Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    CAF123

    User Avatar
    Gold Member

    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

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

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

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

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

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question about creation and annihilation operators?
Loading...