Let us consider the Hamiltonian of a single harmonic oscillator, which is expressed in terms of creation/annihilation operators as $H=\hbar \omega (a^{\dagger}a+1/2)$. The eigenstates of this Hamiltonian are the number states ($n\geq 0$)$$a^{\dagger}a|n\rangle=n|n\rangle.$$

My question is: how to express $|n\rangle\langle m|$ in terms of creation/annihilation operators, where $m$ and $n$ are two arbitrary integers?


3 Answers 3


Although an answer has already been given which includes $| 0 \rangle \langle 0 |$, it is indeed possible to write an expression for $|m \rangle \langle n |$ purely algebraically in terms of $a$ and $a^{\dagger}$. This certainly must be possible in principle, since $a$ and $a^{\dagger}$ furnish a representation of the entire operator algebra of the harmonic oscillator.

There are two steps. First, note that I can project onto any state $|n \rangle$ by simply projecting out every other state. Letting $N = a^{\dagger} a$ be the number operator, I claim that the operator $$ P_n = N \left( \frac{N - 1}{n-1} \right) \left( \frac{N - 2}{n-2} \right) \ldots \left( \frac{N - (n-1)}{n-(n-1)} \right) \left( \frac{N - (n+1)}{n-(n+1)} \right) \ldots = \prod_{j \neq n} \left( \frac{N-j}{n-j} \right) $$ is equivalent to the projector $| n \rangle \langle n|$. It's easy to see that this expression is zero when acting on $|0 \rangle$ due to the first term, zero when acting on $|1 \rangle$ when acting on the first term, and so on. It's also easy to see that $P_n |n \rangle = |n \rangle$.

The second step is much simpler. Once we know we've projected into state $|n \rangle$, then all we need to do is apply raising or lowering operators to get to $|m \rangle$. For example, if $m = n+1$, then all we need to do is apply $a^{\dagger}$. More generally, our total operator is $$ |m \rangle \langle n | = \begin{cases} \frac{(a^{\dagger})^{(m-n)}}{\sqrt{m(m-1) \ldots (n+1)}} P_n, & m > n \\ P_n, & m=n \\ \frac{a^{(n-m)}}{\sqrt{n(n-1) \ldots (m+1)}} P_n, & m < n \end{cases} $$

Of course, this isn't particularly useful for doing an actual calculation. But it's important to know that this is at least in principle possible; indeed, it's extremely important in QFT, where we typically write all of our operators in terms of creation and annihilation operators. This sort of procedure then shows you that this is no restriction, and any operator can indeed be represented in terms of creation and annihilation operators alone.

  • $\begingroup$ The $|0\rangle\langle 0|$ is still present. It is hidden inside the projection operator. $\endgroup$ Commented Mar 13, 2022 at 13:46
  • 1
    $\begingroup$ Sure, what’s your point? It’s pretty clear that the question wanted an answer expressed exclusively in terms of creation and annihilation operators. I showed how to write the projector this way, without reference to explicit kets or bras. $\endgroup$
    – Zack
    Commented Mar 13, 2022 at 14:36
  • $\begingroup$ See Tan Tixuan's first comment to Nickolas Alves' answer. $\endgroup$ Commented Mar 13, 2022 at 17:23

From the properties of the creation operator, one can show that $$| n \rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} | 0 \rangle,$$ which I'm assuming here to be a well-known result. Taking the adjoint of this expression, we find that $$\langle n | = \langle 0 |\frac{(a)^n}{\sqrt{n!}}.$$

Therefore, \begin{align} | n \rangle \langle m | &= \frac{(a^\dagger)^n}{\sqrt{n!}} | 0 \rangle \langle 0 |\frac{(a)^m}{\sqrt{m!}}, \\ &= \frac{(a^\dagger)^n | 0 \rangle \langle 0 | (a)^m}{\sqrt{n! m!}}, \end{align} which I believe is the result you're looking for.

  • $\begingroup$ Actually, I was looking for something in which |0><0| doesn't appear, i.e. purely in terms of ladder operators. $\endgroup$
    – Tan Tixuan
    Commented Mar 9, 2022 at 4:15
  • $\begingroup$ @user143410 $| n \rangle \langle m |$ is an operator and there might be some way to represent as a product or sum of operators constructed from the ladder operators. $\endgroup$
    – Adrian
    Commented Mar 9, 2022 at 4:39
  • $\begingroup$ @Adrian Indeed, from Zack's answer it looks like $|0\rangle\langle 0| = (1-N)\times\frac{2-N}{2}\times\frac{3-N}{3} \times \cdots = \prod_{n=1}^\infty \frac{n-N}{n}$, where $N=a^\dagger a$. $\endgroup$
    – Andrew
    Commented Mar 9, 2022 at 6:15

Combining the existing answers, since $|0\rangle\langle 0|=\prod_{k\ge1}(1-k^{-1}a^{\dagger}a)$ and $|n\rangle =\frac{1}{\sqrt{n!}}a^{\dagger n}|0\rangle$,$$\sum_{mn}c_{mn}|m\rangle\langle n|=\sum_{mn}\frac{c_{mn}}{\sqrt{m!n!}}a^{\dagger m}\prod_k(1-k^{-1}a^{\dagger}a)a^n.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.