What are the units for ΔxΔp in the uncertainty principle?

[KFC]

Hi all,
I am reading an article about uncertainty principle. If we consider a Gaussian wave packet which standard deviation of momentum $\sigma_p$. The uncertainty principle states that the multiplication of variance of x and variance of p is larger or equal to half $\hbar$

$\Delta x\Delta p \geq \dfrac{\hbar}{2}$

I think $\Delta x$ has the unit of meter, $\Delta p$ has the unit of kg.meter/second, so the multiplication of them give the same unit of $\hbar$.

But if we have the Gaussian wave packet, the standard deviation of $\sigma_x$ and $\sigma_p$ should have the unit of meter and kg.meter/second. But reading the expression of variance for Gaussian given by standard deviation

$\Delta x = \sigma_x^2, \Delta p = \sigma_p^2$

So the unit for $\Delta x \Delta p$ becomes kg*meter^4/second^2 ? I am confusing what mistakes I made here.

 

[DEvens]

Just take out the superscript 2 on the sigmas.

 

[KFC]

Thanks for the reply. Yes, if we want the consistency on the unit, we should remove that superscript. But for Gaussian, the variance is defined as square of standard deviation. That's why it is confusing me. Why we need to redefine it by removing the square?

 

[lightarrow]

The uncertainty principle states that the multiplication of variance of x and variance of p is larger or equal to half $\hbar$

No, it' the product of the standard deviations which is that.

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lightarrow

 

[KFC]

Thanks. I think it is the notation confusing me. In the book I am reading, they use $\Delta$ but when I read other context, it uses $\sigma$. If it refers to standard deviation in the expression of uncertainty principle, so $\Delta x$ mean the standard deviation of x not variance, correct? Thanks for pointing that out.

 

[jtbell]

so Δx\Delta x mean the standard deviation of x not variance, correct?

Correct.