Dirac Notation: How $|f\rangle= \sum_{j=1}^{\infty}|\phi_j\rangle\langle\phi_j|f\rangle$?

Question

In a text, it has been written that a function is $|f\rangle= \sum_{j=1}^{\infty}a_j|\phi_j\rangle$, how one can write that -

$ |f\rangle= \sum_{j=1}^{\infty}|\phi_j\rangle\langle\phi_j|f\rangle$,

and

$ |f\rangle= (\sum_{j=1}^{\infty}|\phi_j\rangle\langle\phi_j|)|f\rangle$? plz show the derivation.

The text is given below-

Answer 1

We have $a_j=\langle \phi_j\mid f\rangle$. This immediately gives $$ \sum_{j=1}^{\infty}a_j|\phi_j\rangle=\sum_{j=1}^{\infty}\langle \phi_j\mid f\rangle |\phi_j\rangle $$ (You have an extra $a_j$ on the right-hand side here; that's a typo and not present in your image.)

Now consider the fact that $\langle \phi_j\mid f\rangle$ is just a number, so it commutes with $|\phi_j\rangle$, giving us $$ =\sum_{j=1}^{\infty}|\phi_j\rangle\langle \phi_j\mid f\rangle $$ Finally, Dirac notation multiplication is associative and distributive, so we may group the middle $\langle \phi_j|$ with the left-hand $|\phi_j\rangle$ rather than the right-hand $|f\rangle$, and move the $|f\rangle$ outside the sum.

Answer 2

The equality in the title is wrong. If the sequence $|\phi_i\rangle$ forms a complete orthonormal system, we have that every vector is expressable in the following way: $$|f\rangle=\sum_{i=1}^{+\infty}f_i|\phi_i\rangle$$ Where the equality means convergence in norm induced by the inner product. To get the coefficients, we can do the following: define $|s_n\rangle$ as $$|s_n\rangle=\sum_{i=1}^n f_i |\phi_i\rangle$$ Then we have that $$\langle \phi_j| s_n \rangle = \sum_{i=1}^{n}f_i \langle \phi_j|\phi_i\rangle=\sum_{i=1}^n f_i \delta_{ij}$$ Taking the limit as $n\to +\infty$, we get that $$\lim_{n\to+\infty}\langle\phi_j|s_n\rangle=\langle\phi_j|f\rangle$$ While $$\lim_{n\to+\infty}\sum_{i=1}^n f_i \delta_{ij}=f_j$$ Which means that $$f_j=\langle\phi_j|f\rangle$$ And finaly $$|f\rangle = \sum_{i=1}^{+\infty}\langle\phi_i|f\rangle |\phi_i\rangle$$ Usually the diad $|\phi_i\rangle \langle \phi_i|$ is defined as $$\big(|\phi_i\rangle \langle \phi_i|\big) |v\rangle= \big(\langle\phi_i|v\rangle\big)|\phi_i\rangle \quad \forall v \in H$$ And this allows us to rewrite the sum as you wanted: $$|f\rangle = \bigg(\sum_{i=1}^{+\infty}|\phi_i\rangle\langle\phi_i|\bigg)|f\rangle $$