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Let $\widehat{a}^{+}_{i}$ and $\widehat{a}_{i}$ be the usual bosonic creation and annihilation operators. Consider $$\widehat{q}_{i} = \sqrt{\frac{\hbar}{2m_{i}w_{i}}}(\widehat{a}_{i}+ \widehat{a}^{+}_{i}).$$ We want to calculate time evolution of $\widehat{q}_{i}(\tau)$ in the interaction picture: $$\widehat{q}_{i}(\tau) = e^{i \widehat{H}_{0} \tau/\hbar} \widehat{q}_{i} e^{-i \widehat{H}_{0} \tau/\hbar},$$ where $$\widehat{H}_{0} = \sum_{i} \frac{1}{2m_{i}} \widehat{p}_{i}^{2} + \frac{1}{2} m_{i} w_{i}^{2} \widehat{q}_{i}^{2},$$ where $m_{i}$ and $w_{i}$ is the mass and natural frequency of the $i$th harmonic oscillator, and $\widehat{q}_{i}$ and $\widehat{p}_{i}$ are the canonical position and momentum operator.

I don't understand why the result is $$\widehat{q}_{i}(\tau) = \sqrt{\frac{\hbar}{2m_{i}w_{i}}}(\widehat{a}_{i} e^{-i w_{i} \tau/\hbar} + \widehat{a}_{i}^{+} e^{i w_{i} \tau/\hbar}). $$

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Compute the time dependence of $a$ in the interaction picture first (no hats here, because it's too much typing). Let $a_I$ denote the lowering operator in the interaction picture. Define the following symbols:

  1. $H_i \equiv \hbar \omega \left( a_i^{\dagger}a_i + 1/2 \right)$

  2. $H_0 \equiv \sum_i H_i$

  3. $K_i \equiv H_i/\hbar$.

  4. $K \equiv H_0/\hbar$

  5. $\bar{a}_i \equiv e^{i K \tau}a_ie^{-i K \tau}$, this is the thing you would probably notate as $\hat{a}_i(\tau)$.

Now we compute the time dependence of $\bar{a}_j$

$$\begin{align} \bar{a}_j &= e^{i K \tau} a_j e^{-i K \tau} \\ \dot{\bar{a}}_j &= e^{i K \tau} a_j (-i K) e^{-i K \tau} + (i K)e^{i K \tau} a_j e^{-i K \tau}\\ &= i \left[ K, e^{i K \tau} a_j e^{-i K \tau} \right] \\ &= i \left[ K, \bar{a}_j \right] \quad (*) \end{align}$$

Suppose we guess that $\bar{a}_j = e^{-i \omega_j \tau}a_j$. Plugging this into $(*)$ we get

$$ \begin{align} \dot{\bar{a}}_j &= i \left[ K, a_j \right] \\ -i\omega_j \bar{a}_j &= \frac{i}{\hbar} \sum_i \left[ H_i, \bar{a}_j \right] \\ &= \frac{i}{\hbar}\sum_i e^{-i\omega_j \tau} \left[ \hbar \omega_i \left( a_i^{\dagger} a + 1/2 \right), a_j\right] \\ 1/2\text{ commutes with }a_j \qquad &= i \sum_i \omega_i e^{-i\omega_j \tau} \left[ a_i^{\dagger} a_i, a_j\right] \\ \text{commutation relation}\qquad &= i\sum_i\omega_ie^{-i \omega_j \tau}(-\delta_{ij}a_i) \\ &= -i \omega_j e^{-i\omega_j \tau} a_j \\ &= -i \omega_j \bar{a}_j \end{align} $$ which is consistent with our guess. Therefore, we have proved that $\bar{a}_j = e^{-i\omega_j \tau}a_j$. By complex/hermetian conjugation you find $\bar{a}^{\dagger}_j = e^{i\omega_j \tau} a_j^{\dagger}$.

The expression for $q_i$ is now obtained by writing $q$ in terms of $a$ and $a^{\dagger}$, as you already did, and then substituting the time dependence we just computed.

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  • $\begingroup$ Here $H_{0}$ is actually the sum of the terms $\hbar w (a_{i}^{+} a_{i} + 1/2)$... $\endgroup$ – nchar Sep 12 '14 at 15:33
  • $\begingroup$ @nchar: Yeah, but the existence of multiple modes doesn't change anything in the calculation. $\endgroup$ – DanielSank Sep 12 '14 at 19:28
  • $\begingroup$ @ncha: Oh sorry, didn't realize you were OP. Do you need an explanation of how to deal with the indices on $a_i$? $\endgroup$ – DanielSank Sep 12 '14 at 19:29
  • $\begingroup$ @nchar: Added more detail, in particular showing how the commutator takes care of all those extra indices for you. $\endgroup$ – DanielSank Sep 12 '14 at 21:13

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