Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that $[\frac {d^2}{dx^2}]\psi$ is $\frac {d^2\psi}{dx^2}$ but what about this one $[\frac {d^2\psi}{dx^2}]\psi^*$? Is it this like $\frac {d^2\psi\psi^*}{dx^2}$ or this like $\frac {\psi^*d^2\psi}{dx^2}$?

And what about this one $[\frac {d^2\psi}{dx^2}]\phi$?

share|cite|improve this question

I think the answer is the second one you are suggesting. In words, what you are doing is multiplying the second derivative of $\psi$ by ${\psi^*}$. The double derivative (by your notation, with the way you use the brackets) is only applied to ${\psi}$, and then it's just multiplication.

share|cite|improve this answer
i've seen both of them in QM book! – rabi Dec 10 '12 at 6:56
It's very typical to find what seems like (and I'm sure, sometimes really is) an ambiguity in a textbook. Besides asking your best physicist friend what he thinks, one option is always to try every possible interpretation you can think of, starting with the likeliest-sounding one, and see if it fits. Usually, some thought and some calculation will align one of the interpretations with the context and eliminate the others. – ruggy Dec 10 '12 at 16:48

Brackets can have different meaning depending on context. In the first case, the brackets enclose an operator which acts on the function to the right. In the second case the brackets enclose an operator already acting in a function and the result of the operation is multiplied by the function to the right. Therefore

$$\left[\frac {d^2\psi}{dx^2}\right]\psi^* = \frac {\psi^*d^2\psi}{dx^2}$$


$$\left[\frac {d^2\psi}{dx^2}\right]\phi = \frac {\phi d^2\psi}{dx^2}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.