Derivation of Quantum Momentum Operator in Griffiths

Question

I've been reading through Griffith's Intro to Quantum book and can't understand some of the steps in his derivations. The first one is in his proof of time-independence for the normalization of wave function. Eq 1.25 seems to extract $\frac{\partial}{\partial t}$ from the terms within the parenthesis.

$$ \frac{d}{dt} \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx = \int_{-\infty}^{\infty} \frac{\partial}{\partial t} |\Psi(x,t)|^2 dx \tag{1.21} $$

$$ \frac{\partial}{\partial t} |\Psi|^2 = \frac{i \hbar}{2m} \left( \Psi^* \frac{\partial^2 \Psi}{\partial x^2} - \frac{\partial^2 \Psi^*}{\partial x^2} \Psi \right) = \frac{\partial}{\partial x} \left[ \frac{i \hbar}{2m} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right] \tag{1.25} $$

I'm not sure if this makes sense, as "distributing" $\frac{\partial}{\partial t}$ over the terms would create 4 terms a la product rule right? This is used later when he derives the quantum momentum operator from $\left< x \right>$,

$$ \frac{\partial \left< x \right>}{\partial t} = \int x \frac{\partial }{\partial t} |\Psi|^2 dx = \frac{i \hbar}{2m}\int x \frac{\partial}{\partial x} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi\right) dx \tag{1.29} $$

The next step he cites integration-by-parts and product rule, but it looks like he just cancels out the $dx$ and $x$,

$$ \frac{\partial \left< x \right>}{\partial t} = -\frac{i \hbar}{2m} \int \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) dx \tag{1.30} $$

And then another integration-by-parts that I also don't see,

$$ \frac{\partial \left< x \right>}{\partial t} = -\frac{i \hbar}{m} \int \Psi^* \frac{\partial \Psi}{\partial x} dx \tag{1.31} $$

Basically, how did we get 1.25, and what does the outline for the integration-by-parts for 1.30 look like?

Answer 1

The integration-by-parts for 1.30 & 1.31 are explained in this other post: Integration by parts to derive 𝑑⟨𝑥⟩/𝑑𝑡.

Equation 1.25 is actually exactly as it seems - the product rule results in two terms that cancel out. Taking the RHS of 1.25,

$$ \frac{\partial}{\partial x} \left[ \frac{i \hbar}{2m} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right] \tag{From 1.25} $$

$$ \frac{i \hbar}{2m} \left[ \left(\frac{\partial \Psi^*}{\partial x}\frac{\partial \Psi}{\partial x} + \Psi^*\frac{\partial^2 \Psi}{\partial x^2}\right) - \left(\frac{\partial^2 \Psi^*}{\partial x^2}\Psi + \frac{\partial \Psi^*}{\partial x}\frac{\partial \Psi}{\partial x}\right) \right] $$

Pretty clearly, the first and last term cancel out, therefore: $$ \frac{\partial}{\partial x} \left[ \frac{i \hbar}{2m} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right] = \frac{i \hbar}{2m} \left[ \Psi^* \frac{\partial^2 \Psi}{\partial x^2} - \frac{\partial^2 \Psi^*}{\partial x^2} \Psi \right] $$ Which is Equation 1.25.