Dirac notation

  • #1
n.easwaranand
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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.
 

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  • #2
PeroK
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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?
 
  • #3
n.easwaranand
6
0
What happened to the integral?
This integral ∫Φ*(x')dx' is expressed as <Φ|x'> .
 
  • #4
PeroK
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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>
 
  • #5
n.easwaranand
6
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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'
 
  • #6
PeroK
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In that case, what would be the answer? ψ(x)∫Φ*(x')dx' = <x|Ψ>∫<Φ|x'>dx'
Yes, that's it.
 
  • #7
PeroK
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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.
 
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  • #8
n.easwaranand
6
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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.
 

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