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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.

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  • $\begingroup$ What alternative is there? What other number is associated with an eigenfunction of an operator? $\endgroup$ – G. Smith Oct 15 '20 at 3:56
  • $\begingroup$ @G.Smith Couldn't it just kind of turn out being whatever? It seems tough enough to find an operator that produces the correct basis, let alone one that does so with eigenvalues equal to the possible observed quantities. $\endgroup$ – Jeff Bass Oct 15 '20 at 3:57
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    $\begingroup$ Couldn't it just kind of turn out being whatever? I suppose it could have, but it didn’t. You seem to be missing the point that quantum mechanics works. And we know the operators for observable quantities. Asking why it works is more metaphysics than physics. I have no idea why it works. $\endgroup$ – G. Smith Oct 15 '20 at 3:59
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    $\begingroup$ @G.Smith But that just makes it sound like magic. It's like saying "we don't question why the all-powerful formula works, we just know it does". $\endgroup$ – Jeff Bass Oct 15 '20 at 4:03
  • $\begingroup$ Theoretical physics is about constructing mathematical models that predict how things work. This operator-eigenfunction-eigenvalue model works, and works vastly better than Newtonian mechanics for systems where QM is relevant. $\endgroup$ – G. Smith Oct 15 '20 at 4:04
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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 $|a_i\rangle$ and it is possible to make these states orthonormal such that $\langle a_i|a_j\rangle=\delta_{ij}$. Quantum mechanics is linear so if we have two solutions $|a_1\rangle,|a_2\rangle$ then the state $|\psi\rangle=\alpha|a_1\rangle+\beta|a_2\rangle$ 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 $|\psi\rangle$ 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 $\alpha=\langle a_1|\psi\rangle$ and $\alpha^*=\langle \psi|a_1\rangle$ 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_1\rangle \langle a_1|\psi\rangle a_1+\langle \psi|a_2\rangle \langle a_2|\psi\rangle a_2\\ &=\langle \psi|\left(\sum_i |a_i\rangle\langle a_i|a_i\right)|\psi\rangle \end{align} If we then define $\hat A=\sum_i |a_i\rangle\langle a_i|a_i$ then we get $\mathbb E(A)=\langle \psi|\hat A|\psi\rangle$.

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 $\hat A=\sum_i |\lambda_i\rangle\langle \lambda_i|\lambda_i$ where $\lambda_i$ are its eigenvalues and $|\lambda_i\rangle$ 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 $|a_i\rangle\langle a_i|$ is a projection along $|a_i\rangle$. Any vectors that are orthogonal to $|a_i\rangle$ 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.

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  • $\begingroup$ It seems to make so much more sense this direction, so I don't get why it isn't taught this way. $\endgroup$ – Jeff Bass Oct 15 '20 at 14:01
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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.

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    $\begingroup$ Ahhh it's definitely coming together. My only confusion is that it is obvious how you make this happen when your operator is a matrix, but less obvious when it's something like the momentum operator. $\endgroup$ – Jeff Bass Oct 15 '20 at 4:31
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    $\begingroup$ @JeffBass Setting $\hbar = 1$, the eigenvalues of $- i \partial_x$ are the momenta, but the eigenvalues of $(- i \partial_x)^2$ are the momenta squared, and the eigenvalues of $- 2 i \partial_x$ are twice the momenta, and so on. So the answer to "why $- i \partial_x$?" could just be that it's the one with the right eigenvalues. $\endgroup$ – knzhou Oct 15 '20 at 4:37
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    $\begingroup$ @JeffBass But then you could ask, why does the operator with the right eigenvalues for momentum have such a nice and simple form? I don't think there is a good answer for that. That is the fundamental assumption of wave quantum mechanics. It's a new thing that can't be derived from anything you knew before. Every theory has to start somewhere, and this is the starting point: the momentum corresponds to phase rotation in space, and energy corresponds to phase rotation in time. $\endgroup$ – knzhou Oct 15 '20 at 4:38
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    $\begingroup$ @JohnnyLongsom I don't see the issue with that argument? The momentum operator is the operator whose eigenfunctions are plane waves $\exp(-ikx)$ and whose eigenvalues are proportional to $k$. That seems physical to me. $\endgroup$ – jacob1729 Oct 15 '20 at 10:08
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    $\begingroup$ @knzhou The form of the momentum operator follows directly from the fact that momentum is the generator of spatial translations. This is nothing special for quantum mechanics but works exactly the same in classical mechanics $\endgroup$ – Olof Oct 16 '20 at 10:07
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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.

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

  1. 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.

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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.

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    $\begingroup$ Why is the theory built this way? $\endgroup$ – user253751 Oct 15 '20 at 14:45

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