I don't really know if this question has an anwser but I thought it was worth to try asking. My point here is the following: in Quantum Mechanics, to describe the states of a system we use a Hilbert space $\mathcal{H}$. Then, for each physical quantity we associate one hermitian operator $A \in \mathcal{L}(\mathcal{H}, \mathcal{H})$ and the only possible values of that quantity that can be measured are the eigenvalues of $A$.

If the system is then in the state $|\psi\rangle\in \mathcal{H}$ and if $A$ has discrete spectrum (taken to be non-degenerate for simplicity) with eigenvalues $\{a_n\}$ and corresponding eigenvectors $\{|\varphi_n\rangle\}$ then the probability of measuring the eigenvalue $a_n$ is

$$P(a_n) = |\langle \varphi_n | \psi\rangle|^2.$$

In that case, if the system is in the state $|\varphi_n\rangle$ we are certain to measure the value $a_n$ of the quantity.

Analogously, if $A$ has continuous spectrum, for example $\mathbb{R}$, together with a set of generalized eigenvectors $\{|a\rangle : a\in \mathbb{R}\}$ indexed by the elements of the spectrum, then we can construct a probability density $\rho : \mathbb{R}\to \mathbb{R}$

$$\rho(a)=|\langle a|\psi\rangle|^2$$

such that the probability of finding the value of the quantity on the interval $[a_1,a_2]$ is


Again the state $|a\rangle$ is the state where we are certain to measure the quantity with value $a$.

Now, as is well known, all Quantum Mechanics provides us with are probabilities and probability densities. The natural question to ask, in my opinion, is then: if the system is on the state $|\psi\rangle\in \mathcal{H}$, which is not necessarily eigenvector of any observable of interest, there are two ways to see all of this:

  1. The system doesn't have a definite value of the observables of which its state is not an eigenvector. In that case, however this can be, if the state $|\psi\rangle$ is not eigenvector of the position operator, for instance, the system doesn't have a definite position and if it's not an eigenvector of the Hamiltionian, it doesn't have a definite energy.

  2. The system has always definite values of all physical quantities. So the system does have a definite position, a definite momentum, definite energy and so forth. But both experimentally and theoretically we can't access this data. So, the current mathematical model allows only one statistical approach, while experimentally this might be the case because our measurements disturb the system.

Personally I find quite strange to believe the system doesn't hav definite values of physical quantities and only assuming some value when a measurement is performed.

So which possibiliy is the correct one? System do or do not have definite values of the physical quantities?

Notice that it is quite different being at one place, and knowing that the particle is there.

So, just taking position as example, the particle really is nowhere or it is definitely somewhere which we don't know?

Is there any strong justification for any of the two points of view or we really don't know it?


2 Answers 2


Your second explanation ("the system always has definite values") is very hard to square with the (experimentally confirmed) predictions of quantum mechanics.

Here's why: Take a large number of identically prepared pairs of electrons. Take two observables A and B, where each observable has two possible values, 1 and 0. According to your theory, each electron in each pair has well-defined values for each of these observables. (Those values might vary from pair to pair, because our "identical" preparation might have failed to make them really identical).

So any given pair of electrons has associated with it four values: The A-value of electron 1, the B-value of electron 1, the A-value of electron 2, and the B-value of electron 2. Some fraction of the pairs --- call it $p$ --- will have values $(0,0,0,0)$. Another fraction --- call it $q$ --- will have values $(0,0,0,1)$. Et cetera. There are 16 fractions, which add to 1.

Now make an actual observation on a pair of electrons --- say observation A on electron 1 and observation B on electron 2. What's the probability we'll see $(0,0)$? You can easily write down an expression for this in terms of your sixteen fractions. What's the probability we'll see $(0,1)$? Again you can write down a simple expression. And you can do the same for various other observations, say observation B on electron 1 and observation B on electron 2.

But quantum mechanics already tells us what these probabilities should be. So we have a bunch of equations relating certain sums of your 16 fractions to the probabilities predicted by quantum mechanics.

Now we can try to solve those equations to figure out the values of the fractions $p,q$ etc. And, for many experiments, it turns out that the equations have no meaningful solutions, which is to say they have no solutions in which the fractions turn out to be real numbers between $0$ and $1$. Therefore the fractions don't exist.

But if your particles really had well defined values for both observables, then the fractions would certainly exist --- they're just defined to be the fraction of pairs that have certain well-defined values.

Conclusion: Unless you're willing to countentance some very strange phenomena (like well-defined values for one electron that change instantly depending on the measurement you're making on the other electron), your theory can't work.

The key words to Google for are Bell's Theorem, entanglement, hidden variables, and Aspect experiment.


The real reason for your confusion and angst is simply terminology. You called it a measurement so what happens then sounds weird.

But really you are talking about things that don't commute. And you have interactions that yield eigenvalues.

Let's look at spin: if you interact in the direction $\hat z$ then $\hat z$ then finally $\hat x$ and you do that many times you see that 100% of the time the second $\hat z$ interaction gives an eigenvalue that agrees with the first one. So clearly the first interaction puts it into a state that is the kind that then yields that particular eigenvalue 100% of the time.

But if you instead interact in the direction $\hat z$ then $\hat x$ then finally $\hat z$ and you do that many times you see that only 50% of the time the second $\hat z$ interaction gives an eigenvalue that agrees with the first one. So clearly the $\hat x$ interaction puts into a different state than it was before, one that yields different $\hat z$ interaction results.

This is unavoidable. And it happens from the lack of commutation. And it means that these interactions definitely change the state.

Now, you can try to say that things have secret definite values. But the interactions can't just passively reveal the secret values without changing things because otherwise you can't explain that the order of the interactions matters. And once the interactions are changing things it starts to seem silly to ascribe secret values if they aren't being revealed.

And that is always an option, that there are secret values but these interactions aren't revealing them. But then since the results of the interactions (the eigenvalues) are the things that we care about we have to describe where they come from and since the order matters they don't come about by already existing and then being passively revealed in a way that doesn't change things.

So option two is not going to work in any meaningful way. In particular a spin interaction could have the result be determined by the state of the thing interacting with it as well as type of thing interacting with it or on things that seem unrelated. And they could even be unrelated in the sense of not affecting the rates you get different results but could still be influencing of the particular results each time.

For instance, position could interact with the type of spin interaction device in a contextual manner with a spin state to determine a spin eigenvalue for a particular run of the device.


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