Why do eigenvalues correspond to observable quantities?

Question

It makes sense to me that we can find some operator that gives us eigenfunctions that correspond to definite values for some desired observable. However, I do not see how the eigenvalues happen to give you the actual measurable values. I feel like there is something obvious I'm missing.

AccidentalTaylorExpansion · Accepted Answer

Suppose we don't know quantum mechanics yet and we want to calculate the expecatation value of an observable $A$. Could be momentum, spin whatever. It is given by
$$mathbb E(A)=sum_i a_i,p(a_i)$$
Where $a_i$ are the possible outcomes and $p(a_i)$ are the probabilities of those outcomes. When the outcome is continuous this becomes an integral.
To each of these states we can associate a vector $|a_irangle$ and it is possible to make these states orthonormal such that $langle a_i|a_jrangle=delta_{ij}$. Quantum mechanics is linear so if we have two solutions $|a_1rangle,|a_2rangle$ then the state $|psirangle=alpha|a_1rangle+beta|a_2rangle$ is also a valid solution. How do we interpret this new state? It is a postulate (Born rule) that the probability of finding $a_1$ is given by $p(a_1)=|alpha|^2$. This means we have to normalize $|psirangle$ such that $|alpha|^2+|beta|^2=1$ in order for it to be a valid state.
If we then define Dirac notation as usual we get $alpha=langle a_1|psirangle$ and $alpha^*=langle psi|a_1rangle$ which you can check using orthonormality. After some manipulation we can get the expectation value in the following form
begin{align}
mathbb E(A)&=|alpha|^2a_1+|beta|^2a_2
&=alpha^*alpha a_1+beta^*beta a_2
&=langle psi|a_1rangle langle a_1|psirangle a_1+langle psi|a_2rangle langle a_2|psirangle a_2
&=langle psi|left(sum_i |a_iranglelangle a_i|a_iright)|psirangle
end{align}
If we then define $hat A=sum_i |a_iranglelangle a_i|a_i$ then we get $mathbb E(A)=langle psi|hat A|psirangle$.
So what's the link with eigenvectors/eigenvalues? It turns out that according to the spectral theorem that any Hermitian matrix can be written as $hat A=sum_i |lambda_iranglelangle lambda_i|lambda_i$ where $lambda_i$ are its eigenvalues and $|lambda_irangle$ its eigenvectors. Notably these eigenvectors form an orthonormal basis. This implies that only the eigenvectors of $hat A$ can give the outcome of a measurement. This is because $|a_iranglelangle a_i|$ is a projection along $|a_irangle$. Any vectors that are orthogonal to $|a_irangle$ will be projected out. If a state is orthogonal to all eigenvectors of $hat A$, which means it can't be written as a sum of eigenvectors, then it will automatically give zero contribution in the expectation value because it is projected out.
As a final note I would like to add that my reasoning has been a bit backwards from how you would usually do it but I hope this made it more clear why this eigenvalue/eigenvector construction actually makes a lot of sense.

Dr jh · Answer

If a system is described by the eigenfunction $psi$ of an operator  $A$  then the value measured for the observable property corresponding to  $A$  will always be the eigenvalue  $a$ which can be calculated from the eigenvalue equation
$hat {A} | Psi rangle = a | Psi rangle tag 1$
This is a mathematical  description relating the eigenvalue and eigenfunction of quantum systems. Why this works in reality is probably a philosophical question, but it is true. The fact that this is true is just how nature works.
So, even though equation (1) is a mathematical formulation, it certainly works in reality.
In fact, quantum theory and its mathematical framework, is spectacularly successful at describing nature.

knzhou · Answer

I think the conceptual issue is that you've got it backwards. Given any set of eigenfunctions, you can have operators with those eigenfunctions, whose corresponding eigenvalues are anything you want. For example, if you're in a three-dimensional Hilbert space and the eigenvectors are the standard unit vectors, then
$$A = begin{pmatrix} 1 & 0 & 0  0 & 2 & 0  0 & 0 & 3 end{pmatrix}$$
is an operator that assigns the eigenvalues $1$, $2$, and $3$ to those eigenvectors. But
$$B = begin{pmatrix} 83 & 0 & 0  0 & - pi^e & 0  0 & 0 & 10^{100} end{pmatrix}$$
is an operator with the same eigenvectors but totally different eigenvalues.
Suppose you have a measuring apparatus that yields the values $x$, $y$, and $z$ on those three eigenvectors. Then we define the operator that corresponds to it to be
$$mathcal{O} = begin{pmatrix} x & 0 & 0  0 & y & 0  0 & 0 & z end{pmatrix}.$$
That is, the eigenvalues don't magically match the measured values. Instead, the measured values tell you what the eigenvalues have to be, and that defines what the operator is.

anna v · Answer

Look at the postulates of quantum mechanics, the second page in the link:

With every physical observable q there is associated an operator Q, which when operating upon the wavefunction associated with a definite value of that observable will yield that value times the wavefunction.

Postulates, laws, principles  are an encapsulation of physical observations, data,that have the place of extra (to the mathematical ones) axioms in physics theories. These axioms pick up those solutions of the mathematical differential equations that map the mathematics to data and observables, and so the theory can be used to predict new data and observables. When the predictions are validated , the theory is validated.
At the moment quantum mechanics theories are validated with all new experiments , in the range of variables for the three of the four fundamental forces. For gravity the research for a definitive theory is ongoing.

RobertSzili · Answer

Operators correspond to observables.
Eigenvalues are the observables' possible values.
And the why is:
Well as I learnt , these are axioms of QM. No need for proof, the whole theory is built on these.

Why do eigenvalues correspond to observable quantities?

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