Free particle: expectation of x for all time with Ehrensfest

[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?

 

[kuruman]

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.

 

[kuruman]

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
$= \frac{i}{\hbar} <[P^2/2m,x]>$
$= \frac{i}{2 m \hbar } <[PP,x]>$
$= \frac{i}{2 m \hbar } <P[P,x]+[P,x]P>$
$= \frac{i}{2 m \hbar } <-P[x,P]-[x,P]P>$
$= \frac{i}{2 m \hbar } <-P i \hbar-P i \hbar>$
$= \frac{1}{m} <P>$

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 :) !

 

[kuruman]

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?

 

[kuruman]

$v = \frac{1}{2} m v^2$
$= \frac{m^2v^2}{2m}$
$= \frac{p^2}{2m}$
$\Rightarrow p = \sqrt{2 v m}$
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.

 

[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 :)

 

[kuruman]

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 :)