Quantum Mechanics: Wave Mechanics in One Dimension

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Homework Help Overview

The discussion revolves around calculating the inner product ##\langle\psi|\psi\rangle## in the context of quantum mechanics, specifically using wave mechanics in one dimension. The original poster presents an equation involving integrals and the Dirac delta function, seeking clarification on the calculation process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the manipulation of integrals involving the Dirac delta function and question how it simplifies the expression for ##\langle\psi|\psi\rangle##. There are discussions about the properties of the delta function and its application in the context of the problem.

Discussion Status

Participants are actively engaging with the mathematical expressions and attempting to clarify the role of the delta function in the calculations. Some have provided insights into the properties of the delta function, while others express confusion regarding its application, indicating a productive exchange of ideas without a clear consensus yet.

Contextual Notes

There is mention of the original poster's unfamiliarity with the Dirac delta function, suggesting that the discussion is occurring within the framework of a homework assignment where foundational concepts may not have been fully addressed in their resources.

Robben
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Homework Statement



Let ##\langle\psi| = \int^{\infty}_{-\infty}dx\langle\psi|x\rangle\langle x|.## How do I calculate ##\langle\psi|\psi\rangle?##

Homework Equations



##\int^{\infty}_{-\infty}dxf(x)\delta(x-x_0)=f(x_0)##

The Attempt at a Solution



##\langle\psi|\psi\rangle = \int\int dx'dx\langle \psi|x\rangle\langle x|x'\rangle\langle x'|\psi\rangle = \int\int dx'dx\langle\psi|x\rangle\delta(x-x')\langle x'|\psi\rangle## but then how does that equal to ##\int dx \langle\psi|x\rangle\langle x|\psi\rangle?##
 
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<x|x`> is delta function and x` is as x in above.
 
abbas_majidi said:
<x|x`> is delta function and x` is as x in above.

I am not sure what you mean?
 
Above in OP is equation in 'Relevant equations'
 
Since ##\langle \psi|x\rangle## does not depend on ##x'## you can take it out of the integral over ##dx'##:$$
\int\int dx'dx\langle \psi|x\rangle\langle x|x'\rangle\langle x'|\psi\rangle =\int dx \langle \psi|x\rangle\ \left ( \int dx'\langle x|x '\rangle\langle x'|\psi\rangle \right )
$$
 
BvU said:
Since ##\langle \psi|x\rangle## does not depend on ##x'## you can take it out of the integral over ##dx'##:$$
\int\int dx'dx\langle \psi|x\rangle\langle x|x'\rangle\langle x'|\psi\rangle =\int dx \langle \psi|x\rangle\ \left ( \int dx'\langle x|x '\rangle\langle x'|\psi\rangle \right )
$$

But how does $$\left ( \int dx'\langle x|x '\rangle\langle x'|\psi\rangle \right ) = \langle x|\psi\rangle?$$
 
Robben said:
But how does $$\left ( \int dx'\langle x|x '\rangle\langle x'|\psi\rangle \right ) = \langle x|\psi\rangle?$$

It's what abbas_majidi wrote in post #2. ##<x|x'>=\delta(x-x')##.
 
Dick said:
It's what abbas_majidi wrote in post #2. ##<x|x'>=\delta(x-x')##.

How does ##\delta(x-x')## act on ##\langle x'|\psi\rangle## in order for it to equal ##\langle x|\psi\rangle##, i.e. $$\int dx'\delta(x-x')\langle x'|\psi\rangle?$$
 
Robben said:
How does ##\delta(x-x')## act on ##\langle x'|\psi\rangle## in order for it to equal ##\langle x|\psi\rangle##, i.e. $$\int dx'\delta(x-x')\langle x'|\psi\rangle?$$
You can apply your equation
$$
\int^{\infty}_{-\infty}dxf(x)\delta(x-x_0)=f(x_0)
$$to function values, but also to functions. So at each ##x'## in ##\langle x'|\psi \rangle = \psi(x')## is "replaced by ##x##" .
 
  • #10
BvU said:
You can apply your equation
$$
\int^{\infty}_{-\infty}dxf(x)\delta(x-x_0)=f(x_0)
$$to function values, but also to functions. So at each ##x'## in ##\langle x'|\psi \rangle = \psi(x')## is "replaced by ##x##" .
I see. This is my first time using dirac delta function and i was confused. My book didn't do a good job in explaining. Thank you all!
 

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