Time dependence of scalar product

[Shredface]

How do I show that the scalar product is time independent?

I have: $$\frac{d}{dt}\int\Psi^{*}_{1}(x,t)\Psi_{2}(x,t)dx = 0$$

And have proceeded to take the derivatives inside the integral and using the time dependent Schrodinger eq. ending up with:

$$\frac{i}{\hbar}\int\left(\Psi^{*}_{1}\widehat{H}\Psi_{2}-\Psi_{2}\widehat{H}\Psi^{*}_{1}\right)dx$$

Using the Hermitian properties of $$\widehat{H}$$ I then got to:

$$\frac{i}{\hbar}\int\left(\Psi^{*}_{1}\widehat{H}\Psi_{2} - \Psi_{1}\widehat{H}\Psi^{*}_{2}\right)dx$$

Either these two terms equivalent meaning they cancel and I have my result or I've made a drastic error somewhere. Can anyone help me out?

 

[Meir Achuz]

Your last step is wrong. The hermitian conjugate of $$\Psi^*_1 H|\Psi_2$$ is $$\Psi^*_1 H|\Psi_2$$.
The stars on Psi don't change.

 

[ansgar]

$$\frac{i}{\hbar}\int\left(\Psi^{*}_{1}\widehat{H}\Psi_{2}-\Psi_{2}\widehat{H}\Psi^{*}_{1}\right)dx$$

should read

$$\frac{i}{\hbar}\int\left(\Psi^{*}_{1}(\widehat{H}\Psi_{2})-(\widehat{H}\Psi^{*}_{1})\Psi_{2}\right)dx$$

hermiticy:

$$\int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx$$
gives
$$- \frac{i}{\hbar}\int\left( (\widehat{H}\Psi_{1})^{*}\Psi_{2}-(\widehat{H}\Psi^{*}_{1})\Psi_{2}\right)dx$$