# < f | A f >^2 = < f | A^2 f >

1. May 12, 2005

### TeCTeP

where A is arbitrary operator (not only ermiton), f is function, that < f | f > = 1. How to prove, that f is eigenfunction of operator A?

2. May 12, 2005

### Galileo

Try to rewrite the equation.

3. May 12, 2005

### TeCTeP

i've prooved it when operator is ermiton, but i can't proove for arbitrary operator...

4. May 12, 2005

### dextercioby

What do you know about this baby $|f\rangle\langle f|$...?

Daniel.

5. May 12, 2005

### TeCTeP

i've never used it before.

6. May 12, 2005

### dextercioby

$$\langle f|f\rangle =1 \Rightarrow |f\rangle\langle f|f\rangle =|f\rangle \Rightarrow |f\rangle\langle f|=\hat{1}$$

Does that help...?

Daniel.

7. May 12, 2005

### TeCTeP

Thank you. I've prooved it with your help. The puzzle is solved

8. May 12, 2005

### dextercioby

Some details:

1.$|f\rangle\langle f|$ is called the projector onto the vector $|f\rangle$.I assumed the Hilbert subspace is unidimensional and that the basis is formed by this vector $|f\rangle$.Therefore,the projector is generaly ~ to the unit operator on this subspace,but in this case,due to the orthonormalization,it coincides with the unit operator.

2.It's called HERMITEAN (or symmetric) operator,after the name of the 19-th cent.French mathematician Charles Hérmite.

Daniel.