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?

  • $\begingroup$ The integration-by-parts for 1.30 & 1.31 are explained in this other post Integration by parts to derive π‘‘βŸ¨π‘₯⟩/𝑑𝑑. My question about 1.29 stands though. $\endgroup$
    – Jopplk
    Sep 11, 2023 at 3:22
  • $\begingroup$ Do you mean he extracts $\partial/\partial x$? For $(1.25)$ he uses the Schrodinger equation to relate the time derivative to a spatial derivative and factors out a $\partial/\partial x$. Have you tried doing the differentiation yourself to see how it comes out? $\endgroup$ Sep 11, 2023 at 3:47
  • 1
    $\begingroup$ @KrisWalker He doesn't factor out the $\partial/\partial x$, because OP is right that this should create four terms when pushed back through the parentheses. However, if one computes that, one will find that two of the terms cancel each other out. $\endgroup$
    – march
    Sep 11, 2023 at 4:01
  • $\begingroup$ @march I'm using the word "factor" very loosely here. I'm aware of the cancellation, which is why I suggested the OP do the differentiation themselves. $\endgroup$ Sep 11, 2023 at 4:09
  • $\begingroup$ Embarrassingly, you all are very correct. $\endgroup$
    – Jopplk
    Sep 13, 2023 at 21:53

1 Answer 1


The integration-by-parts for 1.30 & 1.31 are explained in this other post: Integration by parts to derive π‘‘βŸ¨π‘₯⟩/𝑑𝑑.

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.


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