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Proof for commutator $[\hat{H},\hat{a}] = - \hbar \omega \hat{a}

  1. May 4, 2013 #1
    I know how to derive below equations found on wikipedia and have done it myselt:

    \begin{align}
    \hat{H} &= \hbar \omega \left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right)\\
    \hat{H} &= \hbar \omega \left(\hat{a}\hat{a}^\dagger - \frac{1}{2}\right)\\
    \end{align}

    where ##\hat{a}=\tfrac{1}{\sqrt{2}} \left(\hat{P} - i \hat{X}\right)## is a annihilation operator and ##\hat{a}^\dagger=\tfrac{1}{\sqrt{2}} \left(\hat{P} + i \hat{X}\right)## a creation operator. Let me write also that:

    \begin{align}
    \hat{P}&= \frac{1}{p_0}\hat{p} = -\frac{i\hbar}{\sqrt{\hbar m \omega}} \frac{d}{dx}\\
    \hat{X}&=\frac{1}{x_0} \hat{x}=\sqrt{\frac{m\omega}{\hbar}}x
    \end{align}

    In order to continue i need a proof that operators ##\hat{a}## and ##\hat{a}^\dagger## give a following commutator with hamiltonian ##\hat{H}##:

    \begin{align}
    \left[\hat{H},\hat{a} \right] &= -\hbar\omega \, \hat{a}\\
    \left[\hat{H},\hat{a}^\dagger \right] &= +\hbar\omega \, \hat{a}^\dagger
    \end{align}

    These statements can be found on wikipedia as well as here, but nowhere it is proven that the above relations for commutator really hold. I tried to derive ##\left[\hat{H},\hat{a} \right]## and my result was:

    $$
    \left[\hat{H},\hat{a} \right] \psi = -i \sqrt{\frac{\omega \hbar^3}{4m}}\psi
    $$

    You should know that this this is 3rd commutator that i have ever calculated so it probably is wrong, but here is a photo of my attempt on paper. I would appreciate if anyone has any link to a proof of the commutator relations (one will do) or could post a proof here.
     
  2. jcsd
  3. May 4, 2013 #2

    strangerep

    User Avatar
    Science Advisor

    I presume you know the commutation relations between P and X? If so, write them down here.
    From that, you can work the commutator ##[a, a^\dagger]##.
    From that, you can work out your ##[H,a]##, etc. (Hint: apply the Leibniz product rule.)
     
  4. May 4, 2013 #3
    \begin{align}
    \left[\hat{X}, \hat{P}\right] = \left[\tfrac{1}{x_0}\hat{x},\tfrac{1}{p_0}\hat{p}\right]=\tfrac{1}{x_0p_0} \big[\!\!\!\!\!\!\!\!\!\!\!\!\!\ \underbrace{\hat{x},\hat{p}}_{i\hbar \text{... calculated below}} \!\!\!\!\!\!\!\!\!\!\!\! \big]= \frac{1}{x_0 p_0} \hbar i = \sqrt{\frac{m \omega }{\hbar \hbar m \omega}} \hbar i = i
    \end{align}

    \begin{align}
    \left[ \hat{x}, \hat{p} \right] \psi &= \hat{x}\hat{p} \psi - \hat{p} \hat{x} \psi = - x\,i \hbar \frac{d \psi}{d x} - \left(- i\hbar \frac{d}{dx} \left(x \psi \right)\right) = - x\, i \hbar \frac{d \psi}{dx} + i \hbar \underbrace{\frac{d}{dx}\left( x \psi \right)}_{\text{product rule}} =\\
    &= - x\, i \hbar \frac{d \psi }{d x} + i\hbar\left( \frac{d x}{d x} \psi + x \frac{d \psi}{d x} \right) = -x\,i\hbar \frac{d \psi}{dx} + i \hbar \psi + x\, i \hbar \frac{d \psi}{dx} = i \hbar \psi
    \end{align}

    I don't quite know how to start this.

    Lets leave this for later and work out ##[a, a^\dagger]## first.
     
  5. May 4, 2013 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Use that the commutator bracket is linear. So ##[\alpha A + \beta B, C] = \alpha [A,C] + \beta [B,C]## and the same thing in the other variable.
     
  6. May 5, 2013 #5
    You are good to go, note
    [tex]
    a=\frac{1}{\sqrt{2}}(\hat{P}-i\hat{X})
    [/tex]
    and
    [tex]
    a^{\dagger} = \frac{1}{\sqrt{2}}(\hat{P}+i\hat{X})
    [/tex]
    now
    [tex]
    [a,a^{\dagger}] = \frac{1}{2}\left[ (\hat{P}-i\hat{X}),(\hat{P}+i\hat{X}) \right]
    [/tex]
    and bust it out using what you just showed us.
     
  7. May 7, 2013 #6
    I think i wrote this wrong as it is:

    [tex]
    a=\frac{1}{\sqrt{2}}(\hat{X}-i\hat{P})
    [/tex]
    and
    [tex]
    a^{\dagger} = \frac{1}{\sqrt{2}}(\hat{X}+i\hat{P})
    [/tex]
     
  8. May 7, 2013 #7
    Thank you all. I just solved this and it feels great. Now that i know following commutators:

    \begin{align}
    [\hat{H}, \hat{a}] &= -\hbar \omega \hat{a}\\
    [\hat{H}, \hat{a}^\dagger] &= \hbar \omega \hat{a}^\dagger\\
    \end{align}

    On the Wikipedia it is said that these commutators can be used to find eigenstates of Quant. harm. oscilator, but explaination is a bit too fast there. Anyway i strive to be able to derive the equation $W_n = \hbar \omega \left(n + \tfrac{1}{2}\right)$ in full, but first i need to clear why theese two relations hold:

    \begin{align}
    \hat{H}\hat{a} \psi_n &= (W_n - \hbar \omega) \hat{a} \psi_n\\
    \hat{H}\hat{a}^\dagger \psi_n &= (W_n + \hbar \omega) \hat{a}^\dagger \psi_n
    \end{align}

    I can't see any commutators in above relations, so how do the commutators i just calculated help us to get and solve these two relations? And i am sorry for asking so basic questions but i am a selftaught and a real freshman to commutators algebra.
     
  9. May 7, 2013 #8

    strangerep

    User Avatar
    Science Advisor

    Let's stick to the notation used in that Wikipedia page. I.e., let's change your ##W_n## to ##E_n##.
    As explained on the Wiki page, we have
    $$
    H \psi_n ~=~ E_n \psi_n ~.
    $$
    So to evaluate ##H a \psi_n##, use the ##[H,a]## formula to swap the order of ##H## and ##a##, and then use the equation above.

    BTW, are you self-learning from a textbook? If so, which one? (If not, you probably need to invest in one...) Questions at this level should more properly be posted in the homework forums.
     
  10. May 8, 2013 #9
    homework problem. move to other thread.
     
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