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QM - free particle in one dimensional dpace

  1. Dec 1, 2014 #1
    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? o_O
     
    Last edited: Dec 1, 2014
  2. jcsd
  3. Dec 1, 2014 #2

    Orodruin

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    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?
     
  4. Dec 1, 2014 #3
    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?
     
  5. Dec 1, 2014 #4

    Orodruin

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