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Translation operator on ground state

  1. May 9, 2015 #1
    1. The problem statement, all variables and given/known data
    I am working through a time independent perturbation problem and I am calculating the first order correction to the energy, and I am stuck operating the perturbation : v = i b (Exp[i g x]-Exp[-i g x]) on the ground state |0>.

    2. Relevant equations
    <0| v |0> = 1st order correction

    3. The attempt at a solution
    I need to be able to operate the exponential on the ground state.
    I did turn the exponentials in X into terms of creation and annilation operators.
     
  2. jcsd
  3. May 9, 2015 #2

    blue_leaf77

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    What kind of system are you talking about, harmonic oscillator? I just want to clarify, is that perturbation term you have there the only one or are there actually another perturbation?
    If exp(igx) is a translation operator in space then g is the one as an operator while x is just a number.
     
  4. May 10, 2015 #3
    Guess it's x that is the operator and yes it's a harmonic oscillator and the perturbation is a trig function. But if I write it as an exponential I get the x in exponential. I can write it as a combination of creation and Annihilation operators, then separate them. But how do these act on the ground state
     
  5. May 10, 2015 #4
    Guess it's x that is the operator and yes it's a harmonic oscillator and the perturbation is a trig function. But if I write it as an exponential I get the x in exponential. I can write it as a combination of creation and Annihilation operators, then separate them. But how do these act on the ground state
     
  6. May 10, 2015 #5

    blue_leaf77

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    What does the question ask precisely? Is that the only thing the question asks you to calculate? I have got a feeling that that perturbation term doesn't actually contribute anything to the energy of the ground state.
     
  7. May 10, 2015 #6
    the perturbation is in the form of v sin(cx). I need to calculate the lowest non vanishing term to the energy of the ground state.. Im assuming that the nature of sine being an odd function will make first order terms disappear, but not the second order terms.
     
  8. May 11, 2015 #7

    blue_leaf77

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    Alright, now it's clear what the problem actually wants you to show.
    You know the answer already, so in which part of the problem you are still having trouble with?
    If you know the position representation of ##|0\rangle##, you can compute <0|v|0> in the integral form. If the problem ask you to proceed using raising and lowering operator, then it's better to expand v in terms of power series of x.
     
    Last edited: May 11, 2015
  9. May 11, 2015 #8
    right. The First order terms disappear. But the second order terms are the ones I'm worried about now.
    Σ(|<m| v |0>|^2)/(Em-Eo)
     
  10. May 11, 2015 #9

    blue_leaf77

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    I haven't tried to work that one out yet, but if the problem only asked you to identify the first non-vanishing perturbation order, without requiring the closed form answer, you can use merely mathematical reasoning. Notice that all terms in that series is positive, so there is no chance for two or more terms to cancel each other. Second since the series runs over all eigenstates, the ones that contribute are those which have opposite parity as |0>. Thus all terms being summed will give value greater than zero.
    Otherwise, you have to go to the math using the raising and lowering operators.
     
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