# In what range the acceleration value of quantum particle lies?

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 ?

## 1 Answer

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.