Free particle: expectation of x for all time with Ehrensfest

  • #1
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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?
 

Answers and Replies

  • #2
kuruman
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However, these do commute ...
Can you show that this is true?
 
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  • #3
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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.
 
  • #4
kuruman
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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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  • #5
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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 :) !
 
  • #6
kuruman
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Not sure if it's correct, but it feels like it.
Classically, what is v in terms of p?
 
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  • #7
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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?
 
  • #8
kuruman
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## 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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  • #9
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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 :)
 
  • #10
kuruman
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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
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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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