1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Particle in given potential

  1. Dec 1, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle is moving in a one-dimensional potential $$V(x)=-\lambda \delta (x)$$ where ##\lambda >0##. Calculate
    a) Expected value and uncertainty of particle energy.
    b) At ##t=0## we measure the position of the particle. What is the probability that the particle has ##|x|>\frac{\hbar ^2}{m\lambda }## right after the measurement?
    c) At ##t=0## we measure the energy of the particle. What are possible results and their probabilities?
    d) What if we do all those measurements at ##t>0##?

    2. Relevant equations


    3. The attempt at a solution
    Let us firstly find that wavefunction:
    Before the Delta function we expect something like ##\psi _1=A e^{ikx}## and after ##\psi _2=Ae^{-ikx}##.

    Boundary condition $$\psi (-0)=\psi (+0)$$ eliminates all odd functions while the second boundary condition $${\psi (+0)}'-{\psi (-0)}'=-2\frac{m\lambda }{\hbar ^2}\psi (0)$$ leaves me with ##k=-\frac{im\lambda}{\hbar ^2}##.

    With given ##k## and after normalization of the function, we finally get to the desired wavefunction $$\psi (x)=\sqrt{\frac{m\lambda }{\hbar ^2}}e^{-\frac{m\lambda }{\hbar ^2}|x|}$$

    a) I seriously hope there is an easier way than $$<E>=\int _{-infty}^{\infty} \psi(x)[-\frac{\hbar ^2}{2m}\frac{\partial^2 }{\partial x^2}-\lambda \delta (x)]\psi (x)dx=\frac{5m\lambda}{2\hbar ^2}$$ Doing the same horrible integral for ##<E^2>=<\frac{p^4}{4m^2}-\lambda \frac{p^2}{2m}\delta(x)+\lambda ^2 \delta (x)>## will eventually bring me to ##\delta _E=\sqrt{<E^2>-<E>^2} ##

    b) I am quite unsure if this is correct: $$P=\frac{<x>_{part}}{<x>}$$ where ##<x>_{part}=2\int_{0}^{\frac{\hbar^2}{2m}}\psi (x)^2 dx## and ##<x>=2\int_{0}^{\infty }\psi (x)^2 dx##. Or is it?

    Why I am confused, is that if we for example had Harmonic oscillator ##|\psi,0>=\frac{1}{\sqrt 2}(|0>+|1>)##. Than we only have to calculate ##|\psi, t>=p_1|0>+p_2|1>## because ##p_1^2## and ##p_2^2## represent the probability that the particle is in given state. However, I am bit lost in my case. :/

    Ok and no idea about c) and d) :/
     
  2. jcsd
  3. Dec 3, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The problem statement is misleading - in general, a particle in this potential can be free, and have whatever energy it likes to have.
    In a bound state (=what you calculated - it also helps for c and d), what do you expect for ##\delta_E##? And I would expect <E> to be negative.

    (b) I would not call it <x> because it is not the expectation value of the position. The formulas look good.
     
  4. Dec 3, 2014 #3
    Yeah I can see that. That is because I left out the most important part.
    At time ##t=0## the particle is in ground state.

    a) Energy is simply ##E=\frac{\hbar^2}{2m}k^2=-\frac{m\lambda ^2}{2\hbar ^2}##. I was also suspicious because energy wasn't negative, but this seems to be a better solution now. No integration needed.

    b) I was worried the notation may be a bit strange, but as long as we both understand it, it's fine. =)

    Also, I got some answers for c) and d) and I would like to check if they make any sense at all:

    c) Since the particle is in ground state, it can only move to higher energy levels, but for that it needs to somehow get that extra energy. Since the problem doesn't state anything about putting extra energy to the system, I assume it is safe to say that the particle will only (and also always - meaning for ##t>0##) be in ground state.
    According to this, the answer to c) is the energy calculated in b) with probability 1. And the answer to part d) is that nothing changes.
     
  5. Dec 3, 2014 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I would answer the same for (c) and (d). They just look a bit odd.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted