- #1
skrat
- 748
- 8
Homework Statement
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)##
Homework Equations
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?
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