-5
$\begingroup$

Can somebody tell me how the units work out in Heisenberg's principle equation? Mass in $kg$ and velocity in $m/s$ cancel partially with Planck's constant, so what kind of unit is given to $Δx$ to balance the units?

$\endgroup$
0
2
$\begingroup$

$(m) \cdot (kg \cdot \frac{m}{s}) \ge J \cdot s = (kg \cdot \frac{m^2}{s^2}) \cdot s$

So, the unit is $kg \cdot \frac{m^2}{s}$ on both sides.

$\endgroup$
2
$\begingroup$

The position-momentum uncertainty relation is: $\Delta x\Delta p\geq \frac{\hbar}{2}$. Here $\Delta x$ is the 'uncertainty' aka standard deviation of observing a quantum particle at a given point. The standard deviation is defined as $ \Delta x = \sqrt{\left\langle x^2\right\rangle-\left\langle x\right\rangle^2}$. Here x has the dimensions of length and we can then see that $\Delta x$ also has the dimensions of a length. As the uncertainty in momentum is defined in much the same way we see that $\Delta p$ has the dimensions of momentum: $\frac{\text{mass}\cdot\text{length}}{\text{time}}= $. Now $\hbar$ has to have the dimension $ \frac{\text{mass}\cdot\text{length}^2}{\text{time}}=\text{energy}\cdot\text{time}$ or else we wouldn't have a physically or mathematically sensible inequality.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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