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According to Heisenberg's uncertainity principle, the position and velocity of an quantum particle cannot be determined simultaneously. Is it possible to determine position and acceleration simultaneously ? If yes, how it can be determined (or) in what range the acceleration values of quantum particles lie ?

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Is it possible to determine position and acceleration simultaneously ?

An interesting, but very specific, example, is the quantum harmonic oscillator. In Heisenberg representation, position is represented by an operator $X(t)$.

$X(t)$ obeys the equation:

$$ \ddot X(t) + \omega^2 X(t)=0 \tag{0}$$

Of course, from this, it is obvious that $[X(t), \ddot X(t)] = 0$, but it is interesting to look in detail, at the commutation relations at different times.

We have the following relation for the commutator of position operators at $2$ distincts times :

$$ [X(t), X(t')] = -i \frac{\hbar}{m \omega} \sin \omega(t-t') \tag{1}$$

Note that this is coherent with equation $(0)$ and the fact that the commutator of $2$ hermitian operators is anti-hermitian, so $i$, the imaginary unit is needed. This commutator is also anti-symmetric, as needed.

To see the complete coherence of the expression $(1)$, we may derive relatively to $t'$, and multiply by $m$, and we get, with $P(t')= m \dot X(t')$ :

$$ [X(t), P(t')] = i \hbar \cos \omega(t-t') \tag{2}$$

One may check, that, for $t=t'$, we get the usual commutation relations between $X$ and $P$ at equal times :

$$[X(t), P(t)] = i \hbar \tag{3}$$

Now, from $(1)$, we may derive $2$ times relatively to $t'$, and get :

$$[X(t), \ddot X(t')] = i \frac{\hbar \omega}{m } \sin \omega(t-t') \tag{4}$$ From, this, we may deduce the commutator at equal times :

$$[X(t), \ddot X(t)] = 0 \tag{5}$$

So, theorically, it should be possible to determine position and acceleration simultaneously.

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