Specifically, given the system: $$H = \omega_0 (a^\dagger a + 1/2); \qquad q = \frac{a + a^\dagger}{2 \omega_0}; \qquad [a,a^\dagger] = 1; $$ $$q(t) = e^{i Ht} q e^{-i Ht};$$ How would you go about showing that: $$q(t) = \frac{1}{2 \omega_0} \bigg( a e^{-i \omega_0 t} + a^\dagger e^{i \omega_0 t} \bigg)~?$$

I don't really know how to deal properly with exponentials of operators. My first, and only, intuition is just to expand the exponential and hope the commutation makes everything really easy and obvious to see: $$q(t) = \frac{1}{2 \omega_0}\bigg( \sum_{n = 0}^\infty (i \omega_0 t \ a^\dagger a)^n \bigg) (a + a^\dagger) \bigg( \sum_{m = 0}^\infty (i \omega_0 t \ a^\dagger a)^m \bigg)$$ $$= \frac{1}{2 \omega_0} \sum_{n = 0}^\infty \sum_{m = 0}^\infty (i \omega_0 t)^{n+m} (a^\dagger a)^n (a + a^\dagger) (a^\dagger a)^m$$ But it doesn't seem obvious to me at all how the commutator will make this expression simplify; it seems there should be some easy way to see the answer without using this taylor expansion of the exponential.

  • $\begingroup$ Hint: compute $\mathrm e^{iHt}a\mathrm e^{-iHt}$ first. To this end, note that $a$ is an eigenoperator of $H$. $\endgroup$ – AccidentalFourierTransform Mar 24 '18 at 22:47
  • $\begingroup$ IMO this is the correct approach. Just compute the commutator of $(a^\dagger a)^n$ with $a$. As $(a^\dagger a)^n$ is just the number operator and $a$ annihilates a particle, you should already have an intuition of what the result might be. $\endgroup$ – noah Mar 24 '18 at 22:55
  • $\begingroup$ You are familiar with the fundamental Hadamard lema, the 2nd formula in this section, right? $\endgroup$ – Cosmas Zachos Mar 24 '18 at 23:04

You can also work with a basis where the operator $A$ is diagonal, with eigenvalues $\lambda_i$. On that basis $e^A$ is again diagonal, with eigenvalues $e^{\lambda_i}$.

You can use this approach for any function of $A$.


I can offer one method that works often:

Define $q_1 = a/(2\omega_0)$ and $q_2 = a^\dagger/(2\omega_0)$ (with $q = q_1+q_2$) and look for $q_i(t)$. The time derivative of $q_i(t)$ is

$$\frac{dq_i}{dt} = i \text{e}^{i H t} \left[H,q_i\right] \text{e}^{-i H t} \, , \qquad (*)$$

and the commutators are $\left[H,q_1\right] = - \omega_0 q_2$ and $\left[H,q_2\right] = \omega_0 q_1$. Then you can write (*) as

$$ \frac{d}{dt} \left(\begin{array}{c}q_1(t)\\q_2(t)\end{array}\right) = i \omega_1 \sigma_z \left(\begin{array}{c}q_1(t)\\q_2(t)\end{array}\right) \, . \qquad (**) $$

$\sigma_z = \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)$ is the third Pauli matrix. The solution of (**) is

$$ \left(\begin{array}{c}q_1(t)\\q_2(t)\end{array}\right) = \text{e}^{i \omega_0 t \sigma_z} \left(\begin{array}{c}q_1\\q_2\end{array}\right) \, .$$

Finally, you expand the exponential and take advantage of the fact that $\sigma_z^2 = 1$

$$\text{e}^{i y \sigma_z} = \sum_n \frac{(i y \sigma_z)^n}{n!} = \sum_n \frac{(i y \sigma_z)^{2n}}{(2n)!} + \frac{(iy \sigma_z)^{2n+1}}{(2n+1)!} $$ $$= \sum_n \frac{(-y)^{2n}}{(2n)!}+i \sigma_z \sum_n \frac{(-1)^{2n}y^{2n+1}}{(2n+1)!} \, .$$

In the second equality, I break the sum in even ($n=2k$) and odd ($n=2k+1$) values of $n$ and rename $n\rightarrow k$. The two sums after the last equality are the sine and cosine. Putting everything together provides

$$ \left(\begin{array}{c}q_1(t)\\q_2(t)\end{array}\right) = \left[\cos(\omega_0 t) + i \sin(\omega_0 t) \sigma_z \right] \left(\begin{array}{c}q_1\\q_2\end{array}\right) \, .$$

Finally $q(t) = q_1(t) + q_2(t) = \cos(\omega_0 t) (q_1+q_2) + i \sin(\omega_0 t) (q_2-q_1)$. Replacing $q_1$ and $q_2$ provides the final result.


The simplest approach is to expand the exponentials and develop the resulting expression so that \begin{align} e^{A} B e^{-A}&= \left(1+A+\frac{1}{2!}A^2+\ldots \right) B \left(1-A+\frac{1}{2!}A^2+\ldots\right)\, ,\\ &=B+ [A,B]+\frac{1}{2}[A,[A,B]]+\ldots \tag{1} \end{align} which continues as a sequence of nested commutators. In your case, $[H,a]\propto a$ and $[H,a^\dagger]\propto a^\dagger$ so you can actually recompose the series (1) to get the desired result.


Use $$\exp({{\rm ad}_X}) = {\rm Ad}_{\exp X}$$ Where $${\rm Ad}_A:X\mapsto AXA^{-1}$$ $${\rm ad}_A:X \mapsto [A,X]$$

Update: $${\rm ad}_H a = [H,a] = -\omega_0a$$ $${\rm ad}_H a^\dagger = [H,a^\dagger] = \omega_0 a^\dagger$$ So $$ \begin{split} \exp(iHt)q\exp(-iHt) & = {\rm Ad}_{\exp(iHt)}q \\ &= \exp({\rm ad}_{iHt})q \\&= \exp({\rm ad}_{iHt})\frac{a^\dagger+a}{2\omega_0}\\ &=\frac{1}{2\omega_0}\bigg(a\exp(-i\omega_0t) + a^\dagger \exp(i\omega_0t)\bigg) \end{split} $$

  • 1
    $\begingroup$ Please elaborate. $\endgroup$ – QuIcKmAtHs Mar 25 '18 at 4:57

protected by Qmechanic Mar 25 '18 at 5:15

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