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Question about creation and annihilation operators?

  1. Nov 29, 2014 #1
    Hello! I am reading about the creation and annihilation operators and I don't get how you find the creation operator from the annihilation one. The creation one is

    [itex] \hat{a}=\sqrt{\frac{m \omega}{2 \hbar}}\left( \hat{x}+\frac{i \hat{p}}{m \omega}\right) [/itex]

    and the annihilation operator is [itex] \hat{a}\dagger =\sqrt{\frac{m \omega}{2 \hbar}}\left( \hat{x}-\frac{i \hat{p}}{m \omega}\right) [/itex]

    I don't understand how taking the complex conjugate an transpose leads to the minus sign. I thought that [itex]\hat{x}=x[/itex] and [itex]\hat{p}=i\hbar\frac{\partial}{\partial x}[/itex] so [itex]\hat{x}\dagger=\hat{x}=x[/itex] but
    [itex]\hat{p}\dagger=-i\hbar\frac{\partial}{\partial x}=-\hat{p}[/itex] which will cancel with the minus sign from complex conjugating the [itex]i[/itex], so that [itex] \hat{a}=\hat{a}\dagger[/itex].

    I'm sure that this mistake is due to my unfamiliarity with the algebra of operators. I would appreciate it if someone could say which of my assumptions is wrong! My hunch is that it has something to do with taking the dagger of [itex]\left( \hat{x}+\frac{i \hat{p}}{m \omega}\right) [/itex] and this not being equal to taking the dagger of the individual operators...

    Thanks in advance!
  2. jcsd
  3. Nov 29, 2014 #2


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    The problem is where you computed ##p^{\dagger}##. Since this is an operator corresponding to an observable (momentum), you know that it must be hermitian so that ##p^{\dagger} = p##. Therefore your error is in the computation of ##p^{\dagger}##. Find first how the operator ##d/dx## transforms under a hermitian conjugate.
  4. Nov 29, 2014 #3


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    For an operator, [itex]\hat{O}[/itex], the definition of [itex]\hat{O}^\dagger[/itex] is a little technical: If you have two wave functions [itex]\psi[/itex] and [itex]\phi[/itex], then according to Dirac notation:

    [itex]\langle \phi | \psi \rangle = \int \phi(x)^* \psi(x) dx[/itex]

    Then the meaning of [itex]\hat{O}^\dagger[/itex] is given by:

    [itex]\langle \phi | \hat{O} \psi \rangle = \langle \hat{O}^\dagger \phi | \psi\rangle[/itex]

    In the case of a "hermitian" operator, [itex]\hat{O}^\dagger = \hat{O}[/itex], so we can just write:

    [itex]\langle \phi | \hat{O} | \psi \rangle [/itex]

    and not worry about whether [itex]\hat{O}[/itex] is acting to the right, on [itex]\psi[/itex], or acting to the left, on [itex]\phi[/itex].

    In the special case where [itex]\hat{O}[/itex] is the derivative operator, [itex]\frac{d}{dx}[/itex], you can prove[itex]\dagger[/itex]:

    [itex]\langle \phi | \frac{d}{dx} \psi \rangle = \langle (- \frac{d}{dx} \phi) | \psi \rangle[/itex]

    So [itex](\frac{d}{dx})^\dagger = - \frac{d}{dx}[/itex]

    [itex]\dagger[/itex]This is only true when [itex]\phi[/itex] and [itex]\psi[/itex] are well-behaved wave functions. You prove this by integration by parts, and reasoning that [itex]\phi[/itex] and [itex]\psi[/itex] go to zero as [itex]x \rightarrow \pm \infty[/itex].
  5. Nov 29, 2014 #4
    Brilliant! Thank you both! I see now how [itex]\hat{p}\dagger=\hat{p}[/itex], but I'm also wondering whether you can distribute the 'dagger' over the bracket [itex]\left(\hat{x}+\frac{i\hat{p}}{m\omega}\right)[/itex] by simply taking the 'dagger' of the components and [itex]i[/itex]? Or am I thinking about this wrong?

    Thank you! :)
  6. Nov 29, 2014 #5


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    Yes, the [itex]\dagger[/itex] operation distributes.
  7. Nov 29, 2014 #6


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    Technically, [itex](AB + C)^\dagger = B^\dagger A^\dagger + C^\dagger[/itex]
  8. Nov 30, 2014 #7


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    The harmonic oscillator (in 1D) is the simplest possible example of how functional analysis makes the heart of quantum mechanics. A nice propery which enables us to use ladder operators is the fact that [itex]\mathcal{S}(\mathbb{R})[/itex] the Schwartz test functions space is a common dense everywhere domain of essential self-adjointness for both x and p. Hence it makes sense to consider the adjoint of [itex] a = x + \alpha p[/itex] , where [itex]\alpha [/itex] is a complex constant (number, not operator).

    [tex] a^{\dagger} = \left(x + \alpha p \right)^{\dagger} \supseteq x^{\dagger} + (\alpha p)^{\dagger} \supseteq x +\alpha^{*} p [/tex]

    Since [itex] i^{*} = -i [/itex], that's how you end up with a minus before p. If the operators are restricted to [itex]\mathcal{S}(\mathbb{R})[/itex] from [itex]\mathcal{L}^2(\mathbb{R})[/itex], you're safe to use the equality sign.
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