# QM - free particle in one dimensional dpace

1. Dec 1, 2014

### skrat

1. The problem statement, all variables and given/known data
A particle in one dimensional space, $$H=\frac{p^2}{2m}$$ in time $t=0$ has a wavefunction $$\psi (x)=\left\{\begin{matrix} \sqrt{\frac{15}{8a}}(1-(\frac{2x}{a})^2) &,|x|<\frac a 2 \\ 0 & , |x|>\frac a 2 \end{matrix}\right.$$
a) Calculate the expected values of $x$, $x^2$, $p$, $p^2$ and $xp+px$.
b) Calculate commutators $[H,x]$ and $[H,p]$.

Operators $x$ and $p$ in Heisenberg representation are described as $x_H (t)=e^{i\frac H \hbar t}xe^{-i\frac H \hbar t}$ and $p_H(t)=e^{i\frac H \hbar t}pe^{-i\frac H \hbar t}$.
c) Show $[x_H(t)]^2=[x]^2_H(t)$
d) Write operator $x_H(t)$ as combination of operators $x$ and $p$. Hint: Calculate $\frac{d}{dt}x_H(t)$ and $\frac{d}{dt}p_H(t)$

2. Relevant equations

3. The attempt at a solution

a) A lot of work to do here. I won't write down all the integrals and everything. My question here, and a very important one, is if there is a simpler and faster way of getting these results? (Could my $\psi$ in any way be written as a superposition of Harmonic oscillators where I could than use annihilation and creation operators to get those results faster? Or can I maybe use Heisenberg presentation?)

I calculated this using integrals:
$$<x>=\int _{a/2}^{a/2}\psi ^2xdx$$ Note that $\psi$ is not complex. This brings me to $<x>=0$ because all the exponents are odd numbers. I checked all the integrals using Mathemathica so let's just assume I got this part of the problem right.
b)
$[H,x]=\frac{1}{2m}[p\cdot p,x]=\frac{1}{2m}[p[p,x]+[p,x]p]$ where $[p,x]=-i\hbar\frac{\partial }{\partial x}x+i\hbar x \frac{\partial }{\partial x}$. I how NO idea why the second term is 0? Is that because there is nothing to calculate the derivative of?

Well, when that is answered, the result is $[H,x]=-\frac{i\hbar p}{m}$

About $[H,p]=\frac{1}{2m}[p^2,p]$ it is obvious that they commutate, therefore $[H,p]=0$.

c)

$x_H^2(t)=e^{i\frac H \hbar t}x\cdot x e^{-i\frac H \hbar t}=e^{i\frac H \hbar t}xe^{-i\frac H \hbar t}e^{i\frac H \hbar t}xe^{-i\frac H \hbar t}=[x_H(t)]^2$

d)
I don't understand this part at all.

$\frac{\mathrm{d} }{\mathrm{d} t}x_H(t)=\frac{iH}{\hbar}e^{i\frac H \hbar t}xe^{-i\frac H \hbar t}-e^{i\frac H \hbar t}x(-\frac{iH}{\hbar})e^{-i\frac{H}{\hbar} t}=\frac{i}{2\hbar m}(p^2e^{i\frac H \hbar t}xe^{-i\frac H \hbar t}+e^{i\frac H \hbar t}xp^2e^{-i\frac H \hbar t})$

Now what can I do with this?

Last edited: Dec 1, 2014
2. Dec 1, 2014

### Orodruin

Staff Emeritus
It is not zero. In general it will act on a function f. However, this also goes for the other term where you get the derivative of x times f, on which you can use the product rule.

If you express the time derivative of x in terms of the commutator with the Hamiltonian, what do you get?

3. Dec 1, 2014

### skrat

Aaaa, that is some good thinking...

$[p,x]f=-i\hbar \frac{\partial }{\partial x}xf+i\hbar x \frac{\partial }{\partial x}f=-i\hbar(f+x{f}')+i\hbar x{f}'=-i\hbar f$. Simple as that. =)

Aham, $$\dot{x}_H(t)=\frac i H[H,x](t)=\frac{p(t)}{m}$$ and $$\dot{p}_H(t)=\frac i H[H,p](t)=0 \Rightarrow p_H(t)=p_{H_0}=p$$ and than $$x_H(t)=\frac{p}{m}t+x_{H_0}=\frac{p}{m}t+x$$ I guess this is the answer than?

4. Dec 1, 2014

### Orodruin

Staff Emeritus
Yes, note how it corresponds very nicely to the classical $x(t) = vt + x_0$.