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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 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.

Karolex
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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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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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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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