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Posted

I've noticed that there a a few different versions of this formula so are they all correct or is there a correct one.  One of my physics books gives one and another gives a different one.

Is it delta(X).delta(P) = h(bar)/2

or is it 

delta(X).delta(P) = h(bar)

Surely they both can't be correct.  If they are both correct,  why?  Or is one correct and the other incorrect?

Thanks

Posted

One was from a lecture by Walter Lewin on YouTube;  the other was 'College Physics ' Wilson and Buffa and another from Fundamentals of Physics [10th Edition] - Halliday & Resnick.

 

As far as I understand X is the position and p the momentum

Posted

When I've complete my sketch in about 5 minutes, I will post a nice easy derivation, without the difficult maths.

Meanwhile look again and check if your sources use an inequality.

The Uncertainty Principle is not an equation.

Posted

studiot forget the explanation Strange has just put me off carrying on with this query.  

 

x      y   satisfied now Strange

 

Bye.

Posted
3 minutes ago, Morris said:

studiot forget the explanation

Oh, I was looking forward to it!

4 minutes ago, Morris said:

Strange has just put me off carrying on with this query.

But why? I was just answering studiot's question (and guessed the reason he asked was just because it is unusual to see them in upper case).

 

Posted (edited)

So suppose a stream of quantum particles of mass m moves along the x axis with speed v as shown.
Let us try to determine the y coordinate of the particle by placing a slit of height [math]\Delta y[/math] in its path.
 

The De Broglie  wavelength, lambda, is [math]\lambda  = \frac{h}{{mv}}[/math]

It is an observed fact that diffraction will occur as the particles passes through the slit (this is a single slit and not the double slit experiment)

causing a new direction and momentum to the aprticles, within an angular spread[math] \pm \alpha [/math]

where   [math]\sin \alpha  = \frac{{ \pm \lambda }}{{\Delta y}}[/math]

 

The attempt to determine the y positionof the particle has led to an uncertainty [math]\Delta {p_y}[/math] in the y componeent of the momentum.


[math]\Delta {p_y} = 2p\sin \alpha [/math]

 

Thus


[math]\Delta y\Delta {p_y} = 2p\lambda [/math]

or of the order of 2h

However the particle may be considered anywhere within that range so it only provides the outer limits.

This boils down to the greater than or equals sign


[math]\Delta y\Delta {p_y} \ge \frac{h}{{4\pi }}[/math]

Some years alter it was found convenient to introduce a revided version of Plank's constant and add the bar to the h. So


[math]\hbar  = \frac{h}{{2\pi }}[/math]

 

In this format the HUP becomes

 


[math]\Delta y\Delta {p_y} \ge \frac{\hbar }{2}[/math]

 

 

HUP1.jpg.4548ae0977643eb0bb8f26566cc9a5d8.jpg

 

 

 

 

Edited by studiot
Posted (edited)

So the tighter inequaltity with the half is the correct one in the case shown.

But in some circumstances the bigger uncertainty may be appropriate.

However the question of what the undertainty refers to is worth discussing further if anyone less impatient than the OP wishes to do so.

It is quite possible to  generate examples of either with the 1/2 or without in classical situations, which may be easier to understand.

Edited by studiot

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