Hermitian conjugate of differential operator

Question

Help me find $\hat{B^\dagger}$, when we know that $$\hat{B}=i\frac{d}{dr}$$ with the condition that $\hat{B}$ is defined in spherical coordinates. My approach: $$ \langle\psi|\hat{B}\psi\rangle=\int_{0}^{\infty} \psi^* i\frac{d}{dr} \psi dr=\psi^*\psi|_{0}^{\infty}-\int_{0}^{\infty}\psi i\frac{d}{dr}\psi^*dr=\langle\hat{B}\psi|\psi^*\rangle$$ And so I get $ \hat{B^\dagger} = -i\frac{d}{dr} $. Could someone confirm if this is correct?

Answer 1

So yeah, what you need to do is successfully integrate by parts. In spherical coordinates the integral is:$$\langle\phi|\hat B|\psi\rangle = \int_0^\infty dr~\int_0^\pi r~d\phi~\int_0^{2\pi}r\sin\phi~d\theta\;\phi^*(r,\phi,\theta) ~i\left(\frac{\partial\psi}{\partial r}\right)_{\phi,\theta}.$$The integration by parts on the variable $r$ differentiates $r^2 \phi^*(\dots)$ producing $2r \phi^* + r^2 \partial_r\phi^*,$ but we have to pull the $r^2$ out of the front again, back into the integral.

This means that the adjoint of $\hat B$ (the thing that does the same thing as $\hat B$ when acting on the bra-space rather than the ket-space for all inner products) is $$\hat B^\dagger = -i\left(\frac{2}{r} + \frac \partial{\partial r}\right),$$where the minus sign comes from the integration by parts itself.

Answer 2

Hermitian conjugate (also called adjoint) of the operator $A$ is the operator $A^*$ satisfying $$\langle f,Ag\rangle\,=\,\langle A^* f,g\rangle \text{ for all }f,g\,\in H$$ $H$ is so-called Hilbert space and $f,g$ are vectors. Since you are new to QM, you need not be confused with the word "Hilbert space". Just treat it as a special case of vector spaces.

What you want to know is the form of $A^*$ satisfying $$\langle f,i\frac{\mathrm{d}g}{\mathrm{d}x}\rangle\,=\,\langle A^* f,g\rangle \,\text{ for all }f,g\,\in H$$ and you seem to be interested in showing hermiticity of $A$, so assume that the form of $A^*$ is also $i\frac{\mathrm{d}}{\mathrm{d}x}$ and by integrating-by-parts, $$\langle f,i\frac{dg}{dx}\rangle-\langle i\frac{df}{dx} ,g\rangle\,=\,f^*g\,|_{interval}$$

Physically plausible wavefunctions in QM are usually $f( \infty )=0$ (entire space) or $f(2\pi)=f(0)$ (spherical symmetry). With this conditions we now face $\langle f,i\frac{\mathrm{d}g}{\mathrm{d}x}\rangle=\langle i\frac{\mathrm{d}f}{\mathrm{d}x} ,g\rangle$

Here we know that $A=i\frac{\mathrm{d}}{\mathrm{d}x}$ is Hermitian, saying $A$ has its adjoint $A^*=i \frac{\mathrm{d}}{\mathrm{d}x}$.

In the same manner you might see that $B=\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}r}$ is whether Hermitian or not.

In QM, operators which correspond to physical quantities are self-adjoint, not just Hermitian in spite of a lot of basic QM books concentrating on Hermitianity of operators so once you become confident with the theories of operators, you can go forward to see what self-adjointness is.

(edit)

In your approach, what you should consider is

1. volume element of spherical coordinate is $dV=r^2 sin\theta drd\theta d\rho$ not just $dV=drd\theta d\rho$.

2. $\langle B\psi|\psi\rangle=\int d\tau \overline{B \psi(\tau)}\psi$, not just $\langle B\psi|\psi\rangle=\int d\tau {B \psi(\tau)}\psi$

3. Momentum-operator in Cartesian coordinate is $-i \hbar \frac{d}{dx}$ but in spherically coordinated space corresponding momentum operator should change its form, not merely $-i \hbar \frac{d}{dr}$. So we cannot guarantee Hermitianity of $-i \hbar \frac{d}{dr}$.