are completely uncorrelated, i.e. it is as if they are happening in independent copies of the universe. This is, of course, an approximation, but it is one that is supported by observations. – FlatterMann May 12 '23 at 14:16
  • @FlatterMann I don't immediately see the link between the probability function being "linear" over separate runs of the experiment and the linearity of quantum operators over a superposition of states. Hermitian operators do not correspond to probabilities and a system being in superposition should determine probabilities for outcomes of a single run of the experiment (to the extent that such a thing is well defined) – By Symmetry May 12 '23 at 14:44
  • @BySymmetry Operators are always acting on a description of the quantum mechanical ensemble. Nothing in the formalism of quantum mechanics describes a single system, not after we use the Born Rule and not before. Wave functions are already ensemble descriptions. A "single run" does not have a probability (or a wave function) any more than a single throw of dice has a probability. Let's say I tell you the outcome was "4". How many sides do my dice have? Are they fair? These kinds of "single run" questions don't even make sense in classical systems with probabilities. – FlatterMann May 12 '23 at 14:56
  • @FlatterMann Ok, but if we can't talk about single run probabilities, what does your statement "repetitions of the experiment are completely uncorrelated" mean and what are $P(A)$ and $P(B)$ in your equation above? – By Symmetry May 12 '23 at 15:45
  • @BySymmetry We apply them to sub-ensembles, just like in classical probability theory. The "probability of an event" is a mathematical abstraction of the law of large numbers in the frequentist model. To get a frequentist estimate we always have to perform a "large enough" number of experiments. A single experimental outcome tells us nothing useful about either a probabilistic or a quantum system. It is fairly easy to mistake "probability of an event" with "probability of an individual outcome", though. The former is well defined for ensembles, the latter does not make much sense, IMHO. – FlatterMann May 12 '23 at 15:54
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Why are operators in quantum mechanics always linear?

The straightforward (but perhaps disappointing) answer is that linearity is specified in the postulates/axioms of quantum mechanics.

So if you are doing something that you call "quantum mechanics" then you have already admitted that you are working with linear Hermitian operators that represent observables.

Historically, Bohr worked out the "old quantum theory," which described atoms pretty well and then Schrodinger worked out the standard formalism that we teach nowadays. Effectively, Schrodinger introduced an equation like: $$ i\frac{\partial \Psi(\vec x, t)}{\partial t} = \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial \vec x^2}\Psi(\vec x, t) + V(x)\Psi(\vec x,t)\;, \tag{A}$$
which worked really well to describe atomic-scale phenomena.

Why did he want to write such an equation in the first place? Probably because he knew that solving such an equation could result in discrete eigenvalues, and could thus explain the discrete spectra of atoms.

Note that Eq. (A) is linear. So, a linear equation for the wavefunction evolution worked really well. Linear equations for the wave evolution continued to work really well for describing other phenomena. So, why fix what ain't broke?

If you want to describe more than one particle, you again have to solve a linear equation: $$ i\frac{\partial \Psi(\vec x_1,\ldots,\vec x_N t)}{\partial t} = \left(\sum_{i=1}^N\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial \vec x_i^2}\right)\Psi(\vec x_1,\ldots,\vec x_N t) + V(\vec x_1,\ldots,\vec x_N)\Psi(\vec x_1,\ldots,\vec x_N, t)\;, \tag{B}$$

Also note that although Eq. (A) and Eq. (B) are both linear that doesn't mean that we only care about linear equations. For example, if you try to solve an N-particle Schrodinger equation (where N>1) like Eq. (B), you usually make a bunch of approximations and reduce it down to solving a non-linear single particle equation (e.g., via density functional theory or whatever).

hft
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    One more time: the linearity follows, like the Kolmogorov axioms of probability theory, from the fact that all elements of a quantum mechanical ensemble are completely independent. This is the least of the "mysteries" of quantum mechanics. What is not clear to me is why this is not being taught in the first ten minutes of every QM 101 class in the world. – FlatterMann May 12 '23 at 02:40
  • "One more time..." What do you mean "one more time"? – hft May 12 '23 at 13:54