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

  • Thread starter Thread starter maria clara
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the expression < x | f > = f ( x ) in the context of quantum mechanics, specifically relating to the position operator and its eigenvalues and eigenfunctions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the meaning of the expression < x | f > = f ( x ) and question whether it should involve an integral over the product of x and f(x). There is a discussion about the role of eigenvalues and eigenfunctions in this context, with some participants seeking clarification on the notation used in bra-ket notation.

Discussion Status

Some participants express confusion regarding the notation and the implications of the inner product, while others provide insights into the use of delta functions and the reasoning behind the notation of eigenvalues in bra-ket notation. There is an ongoing exploration of these concepts without a clear consensus.

Contextual Notes

Participants are navigating the complexities of quantum mechanical notation and the definitions of eigenstates, with some assumptions about the familiarity with bra-ket notation and the properties of delta functions being evident.

maria clara
Messages
56
Reaction score
0
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...:confused:
 
Physics news on Phys.org
maria clara said:
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...:confused:

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?
 
detla functions... and it certainly works that way, thanks:smile:

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?
 
maria clara said:
detla functions... and it certainly works that way, thanks:smile:

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.
 
everything makes much more sense now, thanks a lot!:smile:
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
8
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K