I came across the following neat property:

For an operator $\hat{A}$ which is a linear combination of creation and annihilation operators, we have: $$ \langle e^{\hat{A}} \rangle = e^{\langle \hat{A^2} \rangle/2}.$$

Any help in approaching the proof of this identity would be appreciated.

  • 1
    $\begingroup$ I haven't seen that identity before, but you could try Taylor expanding the exponential, and seeing if the two sides agree, term-by-term. That's the usual approach for identities like this. $\endgroup$ Jun 17, 2016 at 20:02
  • $\begingroup$ Yeah, I saw it for the first time in a paper as well. Yeah, that's usually the way to go about it. I'm still not sure how A being a linear combination of the ladder operators will force the identity. I haven't worked out the details, though. $\endgroup$ Jun 17, 2016 at 20:04
  • $\begingroup$ If by "a linear combination of creation and annihilation operators", it means $\hat{A} = \alpha \hat{a} + \beta \hat{a}^{\dagger}$, then writing out the power series of both sides will give you the proof. Just note that the expectation value of $\hat{a}$ and $\hat{a}^{\dagger}$ are zero. $\endgroup$ Jun 18, 2016 at 18:03

2 Answers 2


I expect this to be a result of the Wick theorem. If you are considering an equilibrium situation with a quadratic Hamiltonian all odd moments will vanish. Thus you remain with even powers of your operator. If you now count the number of possible contractions you should get the correct result.

P.S.: I just tried and it works indeed. Note the helpful identity

$$ \frac{(2n-1)!!}{(2n)!} = \frac{1}{n! 2^n}$$


What $\left< \cdot \right>$ actually means is

$$ \left< B \right> \equiv \frac{1}{\mathcal{Z}} \text{Tr} \left\lbrace e^{- \beta H } B\right\rbrace$$

for some operator $B$, $\beta = 1/(k_B T)$, and $\mathcal{Z} = \text{Tr}\ e^{- \beta H }$. If your Hamiltonian now takes the form

$$ H = \sum_{nm} a^\dagger_n h_{nm} a_{m},$$

then you can prove the Wick theorem. Let $\alpha_m$ be either a bosonic creation or annihilation operator. I'll spare you the details but you will find that

$$\left< \alpha_1 \alpha_2 \cdots \alpha_{2n} \right> = \sum\limits_{\substack{\text{all possible} \\ \text{pairings}\ \pi}} \left< \alpha_{\pi(1)}\alpha_{\pi(2)} \right>\cdots \left< \alpha_{\pi(2n-1)}\alpha_{\pi(2n)} \right>$$

and $\left< \alpha_1 \alpha_2 \cdots \alpha_{2n+1} \right> = 0 $. This is all you need for the proof:

$$\left< e^{A} \right> \overset{(1)}{=} \sum_{n=0}^\infty \frac{1}{n!} \left< A^n\right> \overset{(2)}{=} \sum_{n=0}^\infty \frac{1}{(2n)!} \left< A^{2n}\right> \overset{(3)}{=} \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!} \left< A^2\right>^n \overset{(4)}{=} \sum_{n=0}^\infty \frac{1}{n!2^n} \left< A^2\right>^n \overset{(1)}{=} e^{\left< A^2\right>/2} . $$

(1) Series expansion.

(2) All odd powers vanish.

(3) There are $(2n-1)(2n-3)(2n-5) \cdots (2n-2n) = (2n-1)!!$ ways of forming $n$ pairs of $\left<A^2 \right>$ by the means of the Wick theorem.

(4) The identity from above.

I didn't check if it holds for fermions as well. There might be a similar identity but the Wick decomposition will change.

  • $\begingroup$ I have no idea about Wick's Theorem; so if you could share your solution with all of us, that'd be great. I have trouble proving identities of these types - exponentiated operators. So, it'd be great if you could share your solution. $\endgroup$ Jun 17, 2016 at 21:42
  • $\begingroup$ I extended my answer. You can find many insightful discussion on the Wick theorem here on Physics Stack Exchange as well. $\endgroup$ Jun 17, 2016 at 22:23
  • $\begingroup$ Great. I have no idea how you were able to express the expectation value of the operator in terms of Z, which looks like the partition function, and the boltzmann distribution. I have very limited knowledge of statistical mechanics and I fail to get it as of now, even though I have tried at looking at some resources online. I asked this question earlier today at: physics.stackexchange.com/q/263088 but I haven't received a response as of yet. If you could help me out over there, that'd great! $\endgroup$ Jun 17, 2016 at 22:29
  • $\begingroup$ Then understanding this formula might be a little over the top for you at the moment... Try to focus on the points you don't understand and get to the core of what you cannot figure out. Then the people of Stack Exchange can help you. If you ask too broad, nobody will reply and your question is very likely to get closed soon. $\endgroup$ Jun 17, 2016 at 22:41
  • $\begingroup$ I know, perhaps that's the case so I was hoping someone here could give me a broad overview of the formula by simply assuming I know what's a partition function etc. No one has done so thus far; if you can, that'd be great. $\endgroup$ Jun 17, 2016 at 22:44

Hint: OP's formula follows from a Wick-type theorem $$ \tag{1} T(f(\hat{A})) ~=~ \exp\left(\frac{1}{2}\hat{C}\frac{\partial}{\partial\hat{A}}\frac{\partial}{\partial\hat{A}} \right):f(\hat{A}): $$ between time-ordering $T$ and normal ordering $::$. Here $$\tag{2} \hat{C}~=~T(\hat{A}\hat{A})~-~:\hat{A}\hat{A}:$$ is a contraction. See e.g. this Phys.SE post and links therein.


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