# Show that a linear operator can be written in terms of its spectral decomposition [closed]

Let $\hat Q$ be an operator with a complete set of orthonomal eigenvectors: $$\hat Q |e_n\rangle=q_n|e_n\rangle\ \ (n=1,2,3,...)$$ Show that $\hat Q$ can be written in terms of its spectral decomposition: $$\hat Q=\sum_n q_n|e_n\rangle\langle e_n|.$$

Firstly in consideration that $|\alpha\rangle$ can be written in the form of $|\alpha\rangle=\sum a_n|e_n\rangle$, so my solution is as follows \begin{align} \hat Q|\alpha\rangle & =\sum a_n\hat Q|e_n\rangle\\ & =\sum a_n(\hat Q|e_n\rangle)\\ &=\sum_n a_nq_n |e_n\rangle \end{align}

I want to derive one equation like that $$\hat Q |\alpha\rangle=\{\text{something(operators in the form of components)}\}|\alpha\rangle$$

So I get $$\hat Q=\text{something}$$ but I don't know what should I do nextly, so I hope someone can give me help and explain the mechanism to me.

• What is the relation between $a_n$ and $\langle e_n|\alpha \rangle$? – Valter Moretti Mar 24 '16 at 10:51
• Isn't the hint basically the answer? $Q= Q\mathbb{1} = Q\sum |n><n|= ...$ – innisfree Mar 24 '16 at 10:51
• @ValterMoretti $a_n=\langle e_n|\alpha\rangle$. Do you mean that \begin{align} \hat Q|\alpha\rangle &=\sum\langle e_n|\alpha\rangle q_n|e_n\rangle\\ &= \{\sum_n q_n |e_n\rangle\langle e_n| \}|\alpha\rangle\end{align} ? – Wang Yun Mar 24 '16 at 10:58
• Please note that homework-like questions and check-my-work questions are generally considered off-topic. – ACuriousMind Mar 24 '16 at 11:25
• @ACuriousMind I don't think my question is off topic, this isn't a 'check- my-work' question. However, I hope someone can help me build a relation between my solution and the hint because I can't understant that – Wang Yun Mar 24 '16 at 12:34

Hint:

if $|e_i\rangle$ is the basis any vector $$|\alpha \rangle= \sum a_i |e_i \rangle$$

So, $\langle e_i|\alpha \rangle = a_i$ (Fourier inverse)

Apply $Q$ on both sides

$$Q|\alpha \rangle=\sum_i a_i Q|e_i \rangle$$

So,

$$Q|\alpha \rangle=\sum_i a_i q_i|e_i \rangle$$

$$Q|\alpha \rangle=\sum_i \langle e_i|\alpha \rangle q_i|e_i \rangle$$

$$Q|\alpha \rangle=\bigg [ \sum_i |e_i \rangle |\langle e_i| q_i \bigg ]|\alpha \rangle$$

If I understood your question correctly there seems to be a confusion with the two summations. Since the basis is orthonormal we can say that, for any two basis vectors $|e_m\rangle$ and $|e_n\rangle$. $$\langle e_m|e_n\rangle = \delta_{mn}$$ i.e $$\delta = 1, m=n$$ and $$\delta = 0, m\neq n$$

Lets expand the last equation in your question $$Q|\alpha \rangle=\bigg [ \sum_i |e_i \rangle |\langle e_i| q_i \bigg ]|\alpha \rangle$$ and write $|\alpha\rangle = \sum_{j} \alpha_j|e_j\rangle$

$$Q|\alpha \rangle=\bigg [ \sum_i |e_i \rangle \langle e_i| q_i \bigg ]\sum_{j} \alpha_j|e_j\rangle$$ $$= \sum_i\sum_j q_i \alpha_j|e_i\rangle \langle e_i|e_j\rangle$$ $$= \sum_{ij} q_i \alpha_j \delta_{ij}|e_i\rangle$$ $$= \sum_i q_i \alpha_i |e_i\rangle$$ because when $i\neq j, \delta_{ij} = 0$ anyway.

First of all, when you introduce the second sum, your expression already contains a sum over 'n'. That means, you'll have to write $\sum_m a_m |e_m\rangle$. Then you can rearrange the terms in such a way that you get $\langle e_n | e_m \rangle$, and if you remember what that is you are as good as done with the calculation.