Quantum Mechanics: Wave Mechanics in One Dimension

[Robben]

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?##

 

[abbas_majidi]

<x|x`> is delta function and x` is as x in above.

 

[Robben]

<x|x`> is delta function and x` is as x in above.

I am not sure what you mean?

 

[abbas_majidi]

Above in OP is equation in 'Relevant equations'

 

[BvU]

Since ##\langle \psi|x\rangle## does not depend on ##x'## you can take it out of the integral over ##dx'##:$$
\int\int dx'dx\langle \psi|x\rangle\langle x|x'\rangle\langle x'|\psi\rangle =\int dx \langle \psi|x\rangle\ \left ( \int dx'\langle x|x '\rangle\langle x'|\psi\rangle \right )
$$

 

[Robben]

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 $$\left ( \int dx'\langle x|x '\rangle\langle x'|\psi\rangle \right ) = \langle x|\psi\rangle?$$

 

[Dick]

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')##.

 

[Robben]

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. $$\int dx'\delta(x-x')\langle x'|\psi\rangle?$$

 

[BvU]

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##" .

 

[Robben]

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!