Heisenberg uncertainty principle

[lyra87]

Sorry for the other post, I clicked post by mistake

Homework Statement

Find the minimum uncertainty in the length of the year

Homework Equations

$$\Delta$$Px >= $$\hbar$$/(2$$\Delta$$x

The Attempt at a Solution

I did:
$$\Delta$$t = ((($$\Delta$$xt)/x)$$^{}2$$+((xm)/$$\Delta$$Px)$$^{}2$$)$$^{}.5$$

and then because of the uncertainty principle:
$$\Delta$$t =((($$\Delta$$xt)/x)$$^{}2$$+((2xm$$\Delta$$x)/$$\hbar$$)$$^{}2$$)$$^{}.5$$

then I took the derivative dt/d$$\Delta$$x to minimize the function. But $$\Delta$$x canceled, and I don't get the correct value for h-bar, so I think I did something wrong.

thanks a lot

 

[Jhero]

Hello, don't worry about the previous post, it happens to all of us! To find the minimum uncertainty in the length of the year, we can use the uncertainty principle, which states that the product of the uncertainties in position and momentum must be greater than or equal to h-bar/2. In this case, we are looking for the uncertainty in time, so we can use the same equation but with time instead of position:

\Deltat\Deltapx >= \hbar/2

To find the minimum uncertainty, we can set the derivative of this equation with respect to \Deltat equal to 0 and solve for \Deltat. This will give us the minimum value for \Deltat, which represents the minimum uncertainty in the length of the year. Let's go through the steps:

1. Take the derivative of \Deltat\Deltapx with respect to \Deltat:

\dfrac{d}{d\Deltat}(\Deltat\Deltapx) = \Deltapx + \Deltat\dfrac{d\Deltapx}{d\Deltat}

2. Set this equal to 0 and solve for \Deltat:

\Deltapx + \Deltat\dfrac{d\Deltapx}{d\Deltat} = 0

\Deltat = -\dfrac{\Deltapx}{\dfrac{d\Deltapx}{d\Deltat}}

3. Now we need to find the derivative \dfrac{d\Deltapx}{d\Deltat}. To do this, we can use the uncertainty principle again, but this time with energy and time:

\Deltat\DeltapE >= \hbar/2

We can rearrange this to solve for \Deltapx:

\Deltapx = \dfrac{\hbar}{2\Deltat\deltapE}

4. Substituting this into our equation for \Deltat, we get:

\Deltat = -\dfrac{\hbar}{2\Deltat\deltapE} \cdot \dfrac{d\Deltapx}{d\Deltat}

5. Now we need to find the derivative \dfrac{d\Deltapx}{d\Deltat}. To do this, we can use the uncertainty principle again, but this time with energy and time:

\Delt