Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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 need to show for $$A = \frac{d}{dx} + \tanh x, \qquad A^{\dagger} = - \frac{d}{dx} + \tanh x,$$ that

$$\int_{-\infty}^{\infty}\psi^* A^{\dagger}A\psi dx = \int_{-\infty}^{\infty}(A\psi)^*(A\psi)dx. $$

Where $\psi$ is a normalized wavefunction. I thought that this was the definition of the Hermitian conjugate of an operator $A$, but the problem asks for me to use integration by parts. I don't really see where the result is going to come from here, surely it is not messy calculation, since it is true in general, right?

I started by by integrating by parts, noting that the surface terms must vanish since $\psi$ is normalized, but after that I just get into a big mess...

share|cite|improve this question
This isn't homework, so I removed the tag. – user27182 Apr 2 '13 at 21:39
The homework tag isn't only for actual homework problems. It's for all questions that are homework-like. (read the tag description for more info) The tag is appropriate here. – Wouter Apr 2 '13 at 21:41
Ok, I'll put it back then. – user27182 Apr 2 '13 at 21:41
Hi Hayeder. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic Apr 2 '13 at 21:42
I did, apologies for my misunderstanding. – user27182 Apr 2 '13 at 21:45

Your set-up isn't quite right.

a) the relationship should be \begin{equation}\int^\infty_{-\infty} \psi^*A^\dagger A \psi dx = \int^\infty_{-\infty} (A\psi)^*( A\psi) dx \end{equation}

b) The Hermitian conjugate of $\frac{d}{dx}$ is $\frac{d}{dx}$. It is not Hermitian, for what it's worth. For that, you need $A= -i\frac{d}{dx}$ (so the Hermitian conjugate is $A^\dagger = i\frac{d}{dx}$).

Hopefully, that will help.

share|cite|improve this answer
Set $D = \frac{d}{dx}$. Integration by parts gives $\langle f, D g \rangle = fg|_{-\infty}^\infty - \langle D f, g\rangle$. So $D^\dagger = -D$. – user1504 Apr 2 '13 at 23:13
Thank you for noticing the error, but $\frac{d}{dx}$ is anti-hermitian – user27182 Apr 3 '13 at 0:40
Yes. That's what I said. No need for 'but'. – user1504 Apr 3 '13 at 0:43
haha, I wasn't talking to you, sorry. I was just letting James know the correct format of the question, rather than speaking on whether or not $\frac{d}{dx}$ is Hermitian. – user27182 Apr 3 '13 at 0:59
Sorry, made a complete hash of that (the lessons of not posting answers just before you go to bed). – James Apr 3 '13 at 12:58

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.