Quantum Mechanics - Finding expectation value

  1. 1. The problem statement, all variables and given/known data

    Find the expectation value of position as a function of time.

    2. Relevant equations

    This is in the latter half of a multi-part question, previously we were given that:

    Eqn 1: Ψ(x, t) = A(ψ1(x)e−iE1t/h¯ + iψ2(x)e−iE2t/h¯)

    and in an even earlier part:

    Eqn 2: ψn(x) = sqrt(2/L)sin(n*pi*x/L)

    Note: h¯ = hbar

    3. The attempt at a solution

    As you can tell, I'm not too awesome at formatting on here so I'm going to quickly explain my method.

    I said:

    <x> = Integral from 0 to L of Ψ*(x, t)Ψ(x, t) dx

    So I told wolframalpha (we're allowed to use it) to simplify Eqn 1, before subbing in Eqn 2.

    This gave me:

    Eqn 3: -2(ψ1ψ2sin(t(E1-E2)/h¯) + ψ12 + ψ22

    I then subbed in Eqn 2 into Eqn 3 and integrated.

    No matter how many times I do this I always end up with a bunch of sines. These sines are all something like sin(n*pi/L) so when I sub in the limits of integration they become sin(n*pi) or sin(0), both of which are 0!

    This means that I'm just left with an answer of 2, which is not time dependent.

    Can anyone see where I've gone wrong?
     
  2. jcsd
  3. Okay so I just found out I've had the formula wrong from the start. FML.

    It's now:

    <x> = Integral from 0 to L of Ψ*(x, t)xΨ(x, t) dx

    However, this still didn't help me.

    I'm now left with:

    A2(L - L/(2pi2) + L/(4pi2) + L/(16pi2))

    And I am thoroughly sick of this.
     
  4. vela

    vela 12,780
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    What happened to the factor of [itex]\sin\left[\frac{(E_1-E_2)t}{\hbar}\right][/itex]?

    After you multiply the integrand out, you should have
    \begin{align*}
    \langle x \rangle &= A^2\int_0^L \left(x\psi_1^2(x) + x\psi_2^2(x) - 2x\psi_1(x)\psi_2(x) \sin [(E_1-E_2)t/\hbar]\right)\,dx \\
    &= A^2\left(\int_0^L x\psi_1^2(x)\,dx + \int_0^L x\psi_2^2(x)\,dx - 2\sin [(E_1-E_2)t/\hbar]\int_0^L x\psi_1(x)\psi_2(x)\,dx\right)
    \end{align*}
    As long as that last integral isn't 0, the expectation value will vary with time.
     
    Last edited: Aug 11, 2011
  5. I love you vela. Thank you so much.
     
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