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Displacement Operator definition

  1. Aug 13, 2012 #1

    Could someone explain how the following two definitions of the displacement operator are equal? The first is the standard one

    1) [itex]e^{\alpha a^{\dagger}-\alpha*a}[/itex]

    But what about this one? This is from a Fock state decomposition of a coherent state.

    2) [itex]e^{\frac{-|\alpha^{2}|}{2}}e^{\alpha a^{\dagger}}[/itex]

    What is the reasoning behind not using the second equation to define the operator that acts on a vacuum state to produce a coherent state? Or if it is fine to use definition 2), then can someone help me show that the two definitions are equivalent?

    Thank you
  2. jcsd
  3. Aug 13, 2012 #2


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    They're not equal as operators, but if you act on the vacuum with them, e.g.,
    $$e^{\alpha a^{\dagger}-\bar\alpha a} \, |0\rangle$$
    you should be able to use one of the BCH theorems (Baker-Campbell-Hausdorff) -- or is it Zassenhaus? -- to convert the exponential into a product of exponentials, where the rightmost one involves only the annihilation operator, and hence leaves the vacuum unchanged.


    The advantage of version (1) is that its formally unitary. Both are useful in different circumstances. Try Mandel & Wolf for lots more on this.
  4. Aug 13, 2012 #3
    Using the BCH equations, the article arrived at [itex]e^{\frac{|\alpha^{2}|}{2}}e^{\alpha a^{\dagger}}e^{\alpha^{*} a}[/itex] for the displacement operator. Now, I expand the last exponential term in powers of a.

    But if the rightmost one only involves the annihilation operator, then we get zero. Isn't it true that [itex]a\left|0\right\rangle = 0[/itex], not [itex]\left|0\right\rangle[/itex]?
  5. Aug 13, 2012 #4


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    It's an exponential whose exponent involves only a. And e0 = 1.
  6. Aug 14, 2012 #5
    Oh of course! I forgot the identity term in the Taylor expansion. Thank you very much Bill_K and strangerep
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