Heisenberg Uncertainty Derivation

AI Thread Summary
The discussion focuses on deriving the quantum-mechanical uncertainty in the position of a particle at temperature T, expressed as Δx ~ √(h²/4mkT). Participants explore the uncertainty relation ΔpΔx ≥ h/2 and the kinetic energy equation K.E. = 0.5 mv² = 0.5 kT. An attempt is made to isolate Δx by substituting Δp with expressions involving mass and velocity, but challenges arise in simplifying the equation correctly. One participant suggests using Δp = √(2mE) with E as the kinetic energy, which equals 1/2 kT. The conversation emphasizes the need for further insights to successfully complete the derivation.
XJellieBX
Messages
39
Reaction score
0

Homework Statement


Using the uncertainty relation for momentum and position, show that the quantum-mechanical uncertainty in the position of a particle at temperture T is
\Delta x~\sqrt{\frac{h^{2}}{4mkT}}
where T is the temperature and k is the Boltzmann's constant.

Homework Equations


\Delta p\Delta x\geq h/2, h being Planck's constant
K.E.=0.5 mv2=0.5 kT


The Attempt at a Solution


I isolated \Delta x and subbed \Delta p=mv=kT/v.
So, \Delta x~h/2\Delta p ~ h/2mv ~ hv/2kT.
I've tried subing in a whole bunch of stuff for v but I can't seem to get the equation. Any insight?
 
Physics news on Phys.org
XJellieBX said:

Homework Statement


Using the uncertainty relation for momentum and position, show that the quantum-mechanical uncertainty in the position of a particle at temperture T is
\Delta x~\sqrt{\frac{h^{2}}{4mkT}}
where T is the temperature and k is the Boltzmann's constant.

Homework Equations


\Delta p\Delta x\geq h/2, h being Planck's constant
K.E.=0.5 mv2=0.5 kT

The Attempt at a Solution


I isolated \Delta x and subbed \Delta p=mv=kT/v.
So, \Delta x~h/2\Delta p ~ h/2mv ~ hv/2kT.
I've tried subing in a whole bunch of stuff for v but I can't seem to get the equation. Any insight?
Delta p[/tex] = sqrt(2mE) where E is the kinetic energy which is equal to 1/2*kT
 
Thanks =)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top