POVM as a function of the number operator¶

Consider the operator

\[\begin{equation} A := \sum_{k=0}^{\infty} x^k \, |k\rangle\langle k|. \end{equation}\]

This operator is diagonal in the Fock basis. The states \(|k\rangle\) are eigenstates of the number operator

\[\begin{equation} \hat{n} = \hat{a}^\dagger \hat{a}, \qquad \hat{n} |k\rangle = k |k\rangle. \end{equation}\]

This property will be used to relate \(A\) to a function of the number operator.

Action of exponentials of the number operator¶

For any function \(f\), the operator \(f(\hat{n})\) can be defined via its power series expansion. In particular, the operator \(x^{\hat{n}}\) can be written as

\[\begin{equation} x^{\hat{n}} = e^{(\ln x)\hat{n}} = \sum_{j=0}^{\infty} \frac{(\ln x)^j}{j!} \, \hat{n}^{\,j}. \end{equation}\]

Since \(|m\rangle\) is an eigenstate of \(\hat{n}\), repeated application of \(\hat{n}\) gives

\[\begin{equation} \hat{n}^{\,j} |m\rangle = m^j |m\rangle. \end{equation}\]

Substituting this into the series expansion yields

\[\begin{equation} \begin{aligned} x^{\hat{n}} |m\rangle &= \sum_{j=0}^{\infty} \frac{(\ln x)^j}{j!} \, m^j |m\rangle \\ &= \left( \sum_{j=0}^{\infty} \frac{(\ln x)^j m^j}{j!} \right) |m\rangle \\ &= e^{m \ln x} |m\rangle \\ &= x^m |m\rangle. \end{aligned} \end{equation}\]

Thus, the operator \(x^{\hat{n}}\) acts diagonally in the Fock basis with eigenvalues \(x^m\).

Action of \(A\) on number states¶

The action of \(A\) on an arbitrary number state \(|m\rangle\) is given by

\[\begin{equation} \begin{aligned} A|m\rangle &= \sum_{k=0}^{\infty} x^k |k\rangle \langle k|m\rangle \\ &= \sum_{k=0}^{\infty} x^k |k\rangle \delta_{k,m} \\ &= x^m |m\rangle. \end{aligned} \end{equation}\]

This shows that \(A\) is also diagonal in the Fock basis and has the same eigenvalues \(x^m\).

Identification of the operators¶

Since both \(A\) and \(x^{\hat{n}}\) act identically on all basis states \(|m\rangle\), they represent the same operator. Therefore,

\[\begin{equation} \sum_{k=0}^{\infty} x^k |k\rangle\langle k| = x^{\hat{n}}. \end{equation}\]

In particular, for \(x = 1-\eta\), this yields

\[\begin{equation} \Pi_0^{(\eta)} = (1-\eta)^{\hat{n}}. \end{equation}\]