# Normal ordered exponential of one-body operators

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*}

Meanwhile,

\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.

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

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}

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

• I think this is actually the same as the proof by @jason-funderberker. Commented Jul 27, 2022 at 12:41
• No, my proof does not use $f^\prime(M) = L^\prime(M)$. Commented Jul 27, 2022 at 12:45
• 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. Commented Jul 27, 2022 at 12:49
• Then they become the same. Commented Jul 27, 2022 at 12:50
• Oh I was sloppy about that point😂 Commented Jul 27, 2022 at 12:52