Quantum Mechanics: Wave Mechanics in One Dimension

1. Mar 30, 2015

Robben

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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 $\int dx \langle\psi|x\rangle\langle x|\psi\rangle?$

2. Mar 31, 2015

abbas_majidi

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

3. Mar 31, 2015

Robben

I am not sure what you mean?

4. Mar 31, 2015

abbas_majidi

Above in OP is equation in 'Relevant equations'

5. Mar 31, 2015

BvU

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

6. Mar 31, 2015

Robben

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

7. Mar 31, 2015

Dick

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

8. Apr 1, 2015

Robben

How does $\delta(x-x')$ act on $\langle x'|\psi\rangle$ in order for it to equal $\langle x|\psi\rangle$, i.e. $$\int dx'\delta(x-x')\langle x'|\psi\rangle?$$

9. Apr 1, 2015

BvU

$$\int^{\infty}_{-\infty}dxf(x)\delta(x-x_0)=f(x_0)$$to function values, but also to functions. So at each $x'$ in $\langle x'|\psi \rangle = \psi(x')$ is "replaced by $x$" .