Trying to understand quantum information. Need some help :( What does this notation $$ \langle\alpha|\hat{n}|\alpha\rangle $$ mean? Here $$|\alpha\rangle$$ is a coherent state and $$\hat{n}$$ is photon-number operator. I know that $$ \langle\alpha|\hat{n}|\alpha\rangle = {|\alpha|}^2 ,$$ but how do I calculate that?


The previous answer gives the meaning, here is the calculation.

The coherent state $\mid \alpha\rangle$ is an eigenvectior of the annihilation operator $\mathsf{a}$, with $$\mathsf{a}\mid \alpha\rangle= \alpha \mid \alpha\rangle .$$ The Hermitian conjugate of this equation is: $$\big(\mathsf{a}\mid \alpha\rangle\big)^+ = \langle\alpha\mid\mathsf{a}^+ = \alpha^* \langle\alpha\mid,$$ where $^*$ denote complex conjugate and $^+$ the adjoint.

Hence $$\langle\alpha\mid \mathsf{n}\mid \alpha\rangle = (\langle\alpha\mid \mathsf{a}^+)(\mathsf{a}\mid \alpha\rangle) = \alpha^* \alpha \langle\alpha\mid \alpha\rangle = | \alpha|^2$$

If you would to follow the path explained in the previous answer, you'd could use the expansion: $$ \mid \alpha\rangle=e^{-|\alpha|^2/2} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} \mid n \rangle $$ And the equalities $$ \mathsf{n}\mid n \rangle = n \mid n \rangle , \quad \langle n'\mid\mathsf{n}\mid n \rangle = \delta_{n,n'} n $$ of the number states (or Fock states).

One has:

\begin{align*} \langle\alpha\!\mid\!\mathsf{n}\!\mid\! \alpha\rangle &=e^{-|\alpha|^2} \sum_{n=0}^\infty \! \sum_{n'=0}^\infty\frac{\alpha^{*n'}\alpha^n}{\sqrt{n'!n!}} \langle n'\!\mid\!\mathsf{n}\!\mid\! n \rangle \\ &= e^{-|\alpha|^2} \sum_{n=0}^\infty n \frac{|\alpha|^{2n}}{n!} = |\alpha|^2 \end{align*}

where the final equality is the standard result on the average of the Poissonian distribution of parameter $|\alpha|^2$.

(Or in elementary terms, taking advantage of the vanishing of the term for $n=0$, one can factorize $|\alpha|^2$, and simplify $n/n!=1/(n-1)!$. Then the sum is $e^{|\alpha|^2}$ which exactly cancels the prefactor.)


If you had many systems in the state $|\alpha\rangle$ and you were to make measurements of the photon number for each of those systems, then $\langle\alpha|\hat n|\alpha\rangle$ is the expectation value of those measurements.

This can be calculated using inner products if you can directly calculate $\hat n|\alpha\rangle$ and $\langle\alpha|\left(\hat n|\alpha\rangle\right)$. Or you can work in some other complete basis $|m\rangle$ so that $$\langle\alpha|\hat n|\alpha\rangle=\sum_m\sum_{m'}\langle\alpha|m'\rangle\langle m'|\hat n|m\rangle\langle m|\alpha\rangle$$

assuming you know components $\langle m|\alpha\rangle$ and matrix elements $\langle m'|\hat n|m\rangle$

  • $\begingroup$ Thank you! Still kind of struggling to understand how the bra-ket notation works. Dumb question, but still. Isn't $$|m><m|=1$$ and $$|m'><m'|=1$$? $\endgroup$
    – Unknown
    Mar 31 '20 at 18:56
  • $\begingroup$ @Unknown No, $\sum_m|m\rangle\langle m|=1$ (Same for $m'$) $\endgroup$ Mar 31 '20 at 18:59
  • $\begingroup$ @Unknown to add to Aaron's answer: the sum $\sum_m |m\rangle\langle m|$ is the identity operator (that is an operator that when acting on any state will return the same state). Not to be be confused with $\langle m | m' \rangle = \delta_{m,m'}$ which for $m=m'$ gives the number 1. A ket-bra like $|m\rangle\langle m'|$ is an operator, while a bra-ket like $\langle m | m' \rangle$ is an inner-product of two states and is a complex number. $\endgroup$
    – user245141
    Apr 1 '20 at 8:53

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