Let $\{a_i\}_{i=1}^N$ be a set of annihilation operators (they are either all bosons, or all fermions) satisfying the canonical commutation or anti-commutation relation. In the book Quantum Theory of Finite Systems by Blaizot and Ripka, Problem 1.6 claims that (summation over repeated indices is implied)

$$ \exp(a^\dagger_i M_{ij} a_j) = N[\exp(a^\dagger_i (e^M-1)_{ij} a_j)] \tag{1} $$

where $M_{ij}$ is an $N \times N$ complex matrix, and $N[A]$ puts creation operators in $A$ to the left, treating all $a^\dagger, a$ in the argument as commuting or anti-commuting numbers. For example, with $\eta = +1$ for bosons, and $-1$ for fermions, we have

$$ N[a_4 a^\dagger_2 a_1 a^\dagger_3] = \eta^{1 + 2} a^\dagger_2 a^\dagger_3 a_4 a_1 = \eta a^\dagger_2 a^\dagger_3 a_4 a_1 $$

I tried to prove eq. (1) by series expansion and comparing terms, but the expansion soon becomes rather complicated. I would appreciate it if someone can provide an elegant and clean proof.

My current attempt: For fermions the exponential function can be greatly simplified. Below I give a proof for fermions when $N = 1$, so that $M$ reduced to a complex number.

The RHS (right hand side) of eq. (1) now actually means

$$ \begin{align*} \text{RHS} &= N[\exp[(e^{M}-1) a^\dagger a]] \\ &= 1 + \sum_{n=1}^\infty \frac{(e^{M}-1)^n}{n!} N\left[(a^\dagger a)^n\right] \end{align*} $$

Normal ordering gives:

$$ \begin{align*} N\left[(a^\dagger a)^n\right] &= N[a^\dagger a a^\dagger a \cdots a^\dagger a] \\ &= \eta^{1 + \cdots + (n-1)} a^{\dagger n} a^n \\ &= \eta^{n(n-1)/2} a^{\dagger n} a^n \end{align*} $$

For fermions, $a^n = 0$ for $n \ge 2$, which is the key to simplify the exponential function:

$$ \begin{align*} \text{RHS} &= 1 + (e^M - 1) a^\dagger a \end{align*} $$


$$ \begin{align*} \text{LHS} &= \exp(M a^\dagger a) = 1 + \sum_{n=1}^\infty \frac{M^n}{n!} (a^\dagger a)^n \end{align*} $$

But with $a a^\dagger = 1 - a^\dagger a$, we notice that

$$ \begin{align*} (a^\dagger a)^2 &= a^\dagger a a^\dagger a = a^\dagger (1 - a^\dagger a) a \\ &= a^\dagger a - \underbrace{ a^{\dagger 2} a^2 }_{= 0} = a^\dagger a \end{align*} $$

which further leads to $(a^\dagger a)^n = a^\dagger a$ for any $n \ge 1$. Therefore

$$ \begin{align*} \text{LHS} &= 1 + \bigg[ \sum_{n=1}^\infty \frac{M^n}{n!} \bigg] a^\dagger a \\ &= 1 + (e^M - 1) a^\dagger a = \text{RHS} \end{align*} $$

But obviously things will be complicated for bosons, since the $a$ operator is no longer nilpotent.

  • 3
    $\begingroup$ Have you tried proving this for a single bosonic or fermionic oscillator? $\endgroup$
    – mavzolej
    Jul 27, 2022 at 9:18
  • 2
    $\begingroup$ @mavzolej I tried $N = 1$ for fermions, and updated the question to include my attempt. $\endgroup$ Jul 27, 2022 at 11:20
  • $\begingroup$ Do you have a source that for fermions normal order introduces a sign change (eta)? I havent seen that. $\endgroup$
    – lalala
    Jul 27, 2022 at 18:11
  • 1
    $\begingroup$ For fermions, this result is Eq. (2.30) of arXiv:1212.6049, which gives a sketch of the proof. $\endgroup$ Jul 27, 2022 at 18:56
  • $\begingroup$ @lalala See Wikipedia $\endgroup$ Jul 28, 2022 at 1:50

2 Answers 2


Hints: First try to show it for a single bosonic oscillator (for fermions this was done by the OP already). To this end, define the following functions: \begin{align} f(M)&:=\exp{a^\dagger a M} \tag{1} \\ L(M)&:=N[\exp{a^\dagger a (e^{M}-1)}] \quad \tag{2}. \end{align} Then show $f(0)=L(0)=\mathbb I$ and that $f$ and $L$ satisfy the same differential equation, which in turn implies $f(M)=L(M)$, proving the claim.

The generalizations for $N>1$ and fermions are left to you.

Here are some useful relations you can use/find/prove: \begin{align} e^{-a^\dagger a M}\, a\, e^{a^\dagger aM} &= e^M\, a \tag{3}\\ N[(a^\dagger a)^n] &= (a^\dagger)^n a^n \tag{4}\\ f^\prime(M) &= a^\dagger a\, f(M) \overset{(3)}{=} e^M a^\dagger\, f(M)\, a \tag{5}\quad . \end{align}

  • $\begingroup$ You have "...Then show that 𝑓(0)=𝐿(0)=𝕀..." you seem to have left out the product! @Jason Funderberker $\endgroup$ Jul 27, 2022 at 11:51
  • $\begingroup$ After the equals sign. $\endgroup$ Jul 27, 2022 at 11:52
  • $\begingroup$ I checked the latex you have: "f(0)=L(0)=\mathbb I" The "I" doesn't come out my end. I assume the equation is correct though. $\endgroup$ Jul 27, 2022 at 11:58
  • $\begingroup$ Ok it looks correct now. I just didn't recognise the identity symbol "I" in that font. $\endgroup$ Jul 27, 2022 at 12:04
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    $\begingroup$ I've corrected my answer, there was a mistake. $\endgroup$ Jul 27, 2022 at 12:52

Let us prove OP's claim for a single bosonic mode:

Proposition: $$ e^{ta^{\dagger}a}~=~:e^{(e^t-1)a^{\dagger}a}: \qquad t~\in~\mathbb{C}.\tag{A}$$

Sketched proof of eq. (A): Let's call the LHS for $U(t)$ and the RHS for $V(t)$. Both sides can be written as a function of the operator $n=a^{\dagger}a$ without the use of $a$ and $a^{\dagger}$. They satisfy the same first-order ODE: $$U^{\prime}(t)~=~a^{\dagger}a e^{tn}~=~a^{\dagger} e^{t(n+1)}a~=~e^t a^{\dagger} U(t)a, \tag{B}$$ $$V^{\prime}(t)~=~e^t a^{\dagger} V(t)a, \tag{C}$$ with the same initial condition $U(0)={\bf 1}=V(0)$. Hence they must be equal. $\Box$

  • 1
    $\begingroup$ I think this is actually the same as the proof by @jason-funderberker. $\endgroup$ Jul 27, 2022 at 12:41
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    $\begingroup$ No, my proof does not use $f^\prime(M) = L^\prime(M)$. $\endgroup$
    – Qmechanic
    Jul 27, 2022 at 12:45
  • $\begingroup$ I just noticed that this was not my intention. I also meant that they satisfy the same initial value problem. I'll fix that. Thanks. $\endgroup$ Jul 27, 2022 at 12:49
  • 2
    $\begingroup$ Then they become the same. $\endgroup$
    – Qmechanic
    Jul 27, 2022 at 12:50
  • $\begingroup$ Oh I was sloppy about that point😂 $\endgroup$ Jul 27, 2022 at 12:52

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