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Do you know what this particular unitary operator is?

  1. Jun 21, 2005 #1
    Recall that the unitary operator [itex] exp(\frac{i}{\hbar}aX) [/itex] transform the operator [itex] P_x +a [/itex] to [itex] P_x [/itex].

    Now, what is the unitary operator that transform the operator [itex] P_x +aX [/itex] to [itex] P_x [/itex] ??

  2. jcsd
  3. Jun 21, 2005 #2
    you mean P_x to P_x + a

    [itex] exp(- \frac{i}{\hbar}aX) [/itex]

    : )


    EDIT: sorry bad reading:
    [itex] exp( \frac{i}{\hbar}aX^2/2) [/itex] is the answer to your question
    Last edited: Jun 21, 2005
  4. Jun 21, 2005 #3
  5. Jun 21, 2005 #4
    There is no typo.
  6. Jun 21, 2005 #5
    See corrected previous post.

  7. Jun 21, 2005 #6
    Thanks. Thats the answer i tempted to throw in too. But i have my doubts.. :confused:
  8. Jun 21, 2005 #7
    Doubts, why? You just have to calculate.
    Hint: try to see what the unitary operator exp(i a(X)) does on the momentum operator.

  9. Jun 21, 2005 #8
    My bad, Seratend. You are absolutely right! Case close :)
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