I want to transform an operator into the interaction Picture. Therefore, I have to calculate the following

$$\exp(i H_E t) (\sum_k g_k a_k )\exp(-i H_E t),$$ where

$$H_E=\sum \omega_k a_k^\dagger a_k.$$

Plugging $H_E$ in yields (I am not 100% sure if the notation that I chose is very clever, so corrections concerning my notation are really appreciated)

$$ \underbrace{\sum_k \sum_n \exp(i \omega_k n) |n\rangle \langle n|}_{\exp(i H_E t)} \, \underbrace{(\sum_l \sum_p g_l c_p |p-1\rangle \langle p| )}_{(\sum_k g_k a_k )} \, \underbrace{\sum_m \sum_j\exp(-i \omega_m j) |j\rangle \langle j|}_{\exp(-i H_E t)}=(*)$$

I chose to represent the creation and annihilation operators such that

$$a_k=\sum_l c_l |l-1\rangle \langle l|$$

$$a_k^\dagger=\sum_l c_l |l+1\rangle \langle l|$$

Besides that, I am not really sure how to Label everything correctly because the annihilation and creation operator only act on states in the correct mode ($\omega_k$), so I don't know how to indicate this in the following expression

$$\sum_k \omega_k a_k^\dagger a_k=\sum_k \omega_k \sum_l l |l \rangle \langle l | .$$

If I continue now my calculation for the representation in the interaction picture, I obtain the following Expression

$$(*)= \sum_k \sum_n \exp(i \omega_k n) \, (\sum_l \sum_p g_l c_p ) \, \sum_m \sum_j\exp(-i \omega_m j) |n\rangle \langle n| |p-1\rangle \langle p| |j\rangle \langle j| $$ $$= \sum_k \sum_n \exp(i \omega_k n) \, (\sum_l \sum_p g_l c_p ) \, \sum_m \sum_j\exp(-i \omega_m) |n\rangle \delta_{n,p-1} \delta_{p,j} \langle j |$$ $$= \sum_k \exp(i \omega_k(p-1)) \, (\sum_l \sum_p g_l c_p ) \, \sum_m \exp(-i \omega_m p) |p-1\rangle \langle p |$$ $$= \sum_k \exp(i \omega_k(p-1)) \sum_m \exp(-i \omega_m p) \, (\sum_l \sum_p g_l c_p |p-1\rangle \langle p | ) $$ $$= \sum_k \exp(i \omega_k(p-1)) \sum_m \exp(-i \omega_m p) \, (\sum_l g_l a_l ). $$

Now, the problem is that this is not correct, since the result is

$$\sum_l g_l a_l \exp(-i \omega_l t).$$

I don't know where the additional $\delta_{k,m}$ and $\delta_{l,m}$ should dome from. Made I started with too many sums and thats the mistake but I don't see why there should be too many of them.

EDIT: Following udrv's advice:

I split up the exponential function

$$ \exp(-i H_E t)= \prod \limits_{i=1} \exp(-i H_i t)$$, where $$ H_i=\omega_i a_i^\dagger a_i$$

Then performing the calculation for a generic $a_k$ yields $$\exp(i H_1 t) \exp(i H_2 t) \dots \exp(i H_k t) \dots g_k a_k \exp(-i H_1 t) \exp(-i H_2 t) \dots \exp(-i H_k t) \dots$$ $$=\exp(i H_1 t) \exp(-i H_1 t) \exp(i H_2 t) \exp(-i H_2 t) \dots \exp(i H_k t)g_k a_k \exp(-i H_k t) \dots$$ $$=\mathbb{I} \cdot \mathbb{I}\cdot \mathbb{I} \dots \sum_l \exp(-i \omega_k l) |l \rangle \langle l | g_k \sum_j c_j |j-1\rangle \langle j| \sum_n \exp(-i\omega_k n ) |n \rangle \langle n |$$ $$=\exp(i \omega_k (j-1)) \sum_j c_j |j-1\rangle \langle j | \exp(- i \omega_k j )$$ $$=\exp(-i \omega_k ) g_k a_k$$.

Now, the Summation over the different $a_k$ should yield the desired result. Correct?

  • $\begingroup$ Hint: Think "second quantization", "operator algebra", and multiple distinct modes. Not first quantization for one single harmonic oscillator. Take one $a_k$ term at a time, $a_k(t) = \exp[iH_E t]\;a_k \exp[-iH_E t]$, take the time derivative, and solve. What do you see? $\endgroup$
    – udrv
    Commented Feb 11, 2017 at 13:53
  • $\begingroup$ I'm sorry, but I don't understand what you mean with the time derivative? Our Tutor told us that it was doable like this, so I would like to do it this way. $\endgroup$
    – anonymous
    Commented Feb 11, 2017 at 14:06
  • 1
    $\begingroup$ Fine. The reason your sums don't work is that you use identical expressions for the ladder operators and Hamiltonian for all modes (no k dependence). The simplest way to fix it is to distribute the $\exp[\pm H_E t]$ over each term in the sum, recall that operators for different modes commute, and simplify each term to $g_k\exp[i\omega_k a_k^\dagger a_k] a_k \exp[-i\omega_k a_k^\dagger a_k]$. Now you can calculate a generic $\exp[i\omega a^\dagger a] a \exp[-i\omega a^\dagger a]$ as you did before, but without having to worry about the different k. Just re-index afterwards and sum. $\endgroup$
    – udrv
    Commented Feb 11, 2017 at 14:41
  • $\begingroup$ Thanks for your answer! I edited the entry post following your instructions. Is it correct now? $\endgroup$
    – anonymous
    Commented Feb 12, 2017 at 10:57
  • 1
    $\begingroup$ Yep, it looks good now. $\endgroup$
    – udrv
    Commented Feb 12, 2017 at 13:23

1 Answer 1


A way to obtain the result is to consider

$$e^{iHt}\sum_k g_k a_k e^{-iHt}=\sum_k g_k\, e^{iHt} a_k \,e^{-iHt}=\sum_k g_k\, e^{i\omega t a^\dagger_k a_k} a_k \,e^{-i\omega t a^\dagger_k a_k}$$

Let $A=i\omega_k t\, a^\dagger_k a_k$ and $B=a_k$. Then, by applying the Baker-Campbell-Hausdorff relation

$$e^A B e^{-A}=B+[A,B]+\frac{1}{2!}[A,[A,B]]+...$$

with commutatior $[a^\dagger_ka_k,a_k]=-a_k$ (Bosons), one finds

$$e^{i\omega_k t a^\dagger_k a_k} a_k \,e^{-i\omega t a^\dagger_k a_k}=e^{-i\omega_kt}\,a_k,$$

and thus

$$e^{iHt}\sum_k g_k\, a_k\, e^{-iHt}=\sum_k g_ke^{-i\omega_kt}\,a_k,$$

corresponding to the result which has already been proven in the comments (completed by the the variable $t$ in the exponent).


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