Troubleshooting a Wrong Multiplying 2nd Equation

[n.easwaranand]

Homework Statement

Write in Dirac notation for
W=ψ(x)∫Φ*(x')dx'

Relevant Equations

1. ψ(x)=<x|Ψ>
2. <Φ|=∫<Φ|x><x|dx

Can't figure out what is wrong with my solution.
Multiplying 2nd equation with |x'> to get <Φ|x'>=∫<Φ|x'><x'|x'>dx' = ∫Φ*(x')dx'
So,
W=ψ(x)∫Φ*(x')dx' = <x|Ψ><Φ|x'>

But this is wrong. Not sure what is final answer.

 

[PeroK]

Homework Statement:: Write in Dirac notation for
W=ψ(x)∫Φ*(x')dx'
Relevant Equations:: 1. ψ(x)=<x|Ψ>
2. <Φ|=∫<Φ|x><x|dx

Can't figure out what is wrong with my solution.
Multiplying 2nd equation with |x'> to get <Φ|x'>=∫<Φ|x'><x'|x'>dx' = ∫Φ*(x')dx'
So,
W=ψ(x)∫Φ*(x')dx' = <x|Ψ><Φ|x'>

But this is wrong. Not sure what is final answer.

What happened to the integral?

 

[n.easwaranand]

What happened to the integral?

This integral ∫Φ*(x')dx' is expressed as <Φ|x'> .

 

[PeroK]

This integral ∫Φ*(x')dx' is expressed as <Φ|x'> .

The Dirac construction $\langle \phi | x' \rangle$ is the inner product of the state bra $\langle \phi |$ with the position eigenstate ket $|x' \rangle$. This corresponds to the complex conjugate of the position space wave-function for the state $|\phi \rangle$, evaluated at the point $x'$:
$$\langle \phi | x' \rangle = \phi(x')^*$$
It's not an integral. In fact, it's just the complex conjugate of the equation from your OP:

Relevant Equations:: 1. ψ(x)=<x|Ψ>

 

[n.easwaranand]

The Dirac construction $\langle \phi | x' \rangle$ is the inner product of the state bra $\langle \phi |$ with the position eigenstate ket $|x' \rangle$. This corresponds to the complex conjugate of the position space wave-function for the state $|\phi \rangle$, evaluated at the point $x'$:
$$\langle \phi | x' \rangle = \phi(x')^*$$
It's not an integral. In fact, it's just the complex conjugate of the equation from your OP:

In that case, what would be the answer? ψ(x)∫Φ*(x')dx' = <x|Ψ>∫<Φ|x'>dx'

 

[PeroK]

In that case, what would be the answer? ψ(x)∫Φ*(x')dx' = <x|Ψ>∫<Φ|x'>dx'

Yes, that's it.

 

[PeroK]

Multiplying 2nd equation with |x'> to get <Φ|x'>=∫<Φ|x'><x'|x'>dx' = ∫Φ*(x')dx'

Let me show you what went wrong here.
$$\langle \phi| = \langle \phi | (\int dx' |x'\rangle \langle x'|) = \int dx' \langle \phi |x'\rangle \langle x'|$$$$\langle \phi|x \rangle = \int dx' \langle \phi |x'\rangle \langle x'|x \rangle = \int dx' \langle \phi |x'\rangle \delta(x' - x) = \langle \phi |x\rangle $$
And you should end up with an identity.

 

[n.easwaranand]

Let me show you what went wrong here.
$$\langle \phi| = \langle \phi | (\int dx' |x'\rangle \langle x'|) = \int dx' \langle \phi |x'\rangle \langle x'|$$$$\langle \phi|x \rangle = \int dx' \langle \phi |x'\rangle \langle x'|x \rangle = \int dx' \langle \phi |x'\rangle \delta(x' - x) = \langle \phi |x\rangle $$
And you should end up with an identity.

I see. I just took the innerproduct with the dummy variable ! Now, I realize my mistake. Thanks a lot, that makes quite clear to me.