Understanding Why < x | f > = f ( x )

[maria clara]

Why is it true that
< x | f > = f ( x )
?
shouldn't it be the result of the integral from minus infinity to infinty on xf(x)?
but then it can't even be a function of x...

 

[George Jones]

Why is it true that
< x | f > = f ( x )
?
shouldn't it be the result of the integral from minus infinity to infinty on xf(x)?
but then it can't even be a function of x...

Here, x is an "eigenvalue" of the position operator. What you need to use in the integral is not the eigenvaule, but the state that is the "eigenfunction" that corresponds to this eigenvalue.

What states are eigenstates of the position operator?

 

[maria clara]

detla functions... and it certainly works that way, thanks

but I'm still confused: the definition of the bra-ket is simply the inner product of the functions you put there:
< f | g > = integral from minus infinity to infinity on f*g (f*=the complex conjugate of f).
So the product < x' | f > should be written as follows:
< delta (x-x') | f >...
why do they write the eigenvalue in the bra, and not the function itself?

 

[Dick]

detla functions... and it certainly works that way, thanks

but I'm still confused: the definition of the bra-ket is simply the inner product of the functions you put there:
< f | g > = integral from minus infinity to infinity on f*g (f*=the complex conjugate of f).
So the product < x' | f > should be written as follows:
< delta (x-x') | f >...
why do they write the eigenvalue in the bra, and not the function itself?

They write the eigenvalue in the bra because they are too lazy to write the eigenfunction. Really, it's just a notational abbreviation.

 

[maria clara]

everything makes much more sense now, thanks a lot!