# Show that for QM operator A: $\int_{-\infty}^{\infty}\psi A^{\dagger}A\psi dx = \int_{-\infty}^{\infty}(A\psi)^*(A\psi)dx$

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...

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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

a) the relationship should be $$\int^\infty_{-\infty} \psi^*A^\dagger A \psi dx = \int^\infty_{-\infty} (A\psi)^*( A\psi) dx$$
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 $A^\dagger = i\frac{d}{dx}$).
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
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