## Uncertainty Principle [ENDORSED]

$\Delta p \Delta x\geq \frac{h}{4\pi }$

mayapartha_1D
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Uncertainty Principle

Hi!!

I have a question regarding the Uncertainty principle. If given a problem where the momentum uncertainty, for example, is + or - 2, does that mean the uncertainty amount is 4? Or do you have to take into account the value of the momentum?

Thanks!

Janice Xiao 1I
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am

### Re: Uncertainty Principle

The uncertainty amount is 4. You don't need to take into account the actual momentum value.

Brandon Fujii 1K
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

### Re: Uncertainty Principle

Also if you use number 1.45 as a practice problem, the solutions manual is incorrect. The actual velocity given is 5.00 $\pm$ 5.0m/s
Thus, $\Delta v$=10.0m/s NOT 5.0m/s as the solution manual incorrectly states. Therefore, your answer is $\Delta x$=6.7x10-37m

Charles Ang 1E
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

### Re: Uncertainty Principle  [ENDORSED]

An easy way to remember this is to think of delta v as the change in v like in math. So, we arent concerned with the value of v, but rather the difference between the two limits to its domain.

nanditasundarapandian1D
Posts: 21
Joined: Sat Jul 22, 2017 3:01 am

### Re: Uncertainty Principle

If it's +/- 4. its easier to write it out like this

(4+4)-(4-4)

Richard Braun 1I
Posts: 65
Joined: Fri Sep 29, 2017 7:05 am

### Re: Uncertainty Principle

The uncertainty is 4. If the problem states +/- 2, that means 2 from 0 both ways, added together, gives you four.