-5
$\begingroup$

It is postulated that the particle is described by a state vector $\Phi$, which is an element of Hilbert space. Let's suppose that there are two Hermitian operators $A$ and $B$, representing two canonically conjugated physical quantities. According to the theory $AB-BA=i * const * I$.

Then, it is possible to prove the uncertainty principle with mathematical precision

$\Delta_A \Delta_B >= constant$

It says that there is no state where $\Delta_A$ (or $\Delta_B$) is zero, since the product is finite. Hence, there is no state where $A$ (or $B$) has a definite value. However, this contradicts another statement that the physical quantities regarding $A$ or $B$ are the eigenvalues of $A$ or $B$. According to the uncertainty principle, the eigenvalue problem for these operators is unsolvable.

This seems to be an inconsistency in the theory. The Hilbert space is not enough to define quantum mechanics. At least we need an extension of the Hilbert space where the eigenvalue problem of $A$ and $B$ is solvable. However, there is no clue about this extension. Is it possible to resolve the above problem? If I missed something, please correct the reasoning.

Improve this question
$\endgroup$
2
  • $\begingroup$ What is "the eigenvalue problem for these operators?" $\endgroup$
    – WillO
    Commented Dec 22, 2021 at 14:11
  • 2
    $\begingroup$ You are misinterpreting the ansertainty principle - it is a purely mathematical statement about variances. See, e.g., here. $\endgroup$
    – Roger V.
    Commented Dec 22, 2021 at 14:12

2 Answers 2

Reset to default
5
$\begingroup$

There are two issues with your question.

  • First the uncertainty principle does NOT state that there is no state where $A$ has a definite value. It states (somewhat loosely) there is no state for which $A$ AND $B$ have simultaneously definite values. The uncertainty principle is a bound on the product of variances of hermitian operators, not on possible eigenvalues or eigenstates of these operators.

  • Second, assumptions of the uncertainty relations are that the states are normalizable AND the operators are self-adjoint. If you are working with $\hat x$ and $\hat p$, their eigenstates are NOT normalizable. If you are working with $\hat \theta$ and $\hat L_z$, then $\hat \theta$ is NOT self-adjoint.

Improve this answer
$\endgroup$
12
  • 2
    $\begingroup$ Sorry this is nonsense. There is nothing to prevent you from determining the eigenstates: they are just not normalizable and thus technically outside the Hilbert space unless you work with a rigged version. $\endgroup$
    – ZeroTheHero
    Commented Dec 22, 2021 at 16:25
  • 2
    $\begingroup$ Why do you say eigenstates cannot be determined/cannot exist when almost every QM textbook does exactly this for many different potentials?? Again, as many people have commented, there is no state that is BOTH an eigenstate for A AND B at the same time. You can find eigenstates for A. You can also find eigenstates for B. But you cannot find an state that is an eigenstates for BOTH A AND B. $\endgroup$
    – Marius Ladegård Meyer
    Commented Dec 22, 2021 at 17:37
  • 2
    $\begingroup$ Let $\hat \theta f(\theta)=\theta f(\theta)$ and $\hat L_zf(\theta)=-i\hbar df(\theta)/d\theta$. Clearly they form a “canonical pair” and the eigenstates of $\hat L_z$ are $g_m(\theta)=e^{i m\theta}/\sqrt{2\pi}$ so the eigenstates exists. The issue here is that $\hat \theta$ is not self adjoint. $\endgroup$
    – ZeroTheHero
    Commented Dec 22, 2021 at 17:52
  • 1
    $\begingroup$ @ZeroTheHero You are right, of course. Till this moment, it never bothered me, of course, that, for a finite spin, if one has an eigenstate for $\hat \sigma_z$, say, the variance for $\hat \sigma_x$ and $\hat \sigma_y$ are still finite. The product of the uncertainties is finite, but though these are non-commuting operators, they are not conjugate. I never thought of $\hat \theta$ which is the conjugate variable, and still has a finite variance. The answer is that it is not self-adjoint. Thanks for giving me the answer before I even asked myself the question... $\endgroup$
    – Alfred
    Commented Dec 23, 2021 at 23:41
  • 1
    $\begingroup$ @Alfred see also physics.stackexchange.com/a/338057/36194 $\endgroup$
    – ZeroTheHero
    Commented Dec 23, 2021 at 23:43
3
$\begingroup$

Then, it is possible to prove the uncertainty principle with mathematical precision. It says that there is no state where $A$ (or $B$) has a definite value

This is not what the uncertainty principle says. However, if we replace your "or" in the parenthesis by "and", then we will get a statement that uncertainty principle does say. The point is that there is no state for which we can determine both the value of the observable $A$ and the value of the observable $B$ with absolute certainty.

In fact, the uncertainty principle gives us even a measure of how uncertain we must be. Mathematically speaking, the uncertainty principle says that if we define the variance in measuring each operator by $\Delta A^2 = \langle A^2 \rangle - \langle A \rangle ^2$ (and similarly for $\Delta B^2$) then $$ \Delta A^2 \Delta B^2 \geq \frac{1}{4} |\langle \left[ A , B \right] \rangle|^2$$

So if our two operators have a constant non-zero commutation relation, the right hand side of the above inequality is a constant, which give us the lower limit for how certain we can be in measuring both $A$ and $B$ on the same state.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ The first member of the uncertainty principle must be squared $\endgroup$
    – Hitman Reborn
    Commented Dec 22, 2021 at 15:58
  • $\begingroup$ See my comment for the first answer. That applies here also. $\endgroup$
    – wawa
    Commented Dec 22, 2021 at 16:12
  • 1
    $\begingroup$ @wawa your comment doesn't make much sense. both $A$ and $B$ have eigenvectors with definite real eigenvalues. the point is that the eigenvectors of $A$ are not those of $B$ and vice versa. that's what the uncertainty principle indicates. maybe it would help if you can give a concrete example $\endgroup$
    – user275556
    Commented Dec 22, 2021 at 17:36
  • 1
    $\begingroup$ @yyy Your answer misses a point. If it were the case that a "valid" operator $B$ had a constant non-zero commutator with $A$, then a vanishing $\Delta A^2$, which is the case for an eigenstate, would imply an infinite $\Delta B^2$. The eigenstates of the operators $\hat x$ and $\hat p$ are not "physical" : they are not normalizable. So uncertainties can be arbitrarily small, and conversely arbitrarily large, but not actually zero and infinite. But if you take a normalisable eigenstate of the angular momentum, say, then $\Delta A^2\equiv 0$. But there is no such $B$, see ZeroTheHero. $\endgroup$
    – Alfred
    Commented Dec 26, 2021 at 0:10
  • 1
    $\begingroup$ Comment too long, part 2 : As ZeroTheHero correctly mentioned, if $A$ is the angular momentum then there is indeed an operator $B$ with non-zero constant commutator, namely the angle. But the latter is not "valid" inasmuch it is not well-defined: it is only defined up to $2\pi$ and that is the problem. Intuitively, if you try to make $\Delta A^2$ two small, the angle will "spread" beyond the range of $2\pi$ and keep expanding "round and round the circle" with uncertainty much larger than $2\pi$, becoming arbitrarily large. For an eigenstate of $A$, $\Delta \theta^2$ is infinite in that sense. $\endgroup$
    – Alfred
    Commented Dec 26, 2021 at 0:28

Not the answer you're looking for? Browse other questions tagged or ask your own question.