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An operator acting on the translated ground state of an SHO

  1. Nov 10, 2015 #1
    I am trying to perform the operation a on a translated Gaussian, ie. the ground state of the simple harmonic oscillator (for which the ground state eigenfunction is e^-((x/xNot)^2). First, I was able to confirm just fine that a acting on phi-ground(x) = 0. But when translating by xNot, so a acting on phi-ground(x-xNot), I am supposed to get a acting on phi-ground(x-xNot) = C*phi-ground(x-xNOT). But I am getting a acting on phi-ground(x-xNot) = + C1 * phi-ground(x-xNot) + C2 * x * phi-ground(x-xNot) ---so I appear to be doing something wrong, as I am getting the second (linearly independent) term, which is the discrepancy between what I was told was the answer. Can anyone help? Thanks!
     
  2. jcsd
  3. Nov 10, 2015 #2
    In case it helps, the problem is described in this video starting at 6:15 into the video through about 8:00 (so only a few minutes to watch that part).
     
  4. Nov 11, 2015 #3
    "a" = annihilation operator btw
     
  5. Nov 11, 2015 #4

    blue_leaf77

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    The second term, was that obtained by acting the term proportional to ##\hat{x}## in ##a## to ##\phi_0(x-x_0)##? Which means the first term is obtained from the action of ##\hat{p}## onto this state? If yes, you should check again your answer for the first term. There should be a resulting term which will cancel the second term (there should be two terms proportional to ##x## but with opposite sign).
     
  6. Nov 11, 2015 #5
    Sorry, yes, I was in error...I rechecked the work and did get one term only, and it was proportional to x (actually x - xNot in this case), so it was actually: C1 (x-xNot) * phi-ground(x-xNot), where C1 = some other constant * (1/xNot -1)...but according to that video the answer is just C1 phi-ground(x-xNot), without the (x-xNOT) factor, I'm assuming?
     
  7. Nov 11, 2015 #6

    blue_leaf77

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    Have you checked whether or not the term proportional to ##x\phi_0(x-x_0)## from the application of ##\hat{p}## cancels the other term proportional ##x\phi_0(x-x_0)## resulting from the application of ##\hat{x}##?
     
  8. Nov 11, 2015 #7
    It does as you say for the unshifted phi-ground, but for the shifted one it yields the x*phi-ground coefficient when I evaluate as you suggest, ie. on the left side, to yield this term on the ride side of the equation. Either I have an error or am misunderstanding something. The annihilation operator on the unshifted ground state yields zero, but is supposed to yield the eigenfunction itself for the shifted ground state. Instead, I get that it yields x times the eigenfunction.
     
  9. Nov 11, 2015 #8

    blue_leaf77

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    We have ##a = \sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} + \frac{i}{m\omega}\hat{p} \right)## and ##\phi_0(x-x_0) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp \left( -\frac{m\omega (x-x_0)^2}{2\hbar} \right)##. What do you get after calculating ##\hat{p}\phi_0(x-x_0)##?
     
  10. Nov 11, 2015 #9
    Thank you for your response. actually in the exponential of the eigenfunction you actually should have 2* hbar in the numerator and m * omega in the denominator--so you have the reciprocal of the coefficient...as a check the exponential should be dimensionless, hence the xNot in the denominator. Also, just to clarify, xNot = sqrt(m*omega/(2*hbar))

    So them for p acting on phi-ground(x - xNot) I get - 2 * i * hbar* ((x-xNot)/xNot^2)*phi-ground(x-xNot) --- and then multiplying by I/pNot, I get --- now I see my previous error-- - (x-xNot)/pNot, which exactly cancels my first term obtained by the x operator/xNot...so now it evaluates to 0.

    So obviously I have made a few errors along the way, but now I get that the translated Gaussian eigenfunction also evaluates to zero? Alan Adams said it should evaluate to a constant coefficient of the eigenfunction itself?
     
  11. Nov 11, 2015 #10

    blue_leaf77

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    I don't think so, you can check the dimensionality in my expression and it should fulfill the requirement that the term in the exponent should be dimensionless.
    ##x_0## is to have a dimension of length, but if it's defined as "xNot = sqrt(m*omega/(2*hbar))", it will be an inverse of a length.
    It will help identify your error if you can write the complete expression of your translated ground state wavefunction, preferably in Latex format.
     
  12. Nov 11, 2015 #11
    Sorry, yes, I need to learn LaTex, but in this case I am now getting zero...just as you seemed to be stating above...whether or not I shift...so it would be great if you could check out this video starting at 6:15 into the video through about 8:00 (I know it is a little pesky but keep clicking a few times and you should be able to queue correctly), as I think I may just have misunderstood something there.

    Shortly stated, I get that the annihilation operator acting on the ground state eigenfunction yields zero whether or not x is shifted.
     
  13. Nov 11, 2015 #12
    Now that I have corrected my errors, this is precisely what I am getting. The question is: why then is Alan Adams getting something else?? Please watch the portion of the video I referenced above and let me know what your think! Thanks in advance!!
     
  14. Nov 11, 2015 #13
    OK, I believe I figured this all out. I had inserted for the position operator, x - xNot, which means I shifted the POTENTIAL along with the parameter of the eigenfunction, and of course annihilated to zero; if you shift everything over you get the same thing! I, instead, was NOT supposed to shift the potential, so the x operator is just x. Then I simply get a 1, which means a constant times the eigenfunction...just like the prof got! And indeed that must be correct, because the term containing the x operator is the term that represents the SHO potential! The term containing the momentum operator is describing the kinetic energy component, which ultimately depends on the potential!
     
  15. Nov 12, 2015 #14

    blue_leaf77

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    Such mistake could have been identified earlier if you had written what you have done in your own effort in your post. Regarding this, this type of question is better posted under Homework section where the poster will be required to show his/her initial attempts in solving the problem, and an identification of errors can be easily traced back.
     
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