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Harmonic Oscillator and Ladder Operators

  1. Jan 5, 2016 #1
    1. The problem statement, all variables and given/known data

    Consider a linear harmonic oscillator with the solution defined by the ladder operators a and a. Use the number basis |n⟩ to do the following.

    a) Construct a linear combination of |0⟩ and |1⟩ to form a state |ψ⟩ such that ⟨ψ|X|ψ⟩ is as large as
    possible.

    b) Suppose that the oscillator is in the state constructed in a) at time t = 0. Write an expression to describe the time dependence of this state state for t > 0. Evaluate the expectation value ⟨ψ(t)|X|ψ(t)⟩ as a function of time for t > 0.

    c) Defining (∆x)2 = ⟨ψ(t)|X2|ψ(t)⟩ − ⟨ψ(t)|X|ψ(t)⟩2. Calculate (∆x)2.

    2. Relevant equations

    Screen Shot 2016-01-05 at 17.21.38.png Screen Shot 2016-01-05 at 17.21.50.png

    3. The attempt at a solution

    I have a hunch that for part a) I need to consider some |ψ⟩ = A|0⟩ + B|1⟩, then differentiate the expression ⟨ψ|X|ψ⟩. However I'm not sure how to implement that, any hints?

    Many thanks
     
  2. jcsd
  3. Jan 5, 2016 #2

    PeroK

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    Why not just go ahead and calculate <X> for arbitrary ##A, B## as your hunch suggests?
     
  4. Jan 6, 2016 #3
    Thanks for the reply.

    Ok so I substituted |ψ⟩ = A|0⟩ + B|1⟩ into ⟨ψ|X|ψ⟩, and from simple manipulation I have:

    <X> = A*A⟨0|X|0⟩ + A*B⟨0|X|1⟩ + B*A ⟨1|X|0⟩ + B*B ⟨1|X|1⟩

    Is this correct? I wonder what can be done to simplify it, should I substitute X in terms of a and a†?
     
  5. Jan 6, 2016 #4

    PeroK

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    Yes, it's correct. You need to know (or calculate) ##<0|X|0>## and ##<1|X|1>##. You have to calculate ##<0|X|1>## as well. The other term is related to this (hint: think about complex numbers).

    You can use the ladder operator or just integrate (if you have a list of standard Gaussian type integrals).
     
  6. Jan 6, 2016 #5
    I've ended up with:

    <X> = (h/2mw)2 [A*B + B*A]

    Now to obtain |ψ⟩ which maximizes <X>, we set <X> = 0 and solve for A and B. This gives: A*B = -B*A. How would one solve for the two unknowns then?
     
  7. Jan 6, 2016 #6

    PeroK

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    I'd double check that factor you've got.

    I'll give you that ##A^*B + B^*A = A^*B + (A^*B)^* = 2Re(A^*B)##. It's useful not to forget that.

    To maximise this, you know that ##|A|^2 + |B|^2 = 1##. Try expressing these in polar form with magnitudes ##a, b## and note that:

    If ##a, b## are real and ##a^2 + b^2 = 1## then ##a = cos(\alpha)## and ##b = sin(\alpha)## for some ##\alpha##. It's useful to remember that as well.

    It's not too hard to maximise ##Re(A^*B)## using this.
     
  8. Jan 6, 2016 #7
    Ah oops, is it: <X> = (h/2mw)1/2 [A*B + B*A]

    I ended up with α = π/4.

    And so |ψ⟩ = A|0⟩ + B|1⟩ = cos(π/4) |0⟩ + sin(π/4) |1⟩

    Not sure about that part as we've solved for the real parts of A and B, but not A and B themselves.
     
  9. Jan 6, 2016 #8

    PeroK

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    I'd move on with with ##A = B = \frac{1}{\sqrt{2}}##

    ##Re(A^*B) = Re(ae^{-i\theta_A} be^{i \theta_B}) = abcos(\theta_B - \theta_A)## was what you missed out.
     
  10. Jan 6, 2016 #9
    Ah I see now, so it's the case that:
    ##|ψ⟩ = \frac{1}{\sqrt{2}}|0⟩ + \frac{1}{\sqrt{2}}|1⟩ ##
     
  11. Jan 6, 2016 #10

    PeroK

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    That's the simplest. I.e. taking ##\theta_A = \theta_B = 0##. Although, any common ##\theta## would have done.
     
  12. Jan 6, 2016 #11
    Is it worth putting ##\theta## in or considering the simplest case for the following parts do you think?
     
  13. Jan 6, 2016 #12

    PeroK

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    Definitely not! It says "construct a state ...". Always take the easiest option.
     
  14. Jan 6, 2016 #13
    Good point :).

    For the first part of b) I've got the following, just want to check it's right.

    ##|ψ⟩ = e^{\frac{-iE_0t}{h}}\frac{1}{\sqrt{2}}|0⟩ + e^{\frac{-iE_1t}{h}}\frac{1}{\sqrt{2}}|1⟩ ## where ## E_n ## is the nth energy eigenvalue given by ##E_n = (n+\frac{1}{2})hw##
     
  15. Jan 6, 2016 #14

    PeroK

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    Yes, that's right.
     
  16. Jan 7, 2016 #15
    Great! So I've calculated ## ⟨ψ|X|ψ⟩ ## and obtained:

    ##⟨ψ|X|ψ⟩= \sqrt{\frac{h}{8mw}} (e^{\frac{it(E_0 - E_1)}{h}} + e^{\frac{it(E_1 - E_0)}{h}}) ##

    As for part c, for ## ⟨ψ|X|ψ⟩^2 ## I square the expression above to get:

    ##⟨ψ|X|ψ⟩^2 = \frac{h}{8mw} (e^{\frac{2it(E_0 - E_1)}{h}} + e^{\frac{2it(E_1 - E_0)}{h}} + 2) ##

    Is that correct so far.
     
  17. Jan 7, 2016 #16

    PeroK

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    You are going to find QM difficult if you keep forgetting about complex conjugates! You need to simplify those expressions.
     
  18. Jan 7, 2016 #17
    Yes, I should've clicked on. It would simplify to:

    ##⟨ψ|X|ψ⟩= \sqrt{\frac{h}{2mw}} cos(\frac{t}{h} (E_0 - E_1)) ##

    And so:

    ##⟨ψ|X|ψ⟩^2= \frac{h}{2mw} cos^2(\frac{t}{h} (E_0 - E_1)) ##
     
  19. Jan 7, 2016 #18

    PeroK

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    You can do something with ##E_0 - E_1## as well. And, remember that ##cos## is an even function.
     
  20. Jan 7, 2016 #19
    Aha!

    ##⟨ψ|X|ψ⟩= \sqrt{\frac{h}{2mw}} cos(wt) ##

    And so:

    ##⟨ψ|X|ψ⟩^2= \frac{h}{2mw} cos^2(wt) ##

    Now it looks splendid.
     
  21. Jan 7, 2016 #20
    Final thing, for ##⟨ψ|X^2|ψ⟩## I got:

    ##⟨ψ|X^2|ψ⟩= \frac{h}{mw} ##, is that correct?
     
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