1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Expected value for momentum and x, intro to quantum mechanics.

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider first a free particle (Potential energy zero everywhere). When the particle at
    a given time is prepared in a state ψ (x) it has

    <x> = 0 and <p> = 0.

    The particle is now prepared in Ψ (x, t = 0) = ψ (x) exp (ikx)

    Give <p> at time t = 0.

    It can be shown that in quantum mechanics <p> is independent of time for a free
    particle, in analogy to Newton’s first law in classical mechanics. Give <x> as a
    function of time.

    2. Relevant equations
    <p> = [itex]\int\Psi*(-i[/itex][itex]\hbar[/itex]d/dx)[itex]\Psi[/itex]dx (Should be a partial derivative
    I just don't know how to write that, also integral is from -infinity to +infinity, same for all
    I write further down too)

    3. The attempt at a solution
    The attempt at a solution:

    Basically substituting straight into the equation I gave above:

    <p> = ∫ψ(x)exp(-ikx).-i[itex]\hbar[/itex].d/dx[ψ(x)exp(ikx)]dx

    I guessed that I have to use the product rule for the second half of that so I get:

    <p> = ∫ψ(x)exp(-ikx).-i[itex]\hbar[/itex].dψ(x)/dx.exp(ikx)]dx+∫ψ(x)exp(-ikx).-i[itex]\hbar[/itex].ψ(x)ik.exp(ikx)]dx

    This is where my complete lack of experience shows if not already. Simplifying the first integral, the exponents add to zero for the exp() terms so they cancel. Also, dψ/dx.dx can be simplified to dψ. This leaves the term in the fist integral as -i[itex]\hbar[/itex]ψ.dψ.

    Simplifying the term in the second integral I use the same exponent rule to cancel those and also --i.i = 1. That leaves, in the second integral, the term [itex]\hbar[/itex]kψ2dx.

    I've been doing educated guessing up to here but now I really start guessing.
    I have:

    <p> = -i[itex]\hbar[/itex]∫ψ(x)dx+[itex]\hbar[/itex]k∫ψ2(x)dx

    And by some math voodoo and because I saw somewhere that the integral of psi2dx = 1, I get zero for the first integral and <p> = [itex]\hbar[/itex]k. I suspect I am wrong at many different points for so many different reasons. I hope this doesn't hurt too many people reading it.

    As for the second part, finding <x> as a function of t, I have no idea where to start so even a pointer or two would be appreciated for that. If anyone could point out the numerous mistakes that are no doubt in my attempt at the first part I would also be very grateful. Thanks!
    Last edited: Apr 6, 2013
  2. jcsd
  3. Apr 6, 2013 #2


    User Avatar
    Homework Helper

    hey, welcome to physicsforums!
    You've got it mostly right. But you need to remember that ψ (x) is a complex number generally, so when you take the complex conjugate of the full wavefunction, this bit is going to become its complex conjugate too. And that is how you get psi2dx in the second integral. (Because this is psi times its complex conjugate, not squared).

    Also, in the first integral, you got to ∫ψ(x).-iℏ.dψ(x)/dx.dx and then you wrote it as -iℏψ.dψ which is fine, but then you never showed why this is zero. To be honest, I don't think it helps to re-write it as -iℏψ.dψ And instead, you should leave it as ∫ψ(x).-iℏ.dψ(x)/dx.dx (again, the psi on the left should actually be the complex conjugate). But anyway, if you look at this integral, then think: does it relate to an expectation value of psi ? (Maybe you realised this and didn't remember to write it down, but it is the key step!)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted