# Free particle: expectation of x for all time with Ehrensfest

• renec112
In summary, the 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.f

#### renec112

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?

However, these do commute ...
Can you show that this is true?

renec112
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?

renec112
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>##

Not sure if it's correct, but it feels like it. Thank you very much for helping me out :) !

Not sure if it's correct, but it feels like it.
Classically, what is v in terms of p?

renec112
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?

## 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.

Last edited:
renec112
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 :)

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>##.

renec112
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 :)