Heisenberg momentum uncertainty

[leehufford]

Homework Statement

An electron is trapped in a one-dimensional well of width 0.132 nm. The electron is in the n = 10 state.
a) What is the energy of the electron?
b) What is the uncertainty in the momentum?
c) What is the uncertainty in the position?

Homework Equations

E_n = h²n²/8mL²
Delta(x)Delta(p) > hbar/2

The Attempt at a Solution

The answers to this problem are in the back of our book, I got 2,160 eV for A, which was correct.
The book gave us a hint for part B: use Delta(x)Delta(p) > hbar/2, but here I have one equation with two unknowns. (I tried using 0.132 nm as delta(x) but that didn't give the right answer. Also, why would they ask for delta(x) in part C if it was given... so

I tried finding the momentum using

p = (1/c)sqrt(E²-(mc²)²). And my answer for momentum is the same as the book's answer for the uncertainty in momentum. I guess I can't see how to relate delta(p) to p. Also, why doesn't the length that the electron is trapped in count for delta(x)? Thanks for reading,

Lee

 

[td21]

Hi!
Uncertainity (S.D.) in momentum and position can be found from the square root of the expectation value of the (p - pbar)^2 and (x-xbar)^2, respectively. Note that pbar is zero and xbar is a/2 in 1 dimensional infinite square well.
Yes, it happens that the S.D. of momentum is equal to momentum in this case. Can you guess the reason behind?

 

[leehufford]

I'm not 100% clear on what you mean.. I've never actually taken a statistics class yet. The formula my book uses is

delta p = sqrt((p²)_avg - (p_av)²). If the average momentum is zero, and the formula reduces to:

sqrt(p²)_avg, would I just use the momentum I found with relativistic dynamics...and that's why the standard deviation is the same as the momentum value? What does that mean physically? Thanks for the reply.

Lee

 

[leehufford]

Just noticed I got moved, I didn't think homework problems from a modern physics class where considered introductory physics, but I guess I will post all of my questions here from now on. Sorry about that.

 

[BvU]

Never mind the moving to introductory. Perhaps they'll move it back once this is indeed about relativistic QM (it's not, I should think)..

We're more used to writing ##p=i\hbar {\partial \over \partial x}## in simple QM, where E=p²/(2m) if V=0. Same difference. But <p> = 0 for sure.

##\Delta p\Delta x \ge \hbar/2## is not an equation but something else (an inequality). Why they provide it as a hint is a mystery to me (at n=10 it's a lot more than ##\hbar/2##).

If your potential well is from 0 to L, I expect <x> to come out L/2 too... Look up the wave function and check. Then do <x²>. You will find that ##\Delta x \propto L##, so L does count!