Free particle: expectation of x for all time with Ehrensfest

AI Thread Summary
The discussion focuses on calculating the expectation value of a free particle's position over time using Ehrenfest's theorem. The initial wavefunction is given, and the participants explore how to apply Ehrenfest's equation to find the time evolution of the expectation value of position, <x(t)>. They clarify the importance of recognizing operators and their commutation relations, particularly between position and momentum. The conversation highlights the need to differentiate and apply the correct identities to derive <x(t)> for all time, emphasizing that the task involves more than just calculating <x> at t=0. Ultimately, the participants are guided to understand the relationship between velocity, momentum, and the expectation value of position over time.
renec112
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Hello physics forums. I'm trying to solve an old exam question. Would love your help.

Homework Statement


A free particle in one dimension is described by:
## H = \frac{p^2}{2m} = \frac{\hbar}{2m}\frac{\partial^2}{\partial x^2}##

at ##t = 0##
The wavefunction is described by:
## \Psi(x,0) = N(a^2-x^2) e^{i k x}## for ##|x| \leq a##
outside ##a##, ## \Psi = 0##. Use Ehrenfest to find the expectation value for all later times ##<x(t))>## of the particles position for all time ## t \geq 0 ##.

Homework Equations


Ehrensfest:
##
\frac{d<Q>}{dt} = \frac{i}{\hbar}<[H,Q]> + <\frac{\partial Q}{\partial t}>
##
Where ##Q## is an operator.

The Attempt at a Solution


We need to find it for all later times, Ehrensfest will show how an operator evolves in time. So set ##Q = x## and use Ehrenfest. Then we know the poisiton for all later time.
##
\frac{d<x>}{dt} = \frac{i}{\hbar}<[H,x]> + <\frac{\partial x}{\partial t}>
##
Since the operator does not change in time we have:
##
\frac{d<x>}{dt} = \frac{i}{\hbar}<[H,x]>
##
Here is where i am stuck. I am trying to do the commutator:
##
\frac{d<x>}{dt} = \frac{i}{\hbar}<(Hx -xH>
##
However, these do commute and hence everything should be zero.

What do you think?
 
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renec112 said:
However, these do commute ...
Can you show that this is true?
 
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kuruman said:
Can you show that this is true?
Actually when you say it like that..
I guess i thought in
## H = \frac{p^2}{2m} = \frac{\hbar}{2m}\frac{\partial^2}{\partial x^2}##
the ##p## is just a constant but off course it's an operator:
##p = i \hbar \frac{\partial}{\partial x}##

I guess the only thing to do is to differentiate a couple of times and then take the inner product :D

##
\frac{d<x>}{dt} = \frac{i}{\hbar}<Hx -xH>
##
##
= \frac{i}{\hbar}<\frac{\hbar}{2m}\frac{\partial^2}{\partial x^2} x -x\frac{\hbar}{2m}\frac{\partial^2}{\partial x^2}>
##
##
= < \Psi | \frac{i}{\hbar}<\frac{\hbar}{2m}\frac{\partial^2}{\partial x^2} x -x\frac{\hbar}{2m}\frac{\partial^2}{\partial x^2} | \Psi>
##

Seems legit?

Thank you for helping me.
 
It's legit, but you need to advance it further. Remember that ##[x,p_x] = i \hbar##. There is also a useful identity about commuting operators, that says$$[AB,C]=A[B,C]+[A,C]B.$$ Prove it then use it. What do you think ##A##, ##B## and ##C## should be identified as?
 
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kuruman said:
It's legit, but you need to advance it further. Remember that ##[x,p_x] = i \hbar##. There is also a useful identity about commuting operators, that says$$[AB,C]=A[B,C]+[A,C]B.$$ Prove it then use it. What do you think ##A##, ##B## and ##C## should be identified as?
Oh I see. That's very smart.
This is what i did:
##\frac{d<x>}{dt} = \frac{i}{\hbar} <[H,x]>##
insert hamilton
##= \frac{i}{\hbar} <[P^2/2m,x]>##
##= \frac{i}{2 m \hbar } <[PP,x]>##
##= \frac{i}{2 m \hbar } <P[P,x]+[P,x]P>##
##= \frac{i}{2 m \hbar } <-P[x,P]-[x,P]P>##
##= \frac{i}{2 m \hbar } <-P i \hbar-P i \hbar>##
##= \frac{1}{m} <P>##

And i already calculated ##<P>## from another task.
Not sure if it's correct, but it feels like it. Thank you very much for helping me out :) !
 
renec112 said:
Not sure if it's correct, but it feels like it.
Classically, what is v in terms of p?
 
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kuruman said:
Classically, what is v in terms of p?
## v = \frac{1}{2} m v^2##
## = \frac{m^2v^2}{2m}##
## = \frac{p^2}{2m}##
## \Rightarrow p = \sqrt{2 v m}##
Like this?
 
renec112 said:
## v = \frac{1}{2} m v^2##
## = \frac{m^2v^2}{2m}##
## = \frac{p^2}{2m}##
## \Rightarrow p = \sqrt{2 v m}##
Like this?
Nope. To begin with, ##\frac{1}{2}mv^2## is kinetic energy ##K##, not speed. What is the classical definition of momentum? Look it up if you forgot.
 
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kuruman said:
Nope. To begin with, ##\frac{1}{2}mv^2## is kinetic energy ##K##, not speed. What is the classical definition of momentum? Look it up if you forgot.
Thanks for helping me :)
Its ##p = mv## off course.
Oh i so the m' cancels and i am left with only ##<x>##... Nice trick ! Thank you :)
 
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renec112 said:
Thanks for helping me :)
Its ##p = mv## off course.
Oh i so the m' cancels and i am left with only ##<x>##... Nice trick ! Thank you :)
Not so fast. So far, you have established that ##\frac{d<x>}{dt}=<v>=\frac{1}{m}<p>##. You have not found ##<x>##.
 
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  • #11
kuruman said:
Not so fast. So far, you have established that ##\frac{d<x>}{dt}=<v>=\frac{1}{m}<p>##. You have not found ##<x>##.
Oh right.. I think i misunderstood the question then.
I need to find ##< x(t)> ## for all later times.
But i can't see, how i can find that.. You are saying ##\frac{d<x>}{dt} \neq < x(t)> ##

I am not sure why i should fine ##< x >## I can see it's a part of the equation, but why does it give me ##<x(t)>## for all later times?

I only know how to find ##<x>## when ##t = 0##. Then it's just: ##< \Psi(x,0) | x | \Psi(x,0)>##

Thank you for the comment! appreciate it :)
 
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