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I study the monograph An Introduction to the Standard Model of Particle Physics written by W. N. Cottingham and D. A. Greenwood. I can't understand the equation about the annihilation operator in the Heisenberg picture which states that: $$ a(t)= \exp(iHt)\, a \exp (-iHt) = \exp (iN \omega t) \,a \exp (-iN \omega t) = exp (-i \omega t) \,a. $$ I am familiar with the idea of Heisenberg picture and with the unitary transformation and the 1st and 2nd term are clear for me. But I do not understand how can I get the last term, the result. Many thanks for hint.

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    $\begingroup$ If you know the commutator $[H,a]$, the Baker-Campbell-Haussdorff formula gives a way to find commutators like $[e^{iHt},a]$. $\endgroup$
    – NewUser
    Mar 20, 2020 at 0:36

2 Answers 2

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The easiest way to derive your solution is probably to convert the first equality into a differential equation:

$$ a(t)=e^{iHt}ae^{-iHt}\qquad\Longrightarrow\qquad \frac{d}{dt}a(t)=e^{iHt}(iHa-aiH)e^{-iHt}=ie^{iHt}[H,a]e^{-iHt} $$

where I used

$$ \frac{d}{dt}e^{iHt}=iHe^{iHt}=e^{iHt}iH\qquad\qquad \frac{d}{dt}e^{-iHt}=-iHe^{-iHt}=e^{iHt}(-iH) $$

It follows that, with $[H,a]=-\omega a$

$$ \frac{d}{dt}a(t)=-i\omega\, e^{iHt}ae^{-iHt}=-i\omega\, a(t) $$

The latter, together with the initial condition $a(0)=a$, has of course solution $a(t)=ae^{-i\omega t}$.

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  • $\begingroup$ Hi, I was wondering if you could please verify explicitly why we can commute $iHe^{iHt}$? $\endgroup$
    – wrb98
    Oct 26, 2021 at 15:05
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    $\begingroup$ Hi there. Actually, it needs no hard proof. Observe that $e^{iHt}=\sum_{n}\frac{(it)^{n}}{n!} H^{n}$, and $[H,H^{n}]=0$ since of course $[H,H]=0$. $\endgroup$ Nov 4, 2021 at 15:10
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Like the comment says, you know $$ [H,a] = \omega\,\left[a^\dagger a + \tfrac12, a\right] = -\omega\, a\,. $$ Moreover, you also know the $n$fold nested commutator of $H$. Calling $\mathrm{ad}_H(X) \equiv [H,X]$ one has $$ \mathrm{ad}_H^n(a) \equiv\; \stackrel{n\text{ times}}{\overbrace{[H,[H,[\cdots[H}},a]\cdots]]] = (-\omega)^n a\,. $$ This is useful because there is the following formula: given two operators $X$ and $Y$ then $$ e^{X}Y e^{-X} =e^{\mathrm{ad}_X}Y\,. \tag{1}\label{1} $$ I will prove it later. Then you can use it to this end: $$ \begin{aligned} e^{iHt} a\,e^{-iHt} &=e^{\mathrm{ad}_{iHt}}a \\&= \sum_{n=0}^\infty \frac{(i t)^n}{n!}\,\mathrm{ad}_{H}^n(a) \\&= \sum_{n=0}^\infty \frac{(it)^n}{n!} (-\omega)^na = \\&=e^{-i\omega t}a\,. \end{aligned} $$


Proof of \eqref{1}

Now let's prove \eqref{1}. This is taken by a previous answer of mine.

For bookkeeping purposes let us define $X\to tX$ where $t \in \mathbb{R}$ and will be set to $1$ at the end. The strategy is to do a Taylor expansion in $t$ and verify that it matches order by order. The subtleties related to whether the Taylor series represents the full function or not will not be addressed in this answer. For order zero we have the obvious equality $$ e^{0}Y e^{0}= e^{0}Y\,. $$ Now take a derivative with respect to $t$ $$ e^{tX} \,(XY - YX)\, e^{-tX} = e^{t \,\mathrm{ad}_X} \mathrm{ad}_X(Y)\,, $$ which is true for $t=0$ by definition of $\mathrm{ad}$. Now call $\mathrm{ad}_X(Y) \equiv Y'$. The second derivative of \eqref{1} just amounts to $$ e^{tX} \,(XY' - Y'X)\, e^{-tX} = e^{t \,\mathrm{ad}_X} \mathrm{ad}_X(Y')\,, $$ and clearly they match again. We could continue forever by defining $Y''$ and so on. So the equation \eqref{1} holds order by order and one just has to set $t=1$. Another way to say it is that both sides of the equality satisfy the same first order ODE in $t$ with the same initial conditions, therefore they are equal: $$ \phi(t) = e^{tX} \,Y\, e^{-tX},\,\qquad \tilde{\phi}(t) = e^{t \,\mathrm{ad}_X} Y\,. $$ with $$ \begin{aligned} \phi(0) &= \tilde{\phi}(0) = Y\,,\\ \phi'(0) &= \tilde{\phi}'(0) = [X,Y]\,. \end{aligned} $$

There is also a direct computation that proves it. It starts from an explicit expansion of the exponential as $$ e^X\,Y\,e^{-X} = \sum_{n,m=0}^\infty \frac{(-1)^m}{n!m!} X^n Y X^m\,, $$ and then uses the following identity $$ (\mathrm{ad}_X)^m(Y) = \sum_{k=0}^m \binom{m}{k} X^kY(-X)^{m-k}\,, $$ which you can also prove explicitly. If you want more details, check out Chapter 2 of $[1]$.


$[1]\;$ Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Graduate Texts in Mathematics, 222 (2nd ed.), Springer

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