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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.

wawa
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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.

ZeroTheHero
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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.