Does the binomial formula $(a+b)^n=\sum_i C_n^ia^ib^{n-i}$ still work when $n$ is replaced by operator $\hat{n}$(an operator), where $a$ and $b$ are numbers? Since it's not the normal binomial formula mentioned in a lot of cases, and also can't be easily explained by the operator function, which is a definition in a Taylor-expansion way. So, does the formula work right? It seems okay to me.

Edit $C_{\hat{n}}^i$ will stands for $\frac{1}{i!}\hat{n}(\hat{n}-I)(\hat{n}-2I)...(\hat{n}-iI)$, and the original problem I met is to simplify the expression

$$ \begin{align} \sum_{l=0}^\infty l\frac{(1-\eta)^l}{l!}(a^\dagger)^l\eta^{\hat{n}}a^l \tag{1} \end{align} $$ into $$ \begin{align} (1-\eta)\hat{n} ,\tag{2}\end{align}$$ where $\hat{n}$ is photon number operator, $a^\dagger$ is the creation operator, $a$ is the annihilation operator, $\eta$ and $l$ are real numbers. My way is to think about the expectation value of eq. (1) with respect to the Fock state $|n\rangle$, so the sum over $\infty$ will become the sum over $n$, and we can have real number to be calculated and use the calculation skill where we used in the expectation value of binomial distribution, i.e., $\sum_{i=0}^n iC_n^ia^i(1-a)^{n-i}=na$.

But I think if we can use $C_\hat{n}^i=\frac{1}{i!}\hat{n}(\hat{n}-I)(\hat{n}-2I)...(\hat{n}-iI)$, then the formula $(a+b)^{\hat{n}}=\sum_i C_{\hat{n}}^ia^ib^{\hat{n}-iI}$ should be reasonable. I know Taylor expansion to treat $a+b$ together as a number can get the polynomial form of operator $\hat{n}$, but can the formula with binomial expansion alike get the same result while different form? Anyway, the original motivation for me to do so is to simplify eq. (1) into eq. (2).

  • $\begingroup$ How do you define the summation index in the operator version? What is $\hat{n} - i$ when $\hat{n}$ is an operator? $\endgroup$
    – Rammus
    Jul 1, 2021 at 14:39
  • $\begingroup$ Please ask a more precise question. First of all, what are the binomial coefficients $C^i_n$ when $n$ is an operator? What is $n-i$, should $i$ be the identity operator? Notice that you can define $a^A$, where $a$ is a number and $A$ is a linear operator. Simply use $a = e^{\log a}$ to write $a^A = e^{(\log a) A}$ and expand the exponential in a Taylor series. Is that what you're looking for? $\endgroup$ Jul 1, 2021 at 15:07
  • $\begingroup$ I don't know about the case of n (and what that even means), but if a and b are operators, the answer, in general, is no because, for example, you cannot have $na^{n-1}b$, but need $a^{n-1}b+a^{n-2}ba+a^{n-3}ba^2+\ldots$ $\endgroup$
    – DaftWullie
    Jul 1, 2021 at 15:08
  • $\begingroup$ @Rammus $\hat{n} - iI$, where $I$ is the identity matrix. $\endgroup$
    – narip
    Jul 1, 2021 at 15:11

1 Answer 1


In most cases the answer should be yes, if you treat things correctly. If we act on an eigenstate $|n\rangle$ of $\hat{n}$, we get the result $$ (a+b)^{\hat{n}}|n\rangle=(a+b)^n|n\rangle. $$ This means that we can write the general operator in a complete basis as $$ (a+b)^\hat{n}=\sum_{n=0}^\infty(a+b)^n|n\rangle\langle n|=\sum_{n=0}^\infty\left[\sum_{i=0}^n {{n}\choose{i}}a^ib^{n-i}\right]|n\rangle\langle n|. $$ This is the sense in which your expression is correct.

Edit: Now you are trying to solve a question in which $a$ and $b$ are also operators, so the above cannot work! This is even more different, but we can do a follow-the-nose derivation...

We know that we can rewrite the operator $$ \eta^\hat{n}=\sum_{k=0}^\infty \frac{(\ln \eta)^k \hat{n}^k}{k!}=\sum_n |n\rangle\langle n|\sum_{k=0}^\infty \frac{(\ln \eta)^k n^k}{k!}. $$ We can directly act with the creation and annihilation operators from either side, because they each change $|n\rangle$ by the same amount, but I think it would be nicer to act on this with both the creation and annihilation operators being on the same side.

We start by writing $[a,\hat{n}]=a$. From this, we know that $[a^l,\hat{n}]=la$, with which we can write $$ \hat{n}a^l=a^l(\hat{n}-l). $$ Repeating this process with $k$ different copies of $\hat{n}$, we find that $$ \hat{n}^k a^l=a^l(\hat{n}-l)^k. $$ This means that $$ \eta^\hat{n} a^l=a^l \sum_{k=0}^\infty \frac{(\ln \eta)^k (\hat{n}-l)^k}{k!}=a^l \eta^{\hat{n}-l}=a^l \eta^\hat{n}\eta^{-l}. $$ Substituting this into your expresion yields $$ \sum_{l=0}^\infty l\frac{\eta^{-l}(1-\eta)^l}{l!} a^{\dagger \,l}a^l\eta^{\hat{n}}. $$

Ok great. Now we can use that $a^{\dagger\,l}a^l|n\rangle=\frac{n!}{(n-l)!}|n\rangle$ for $n\geq l$ and $0$ otherwise to equate your expression with \begin{align} \sum_{l=0}^\infty \sum_{n\geq l}l\eta^{-l}(1-\eta)^l \binom{n}{l}|n\rangle\langle n| \eta^n &=\sum_{n=0}^\infty |n\rangle\langle n|\sum_{l=0}^n l\eta^{n-l}(1-\eta)^l \binom{n}{l} \\ &=\sum_{n=0}^\infty |n\rangle\langle n| [n(1-\eta)]\\ &=\hat{n}(1-\eta) \end{align} as desired. Never did we write a binomial coefficient as an operator.


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