- #1

maria clara

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< 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...

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- Thread starter maria clara
- Start date

- #1

maria clara

- 58

- 0

< 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...

- #2

George Jones

Staff Emeritus

Science Advisor

Gold Member

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- 1,475

< 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?

- #3

maria clara

- 58

- 0

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?

- #4

Dick

Science Advisor

Homework Helper

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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.

- #5

maria clara

- 58

- 0

everything makes much more sense now, thanks a lot!

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