# Finding position uncertainty

1. Feb 19, 2016

### ognik

1. The problem statement, all variables and given/known data
I don't understand how the following works:

2. Relevant equations
I follow $\Delta \nu_0 = \frac{\hbar }{2 m \Delta x_0}$ easily, but then

$\Delta x = \frac{\hbar }{2 m \Delta x_0} t$ ;leaves me puzzled - I understand multiplying through by t, but how does $\nu_0 t = \Delta x$ instead of $= \Delta x_0$ ?

3. The attempt at a solution

2. Feb 20, 2016

### drvrm

please enter the physical picture- what you are attempting at so that uncertainties in measurements can be defined-physics is not purely mathematical formula -as i understand you are using position / momentum uncertainty realation or energy/ time which are canonially conjugate observables and represented by your relationship !

3. Feb 20, 2016

### ognik

Frustratingly I can't find the source I found this in, but this was in a section on wave packets and the uncertainty principle - and my apologies for using \nu instead of v ...

$\Delta x_0 \Delta p_0 \ge \frac{\hbar}{2}$
$\therefore \Delta v_0 \ge \frac{\hbar}{2 m \Delta x_0 }$
$\therefore \Delta x \ge \frac{\hbar}{2 m \Delta x_0 } t$
So my question is just why $\Delta v_0 t = \Delta x$ and not $\Delta x_0$?

4. Feb 20, 2016

### drvrm

then its a 'riddle' in the sense that your uncertainty in x(0) and p(0) is related by your given equation -
if its at time say t=t(0) then as time advances -goes to t=t they wish to calculate the delta change in x i.e. equivalent to ( x- x(0))
its an infinitesimal change in x so its written in that manner- this has been effected by the uncertainty in v(0) times the time elasped , so if somebody has plotted a shape of the packet with time with t it may be seen as spread of the wave packet in x and naturally its momentum will be /may be more sharp-
but the above is a wild guess only and any question /problem should be framed in a context- if you look up any text on QM and see the discussion on time development of a packet -then perhaps you can get open the riddle yourself
we know how a gaussian packet spreads with time - as the graphs are given in text books

For instance, if an electron is originally localized in a region of atomic scale (i.e., [PLAIN]http://farside.ph.utexas.edu/teaching/qmech/Quantum/img395.png) [Broken] then the doubling time is only about [PLAIN]http://farside.ph.utexas.edu/teaching/qmech/Quantum/img396.png. [Broken] Evidently, particle wave packets (for freely moving particles) spread very rapidly..

for details see <http://farside.ph.utexas.edu/teaching/qmech/Quantum/node26.html> [Broken]

Last edited by a moderator: May 7, 2017
5. Feb 20, 2016

### ognik

6. Feb 20, 2016

### drvrm

thanks the ref given earlier. is not good/working
you can instead look up the following
<The spreading of the wave packet makes sense in terms of the Uncertainty Principle.
Suppose a particle has an uncertainty in position of ∆x0 at t=0. The momentum at t=0 is p0 . The uncertainty in momentum at this time is at least (1) Thus, we can determine the uncertainty in speed as (2)
From this we can determine the uncertainty in position and at a later time. (3) Thus, we see that the uncertainty in position at any later time is inversely proportional to the initial position. The better we know it now, the worse we will know it later.
Hopefully this principle does not apply to your knowledge of quantum mechanics.>

ref.https://web.phys.ksu.edu/vqm/vqmnextgen/qmbasics/wavepackets.pdf