Why is the measured value of some observable $A$, always an eigenvalue of the corresponding operator?

Question

Explain why when we make a measurement of some observable $A$ in QM, the measured value is always an eigenvalue of the operator $A$.

Answer 1

You're asking what could be a very deep question. But let me take a not-so-deep approach to answering that gets at the nature of science.

It is experimental fact that we only measure particular values of quantities at the quantum level. That's it. Nature doesn't care about eigenvalues or operators; only humans do. That being said, physicists have, based on such observations, developed a model that can predict outcomes of experiments. Part of this model involves operators and eigenvalues because it describes nature and can make predictions.

Answer 2

Mathematics is a very powerful discipline. One can have a very large number of theories, starting from algebra, going to topological theories, and they are based on an internal very strict self-consistency. They start by axioms, develop functional dependences and the whole mathematical construct is self-contained. That is why one has proofs in mathematical propositions, propositions can be proven to be either true or false, and there can be no argument about it because mathematics is a beautiful self-consistent discipline.

Physics from about the time of Newton and on, uses mathematics as a modeling tool. Observations and experimental data that are accumulated are fitted mathematically, for example the fall of projectiles by a parabola. In the best of cases one has a theory, classical mechanics for example, that is described by a complete mathematical model. In order for the mathematical model to be predictive and relevant to data, one has to impose extra types of axioms, called postulates now and "laws" before, as Newton's three laws for example. It is how the beautiful mathematical models can become relevant to the measurements and the observations in the physical world. This means that in contrast to mathematics, a physics theory cannot be proven correct, it can only be validated or falsified by wrong predictions. If an inconsistency in the predictions comes up, the problem will not be with the mathematical framework, but with the postulates which will have to be changed. Example going from classical mechanics to relativistic mechanics, because there were data that were inconsistent with classical mechanics.

In a similar way, quantum mechanics became necessary when observations reached the nanometer and smaller length scales, and the physics theory depends on a strictly constructed mathematical model ( various versions of mathematics exist) and postulates that relate the mathematical numbers to observations.

Here is a list of the postulates for quantum mechanics as mathematically expressed by the Schrodinger equation.

1. Associated with any particle moving in a conservative field of force is a wave function which determines everything that can be known about the system.
2. 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.
3. Any operator Q associated with a physically measurable property q will be Hermitian.
4. The set of eigenfunctions of operator Q will form a complete set of linearly independent functions.
5. For a system described by a given wavefunction, the expectation value of any property q can be found by performing the expectation value integral with respect to that wavefunction.
6. The time evolution of the wavefunction is given by the time dependent Schrödinger equation.

You state:

Explain why when we make a measurement of some observable A in QM, the measured value is always an eigenvalue of the operator A.

It is because of the postulates 2) and 5). These are the postulates that allow a physics interpretation of the mathematical relationships. So the answer to your "why" is "because" these postulates are necessary so that the mathematics of quantum mechanics can describe data in one to one correspondence, and predict new behaviors. Ultimately, these postulates are necessary in order to model the behavior of matter in the microcosm.

Answer 3

I'm a beginner in quantum probability and I have no background in physics, only in mathematics.

I think it is good to know the following point, but I do not pretend it is a good answer to the question.

Your question is about the postulate:

When the quantum system is in state $\psi$ and a measurament of an observable $A$ (a self-adjoint matrix) is made, then Nature reveals the $k$-th eigenvalue $\lambda_k$ of $A$ with probability ${|\psi_k|}^2$ (the squared norm of the the projection of $\psi$ on the line spanned by the $k$-th eigenvector).

Assuming this principle, one can check that the expected value of the measurement is $$ \boxed{\mathbb{E}_\psi(A) = \langle \psi, A\psi \rangle}. $$

In finite dimensional quantum probability, one rather starts by defining the "expectation" of an observable $A$ under the state $\psi$ by the above formula.

This looks like an expectation in classical probability because this is linear in $A$, playing the role of a random variable. The huge difference lies in the fact that $\mathbb{E}_\psi$ is not defined on a set of classical random variables, but on the set of the self-adjoint matrices in the underlying Hilbert space.

But following this analogy with classical probability, it is natural to say that the Fourier transform of $A$ is the function of $t$ $$ \mathbb{E}_{\psi}(e^{-itA}) = \langle \psi, e^{-itA}\psi \rangle. $$ Note that the exponential matrix $e^{-itA}$ is well defined whenever $A$ is self-adjoint and it is self-adjoint too.

Now, one can check that $$ \langle \psi, e^{-itA}\psi \rangle = \sum_k e^{-it\lambda_k} {|\psi_k|}^2, $$ and it turns out that this is exactly the (classical) Fourier transform of the probability which picks $\lambda_k$ with probability ${|\psi_k|}^2$.

This is why my answer is not a satisfactory one: to sum up, I just claim that if, instead of postulating the rule involving eigenvalues, one postulates that the average measurement of $A$ under $\psi$ is $$ \mathbb{E}_\psi(A) = \langle \psi, A\psi \rangle, $$ then it is mathematically nice to say that the measurement reveals the $k$-th eigenvalue $\lambda_k$ of $A$ with probability ${|\psi_k|}^2$.

Answer 4

Let's go to the mathematics. The way that we describe this things is in terms of linear operators acting on a linear space. Each element of the linear space $|\psi\rangle$ encode possibile preparations of the measurement. And each hermitian operator $A$ is associated with something that we can measure. The expectation value of the observable are defined to be:

$$ \langle A\rangle=\langle\psi|A|\psi\rangle $$

If we pick an eigenvector of $A$ we have that:

$$ \langle A\rangle ^2 = \langle A^2\rangle $$

so, we do not have fluctuations over the outcome. So, we can determine the eigenvalue of $A$ as the "true" value of the observable at this state. This eigenvectors and eigenvalues need to exist because if you measure the energy, and then the energy again and again, the outcome don't fluctuates (this is a fact of Nature).

Others fact of Nature can guide us for other conclusions about how should be your description in terms of linear operators on linear spaces. If you measure something in one outcome, all the others outcomes are excluded to be a possible outcome of a repeated measurement. With this you may see why the operators need to be hermitian (orthogonal eigenvectors + real eigenvalues). Each eigenvalue is a possible expectation value with zero fluctuation. So, they are the possible things that we can measure about the observable.

We can build something like that: each physical observable $A$ are descibed by:

$$ \sum_{outcomes\,\, \alpha}\alpha|\alpha\rangle\langle\alpha| $$

with $|\alpha\rangle$ being the preparation of the system after the measurement of $A$ at the value $\alpha$. And then, the uncertainty principle can be encoded by existence of some observable $B$ with $|\beta\rangle$ not paralllel to each of this $|\alpha\rangle$ states. Or, in other words, $[A,\,B]\neq 0$.

Answer 5

If you measure the same observable twice, you get the same answer. Quantum mechanics's way of enforcing this is that after the first measurement, the system is left in a state that has a definite value of the observable.

Answer 6

@Virgo comment is actually part of the answer. It is an axiom of quantum mechanics. Thus the correct main stream answer is that nobody knows.

Answer 7

I had the same doubt when I started learning QM. Although my explanation to this is vaque, but it gives a general idea.(I found this on Introduction to Qunatum Mechanics, David J Griffith) Ordinarily, when you measure an observable Q on an ensemble of identically prepared systems, all in the same state $\Psi$, you do not get the same result each time. Let us hypothetically prepare a state such that every measurement of Q is certain to return the same value (call it q). This state is called the determinate state. The standard deviation of Q, in this state would be zero, $$\sigma^2 = \left \langle (Q-\left \langle Q \right \rangle)^2 \right \rangle = \left \langle \Psi|(Q-q)^2\Psi \right \rangle=\left \langle (Q-q)\Psi|(Q-q)\Psi \right \rangle=0$$ Where i have used the fact that $\left \langle Q \right \rangle=q$ and Q (and also $(Q - q)$) is Hermitian to move one factor over to the first term in the inner product.
But the only vector whose inner product with itself vanishes is 0, so $$Q\Psi=q\Psi$$ Generally, your $\Psi$ wont be on a Determinate state. But regardless of any state your wave function is in, you will be able to write it in terms of Linear combinations of determinate state (since the eigen functions of an observable are orthogonal and complete).