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!

Griffiths Introduction to Quantum Physics 2.13.B

  1. Aug 6, 2017 #1
    1. The problem statement, all variables and given/known data
    construct ψ(x,t)^(2) where ψ(x,t) = 1/5(3ψ_0(x)e^(-iE_0t/ħ)+4ψ_1(x)e^(-iE_1t/ħ). I know we square it but we have to find E_0 and E_1 and put it in.

    2. Relevant equations
    E_n = (ħ^(2)k_n^(2))/2m = (n^(2)π^(2)ħ^(2))/2ma^(2)

    3. The attempt at a solution
    E_0 = 0 and E_(1) = (1π^(2)ħ^(2))/2ma^(2)
    Proper solution: E_0 = ħω/2 and E_1 = 3ħω/2
  2. jcsd
  3. Aug 6, 2017 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    You have said nothing of the potential that is assumed for the given problem. This is necessary information as it will affect the energy eigenvalues and eigenstates.
  4. Aug 6, 2017 #3
    The full problem is:
    A particle in the harmonic oscillator potential starts out in the state Ψ(x,0) = A[3ψ_0(x)+4ψ_1(x)]
    A. = Find A, A = 1/5
    B. Construct Ψ(x,t) and |Ψ(x,t)|^2:
    Ψ(x,t) = 1/5[3ψ(x)e^(-((iE_0 t)/ħ)+4ψ(x)e^(-(iE_1 t)/ħ)]

    What I understand is that the the harmonic states is 0 and 1 so n = 0 and n = 1
  5. Aug 6, 2017 #4
    However, I just found the equation
    E_(n) = (n+1/2)ħω

    so it makes sense now, however, I have a question about part C. which is find <x> and <p>.
    so x = (ħ/(2mω))^(½)(a_(+)-a_(-)) and <x> = (1/25)(ħ/(2mω))^(½)(∫Ψ^*(a_+a_-)Ψdx but I am alittle confused on what to do next.
    I believe what happens is a_+3ψ_0 + a_4ψ_1 = 3ψ_(n+1)^(2)e^(-iωt)+4ψ_(n-1)^(2)e^(iωt)= 3(4ψ_1^(2))e^(-iωt)+4(3ψ_(0)^2)e^(iωt) is this the correct understanding?
  6. Aug 6, 2017 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your post is really hard to read. Please take a few minutes to learn LaTeX.

    It's pretty clear the two sides around the first equal sign aren't, in fact, equal. You're not doing yourself any favors being sloppy with the math. Clean up your notation.
  7. Aug 7, 2017 #6
    In part C. I need to find $$\left<x\right>$$ and $$\left<p\right>$$ using Ladder Operators:
    $$\Psi\left(x,t\right) = \frac{1}{5}\left(3\psi_{0}\left(x\right)e^{-\frac{i\omega t}{2}}+4\psi_{1}\left(x\right)e^{-\frac{3i\omega t}{2}}\right)$$
    $$x = \sqrt{\frac{\hbar}{2m\omega}}\left(a_{+}+a_{-}\right)$$
    $$\left<x\right> = \sqrt{\frac{\hbar}{2m\omega}}\int \Psi^{*}\left(a_{+}+a_{-}\right)\Psi_{n} (Eq1)$$, I know that the next step in the solution is $$\left<x\right>=\sqrt{\frac{\hbar}{2m\omega}}\int\left(12\psi_{0}^{2}e^{-i\omega t}+12\psi_{1}^{2}e^{i\omega t}\right)dx(Eq2)$$ and I am confused on how you get from Eq1 to Eq2. My idea of the step in between is
    $$\left<x\right>=\sqrt{\frac{\hbar}{2m\omega}}\int \left(3a_{+}\psi_{0} +4a_{-}\psi_{1}\right)$$
  8. Aug 7, 2017 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The usual distributive rules apply here. Multiply everything out as you usually would to calculate ##(a_+ + a_-)(c_1 \psi_0 + c_2\psi_1)##.
  9. Aug 7, 2017 #8
    from that distribution we get $$ \left(3a_{+}\psi_{0}\left(x\right)e^{-\frac{i\omega t}{2}}+4a_{-}\psi_{1}\left(x\right)e^{-\frac{3i\omega t}{2}}+3a_{-}\psi_{0}e^{-\frac{i\omega t}{2}}+4a_{-}\psi\left(x\right)e^{-\frac{3i\omega t}{2}}\right)$$
    $$a_{+}=\sqrt{n+1}\psi_{n+1}, a_{-}=\sqrt{n}\psi_{n-1}$$
    and using that operate you get:
    $$ 3\sqrt{n+1}\psi_{1}\left(x\right)e^{-\frac{i\omega t}{2}}+4\sqrt{n+1}\psi_{2}\left(x\right)e^{-\frac{3i\omega t}{2}}+3\sqrt{n}\psi_{-1}e^{-\frac{i\omega t}{2}}+4\sqrt{n}\psi_{0}\left(x\right)e^{-\frac{3i\omega t}{2}}$$
    which is
    $$ 3\sqrt{0+1}\psi_{1}\left(x\right)e^{-\frac{i\omega t}{2}}+4\sqrt{1+1}\psi_{2}\left(x\right)e^{-\frac{3i\omega t}{2}}+3\sqrt{0}\psi_{-1}e^{-\frac{i\omega t}{2}}+4\sqrt{1}\psi_{0}\left(x\right)e^{-\frac{3i\omega t}{2}}$$
    So you get:
    $$ 3\psi_{1}\left(x\right)e^{-\frac{i\omega t}{2}}+4\sqrt{2}\psi_{2}\left(x\right)e^{-\frac{3i\omega t}{2}}+4\psi_{0}\left(x\right)e^{-\frac{3i\omega t}{2}}$$
    Then multiplying by $$\Psi^{*}$$ we get:
    $$9\psi_{0}\psi_{1}+12\sqrt{2}\psi_{2}\psi_{0}e^{-i\omega t}+12\psi_{0}^{2}e^{-1i\omega t}+12\psi_{1}^{2}e^{i\omega t}+16\sqrt{2}\psi_{2}\psi_{1}+16\psi_{0}\psi_{2}$$
    and since $$\psi_{0}$$, $$\psi_{2}$$ and $$\psi_{1}$$ are all orthogonal to each other we get:
    $$12\psi_{0}^{2}e^{-i\omega t}+12\psi_{1}^{2}e^{i\omega t}$$
    Is that the correct process of thinking, I get the solution but I just want to make sure I got the right methodology.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Griffiths Introduction to Quantum Physics 2.13.B
  1. Griffiths Quantum 2.51 (Replies: 4)

  2. Griffiths Quantum 4.58 (Replies: 5)