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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?

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  • $\begingroup$ How can spherical coordinates after transformation to Cartesian ones give rise to an operator in one dimension only? $\endgroup$ – Gert Sep 26 '15 at 11:32
  • $\begingroup$ B is defined in the spherical coordinate system (in my problem). The sample d/dx, that I started to prepare for my main problem is separate thing, where we solve only in one dimension. As to why B is defined in the spherical, I don't know, but it is not an easy problem. $\endgroup$ – FringeEvent Sep 26 '15 at 11:38
  • $\begingroup$ Here you can find a full solution to the problem. Utpal Roy, Suranjana Ghosh. Kaushik Bhattacharya, Some intricacies of the momentum operator in quantum mechanics. $\endgroup$ – RivuxG Dec 3 '17 at 18:44
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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.

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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.

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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 $-i \hbar \frac{d}{dr}$.

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  • $\begingroup$ Yes, I read the theory, I know the definition of Hermitian conjugate, but now I am trying to find these conjugates for specific cases. I need to find it for B like I found it for d/dx $\endgroup$ – FringeEvent Sep 26 '15 at 12:42
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    $\begingroup$ Discovery: hmm, acc. this source quantummechanics.ucsd.edu/ph130a/130_notes/… the Hermitian conjugate of $\frac{d}{dx}$ is $-\frac{d}{dx}$. $\endgroup$ – Gert Sep 26 '15 at 12:52
  • $\begingroup$ @Gert Uhh... I mistook and was intended to show the Hermitianity of $i\frac{d}{dx}$. Thanks for pointing out. $\endgroup$ – Discovery Sep 27 '15 at 22:44
  • $\begingroup$ @FringeEvent Since $\langle B\psi|\psi\rangle=\int d\tau \overline{B \psi(\tau)}\psi$, we should pay attention to "conjugate" symbol when we calculate QM-inner product. I edited my post. $\endgroup$ – Discovery Sep 27 '15 at 22:47
  • $\begingroup$ @Discovery: Welcome. But I thought a volume element $dV$ in spherical coordinates was given simply by $dV=4\pi r^2dr$? (concentric shell, $dV$ thick). $\endgroup$ – Gert Sep 27 '15 at 22:53

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