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This is really straight forward, once you get used to the notation. (Don't you hate it when people say that?) $$[\pi (\vec{x},t), (-\nabla^{2} +m^{2})\phi (\vec{y},t)] ,$$ Here you need to remember that $\nabla^2$ acts on the $\phi(\vec{y},t)$ only, so $\pi$ can pass right through this wave operator. Now when you evaluate the commutator you'll end up with ...


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Any, the Hamiltonian whos potential nergy is a function of $r$ only has rotational invariance. Since the components of $\mathbf{L}$ are generators of rotation, it can be shown that $[\mathbf{L},V]=[L^2,V]=0$


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Angular momentum is defined as, $$ L=r\times p\equiv r\times\frac{\hbar}{i}\nabla $$ From this it follows that $$ \left[L_x,\,x\right]=\left[L_x,\,p_x\right]=0 $$ where $$ L_x\sim yp_z-zp_y $$ Noting that $$ \left[A,BC\right]=\left[A,B\right]C+B\left[A,C\right] $$ the expected results can be determined.


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Yes, for the hydrogen atom Hamiltonian, $L^2$ commutes with the Hamiltonian because it commutes with $r^2$.


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Let $\mathcal{H}$ be the Hilbert space for one particle. Then, $S_{x}\in\mathcal{B}(\mathcal{H})$ is a bounded, self-adjoint operator. Now, if you want to have the Hilbert space for two particles, remember that this is the tensor product, i.e. $\mathcal{H}=\mathcal{H}_1\otimes \mathcal{H}_2$ (where $\mathcal{H}_1$ is the Hilbert space of the first and ...


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You can think of the continuos formalism as being the limiting case of the discrete-momentum one: if the momentum is taken as a discrete variable (which amounts to constraining the particles to be in some finite volume $V$) the fourier expansion of the (real, scalar) field is: $$ \tag{1} \varphi(x) = \sum_{\textbf{k}} \frac{1}{\sqrt{2V \omega_{\textbf{k}}}} ...



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