If the system is in an eigenstate of $B$, then $\Delta B = 0$, but $\frac{|[A,B]|^2}{ 4} \not = 0$, so how can uncertainty relation hold?

Question

Let $A,B$ be two non commuting observables, then we do know that

$\langle (\Delta A)^2 \rangle \cdot \langle (\Delta B)^2 \rangle \geq \frac{|[A,B]|^2}{ 4}.$ Also, we do know that, when the system is in an eigenfunction state of $B$, then $\Delta B = 0$, but this state in general does not have to be an eigenstate of $A$, so $\Delta A \not =0$, but the product is clearly $\langle (\Delta A)^2 \rangle \cdot \langle (\Delta B)^2 \rangle = 0$ even though the RHS is not zero.

I'm sure I'm missing something, but couldn't figure out what it is. I've looked some question similar to this one in this site, but all of them were about the position and the momentum operators, which are pathological.

Answer 1

Your claim that the r.h.s. is zero is simply incorrect. The general Robertson-Schrödinger uncertainty relation is $$ \sigma_A(\psi)\sigma_B(\psi) \geq \langle \frac{1}{2}[A,B]\rangle_\psi,$$ where $\sigma_A(\psi)$ is the standard deviation of $A$ in the state $\psi$ (also ofen denoted $\Delta A$) and $\langle ...\rangle_\psi$ is the expectation value in the state $\psi$. Note that the value of the r.h.s. is dependent on the state: Even if the commutator in general is non-zero, the r.h.s. of the uncertainty relation can be.

Now, if $\psi$ is an eigenstate of $B$ with eigenvalue $b$, then the r.h.s. works out as follows: $$ \langle \psi \vert \frac12 [A,B]\vert \psi\rangle = \frac12 \left(\langle \psi \vert AB\vert\psi\rangle -\langle \psi\vert BA\vert \psi\rangle \right) = \frac12 \left(\langle \psi \vert Ab\vert\psi\rangle -\langle \psi\vert bA\vert \psi\rangle \right) = \frac{b}2 \left(\langle A\rangle_\psi - \langle A\rangle_\psi \right) = 0.$$ That is, for every eigenstate of one of the two observables in the uncertainty principle, the r.h.s. is always zero and therefore the uncertainty relation is trivially fulfilled regardless of the value of the standard deviation of the other observable.

Answer 2

To see this, you have to be a Little more mathematical precise. Consider the commutator

$[A,B]=AB-BA$.

Take the norm $||...||_n$ (with natural number $n$ as a parameter) defined by $||X||_n=E_n(X^2)$ for any Operator $X$ and the Quantum expectation value $E_n$ given by the relation:

$E_n(X) = <\psi^{(n)}|X|\psi^{(n)}>$.

Now I clarify the Parameter $n$: We can define a sequence $|\psi^{(n)}>$ of wave functions that converge to the correct wave function $|\psi^{(n)}>$ in the Limit $n \mapsto \infty$. So we have only the Quantum state as a Limit of a sequence! Now proceed with computing the norm and using triangle inequality

$||[A,B]||_n = ||AB-BA||_n \leq ||AB||_n+||BA||_n \leq ||A||_n||B||_n+||B||_n||A||_n = 2 ||B||_n||A||_n$.

Let $B'$ be the Operator where the state $|\psi>$ is eigenstate. Then, the Operator

$B = B' - <B'>$ has the following norm:

$||B'-<B'>||_n \mapsto 0$ as $n \mapsto \infty$.

But: If $n$ is not at infinity, we have a very very Tiny remainder that we can denote by $\epsilon_n>0$. So in General we have

$||B'-<B'>||_n=\epsilon_n$. Thus we will have for the value $A$:

$||A||_n \geq \frac {||[A,B]||_n }{2 \epsilon_n}$.

Letting $\epsilon_n$ to Zero in the Limit $n \mapsto \infty$, we will see that the uncertainty of the other Operator $||A||_n$ tends to infinity. Uncertainty for one Operator goes only in Limit to Zero, while the other goes to infinity in the limit.