# Proof that rotational symmetric potential operators are scalar operators

Defintion: A scalar operator B is an operator on a ket space that transforms under rotations $$\begin{equation}\left| \xi ' \right >=\exp{(\frac{i}{h} \mathbf{\phi \cdot J})}\left| \xi \right >\end{equation}$$ in such a way that $$\begin{equation}\left< \xi ' |B| \psi'\right>=\left< \xi |B| \psi\right>\end{equation}$$

I demonstrated that a operator B is a scalar operator if and only if $$0=[J_i,B]$$

What I'd like to show next is that the Hamiltonian $$H=\frac{\mathbf{P}^2}{2m}+V$$ is a scalar operator for "rotational symmetric potential operators".

Sadly I have conceptual difficulties with this potential operator and find the treatment in all textbooks I've read so far very bad. Most don't talk about the potential operator acting on kets but instead about a basis representation of this operator acting on a wavefunction - Without even using different notation for both. Furthermore I can't extent the concept of rotational symmetrie I know from classical mechanics to this abstract operator V. According to my exercise sheet the above result should be right though.

This question is linked to the unanswered question Is potential energy a scalar operator?

• You need to compute a commutator of the form $[\mathbf{r} \times \mathbf{p}, V(\mathbf{r})]$. Do you know how to decompose such commutators? – knzhou Jul 9 '19 at 19:38
• Since I do not know how the operator V commutes with either the momentum nor the position operator, no. And by saying that V is dependet on $\mathbf{r}$ do you mean that V is dependet on the position operator? – TheoreticalMinimum Jul 9 '19 at 19:43
• No, $V(\mathbf{r})$ is a function of the operator $\mathbf{r}$. For example, if the function $V$ is the identity function, then $V(\mathbf{r})$ is the operator $\mathbf{r}$. – knzhou Jul 9 '19 at 19:45
• It's important to distinguish between when $\mathbf{r}$ is a parameter (i.e. a set of three numbers), and when it's a set of three operators. In this case it is a set of three operators. If it were three numbers, then $V(\mathbf{r})$ would merely be a number, so of course its commutator with anything else would vanish. – knzhou Jul 9 '19 at 19:45
• If this question is open a bit later, when I have time, I'll type up how to do it without position representation. – knzhou Jul 9 '19 at 20:08

Thanks to the assuring comments by user @knzhou I figured out how to do this last night. Using the above theorem it's enough to prove $$\begin{equation} 0=[\mathbf{J},H]=[\mathbf{J},\frac{\mathbf{P}^2}{2m}+V(\mathbf{X})] \end{equation}$$ The firs term in the commutator is trivially zero. But one has to prove, that $$\begin{equation} 0\stackrel{\text{!}}{=}[\mathbf{J},V(\mathbf{X})]=[\mathbf{X}\times\mathbf{P},V(\mathbf{X})] \end{equation}$$ Using the bijective map defined in Applying an operator to a function vs. a (ket) vector, it is enough to show that for any State $$\left|\boldsymbol{\psi}\right>$$ $$\begin{equation} 0=\left<\mathbf{x}|[\mathbf{X}\times\mathbf{P},V(\mathbf{X})]|\boldsymbol{\psi}\right> =\left<\mathbf{x}|\mathbf{X}\times\mathbf{P}V(\mathbf{X})-V(\mathbf{X})\mathbf{X}\times\mathbf{P}|\boldsymbol{\psi}\right>\\ =\int \mathbf{dx'}\left(\left<\mathbf{x}|\mathbf{X}\times\mathbf{P}|\mathbf{x'}\right>\left<\mathbf{x'}|V(\mathbf{X})|\boldsymbol{\psi}\right>-\left<\mathbf{x}|V(\mathbf{X})|\mathbf{x'}\right>\left<\mathbf{x'}|\mathbf{X}\times\mathbf{P}|\boldsymbol{\psi}\right>\right) \end{equation}$$ Now we assume that any potential $$V(\mathbf{X})$$ we are dealing with can be expanded in terms of the position operator $$\mathbf{X}$$ as $$\begin{equation} V(\mathbf{X})=\sum v_i \mathbf{X}^i \end{equation}$$ Furthermore the coefficients of the potential are supposed to be real, so that using that the momentum operator $$\mathbf{X}$$ is hermitian one can deduce that $$V(\mathbf{X})$$ is hermitian as well. Now it's easy to see that $$\begin{equation} \left<\mathbf{x}\right|V(\mathbf{X})=\sum v_i \left<\mathbf{x}\right|\mathbf{X}^i=\sum v_i \left<\mathbf{x}\right|\mathbf{x}^i=V(\mathbf{x}) \end{equation}$$ This is very useful in evaluating the integral above, continuing this equation $$\begin{equation} 0\stackrel{\text{!}}{=}\int \mathbf{dx'}\left(\left<\mathbf{x}|\mathbf{X}\times\mathbf{P}|\mathbf{x'}\right>V(\mathbf{x})\left<\mathbf{x'}|\boldsymbol{\psi}\right>-V(\mathbf{x})\left<\mathbf{x}|\mathbf{x'}\right>\left<\mathbf{x'}|\mathbf{X}\times\mathbf{P}|\boldsymbol{\psi}\right>\right)\\ =\int \mathbf{dx'}\epsilon_{ijk}\left(\left<\mathbf{x}|X_jP_k|\mathbf{x'}\right>V(\mathbf{x})\psi(\mathbf{x'})-V(\mathbf{x})\left<\mathbf{x}|X_jP_k|\mathbf{x'}\right>\left<\mathbf{x'}|\boldsymbol{\psi}\right>\right)\\ =\int \mathbf{dx'}\epsilon_{ijk}\left(\left<\mathbf{x}|X_jP_k|\mathbf{x'}\right>V(\mathbf{x})\psi(\mathbf{x'})-V(\mathbf{x})\left<\mathbf{x}|X_jP_k|\mathbf{x'}\right>\psi(\mathbf{x'})\right) \end{equation}$$ Where we used the definition of the wave function $$\psi(\mathbf(x'):=\left<\mathbf{x'}|\boldsymbol{\psi}\right>$$ and the scalar product and completeness relation of the position eigenkets. Now we note $$\begin{equation} \left<\mathbf{x}|X_jP_k|\mathbf{x'}\right>=x_j\left<\mathbf{x}|P_k|\mathbf{x'}\right>=x_jih\frac{\partial}{\partial x'_k}\delta(x-x') \end{equation}$$ After doing "partial integration" one concludes $$\begin{equation} 0\stackrel{\text{!}}{=}\epsilon_{ijk}x_j\left(\frac{\partial}{\partial x_k}(V(\mathbf{x})\psi(\mathbf{x}))-V(\mathbf{x})\frac{\partial}{\partial x_k}\psi(\mathbf{x})\right)\\ =\epsilon_{ijk}x_j\psi(\mathbf{x})\frac{\partial}{\partial x_k}V(\mathbf{x}) \end{equation}$$ Now we use that our potential is not arbitrary but $$V(\mathbf{x})=V(||\mathbf{x}||_2)$$. $$\begin{equation} 0\stackrel{\text{!}}{=}\epsilon_{ijk}x_j\psi(\mathbf{x})\frac{\partial}{\partial x_k}V(||\mathbf{x}||_2)=\epsilon_{ijk}x_jx_k\psi(\mathbf{x})\frac{V(||\mathbf{x}||_2)}{||\mathbf{x}||_2}=0 \end{equation}$$ Which concludes the proof.