Question 3.27: Let Q be an operator with a complete set of orthonormal eige...

Let \hat{Q} be an operator with a complete set of orthonormal eigenvectors:

\hat{Q}\left|e_{n}\right\rangle=q_{n}\left|e_{n}\right\rangle \quad(n=1,2,3, \ldots) .

(a) Show that \hat{Q} can be written in terms of its spectral decomposition:

\hat{Q}=\sum_{n} q_{n}\left|e_{n}\right\rangle\left\langle e_{n}\right| .                  (3.103)

Hint: An operator is characterized by its action on all possible vectors, so what you must show is that

\hat{Q}|\alpha\rangle=\left\{\sum_{n} q_{n}\left|e_{n}\right\rangle\left\langle e_{n}\right|\right\}|\alpha\rangle .

for any vector |\alpha\rangle .

(b) Another way to define a function of \hat{Q}   is via the spectral decomposition:

f(\hat{Q})=\sum_{n} f\left(q_{n}\right)\left|e_{n}\right\rangle\left\langle e_{n}\right| .      (3.104).

Show that this is equivalent to Equation 3.100 in the case of e^{\hat{Q}} .

e^{\hat{Q}} \equiv 1+\hat{Q}+\frac{1}{2} \hat{Q}^{2}+\frac{1}{3 !} \hat{Q}^{3}+\cdots       (3.100).

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(a)

|\alpha\rangle=\sum_{n} c_{n}\left|e_{n}\right\rangle \Rightarrow \hat{Q}|\alpha\rangle=\sum_{n} c_{n} \hat{Q}\left|e_{n}\right\rangle=\sum_{n}\left\langle e_{n} \mid \alpha\right\rangle q_{n}\left|e_{n}\right\rangle=\left(\sum_{n} q_{n}\left|e_{n}\right\rangle\left\langle e_{n}\right|\right)|\alpha\rangle \Rightarrow \hat{Q}=\sum_{n} q_{n}\left|e_{n}\right\rangle\left\langle e_{n}\right| .

(b) Noting that \hat{Q}\left|e_{n}\right\rangle=q_{n}\left|e_{n}\right\rangle \Rightarrow \hat{Q}^{2}\left|e_{n}\right\rangle=q_{n} \hat{Q}\left|e_{n}\right\rangle=q_{n}^{2}\left|e_{n}\right\rangle \cdots \Rightarrow \hat{Q}^{j}\left|e_{n}\right\rangle=q_{n}^{j}\left|e_{n}\right\rangle , we have

e^{\hat{Q}}|\alpha\rangle=\sum_{j=0}^{\infty} \frac{1}{j !} \hat{Q}^{j}\left(\sum_{n}\left|e_{n}\right\rangle\left\langle e_{n}\right|\right)|\alpha\rangle=\sum_{n}\left(\sum_{j} \frac{1}{j !} q_{n}^{j}\right)\left|e_{n}\right\rangle\left\langle e_{n}|| \alpha\right\rangle=\left(\sum_{n} e^{q_{n}}\left|e_{n}\right\rangle\left\langle e_{n}\right|\right)|\alpha\rangle .

so

e^{\hat{Q}}=\left(\sum_{n} e^{q_{n}}\left|e_{n}\right\rangle\left\langle e_{n}\right|\right) .    QED

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