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I got this problem from a book called Introduction to quantum mechanics, griffin 2nd edition. and I did not get why the solution says

first term integrates to zero, integration by parts twice?!

Please see the solution below! Thanks everyone for helping!

  • $\begingroup$ This question appears to be off-topic because it is about interpreting a book, and not about physics. $\endgroup$ Sep 8, 2013 at 8:15

1 Answer 1


He uses

$$\int\left(\frac{\partial}{\partial x}f(x)\right)\ g(x)\ \text dx=\int\ f(x)\left(-\frac{\partial}{\partial x}g(x)\right)\ \text dx,$$

to get

$$\int\left(\frac{\partial^2}{\partial x^2}\Psi^*\right)\frac{\partial}{\partial x}\Psi\ \text dx =\int\left(\frac{\partial}{\partial x}\Psi^*\right)\left(-\frac{\partial^2}{\partial x^2}\Psi\right)\ \text dx=\int\Psi^*\frac{\partial^3}{\partial x^3}\Psi\ \text dx.$$

The fact that you can shift the sqare of the derivative $\frac{\partial^2}{\partial x^2}$ around like this is already suggested by $\Delta=\sum_{n=1}^3\frac{\partial^2}{\partial x_n^2}$ being an observable in several contexts. If it's an observable, then it's hermitean and $\langle \Delta\Phi|\Phi\rangle=\langle \Phi|\Delta\Phi\rangle$.

  • $\begingroup$ Sorry,Nick.Can you explain this a bit further since I'm a newbie in this field and just started to learn QM. I don't get what you mean by an observable. $\endgroup$ Sep 7, 2013 at 14:24
  • $\begingroup$ @JamesNgaiChunTat: Well just check wikipedia for starters: wikipedia.org/Observable. The first two lines I wrote are mathematical identities, the last sentence is just for your information. $\endgroup$
    – Nikolaj-K
    Sep 7, 2013 at 16:22

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