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For each $r>0$, the divergence of the magnetic field of the monopole is zero as you have already checked; \begin{align} \nabla\cdot\mathbf B(\mathbf x) = 0, \qquad \text{for all $\mathbf x\neq \math …

Here's the most logical way to proceed if you ask me. Given any $a\in\mathbb R$, we define the translation operator $T_a$ by its action on position basis vectors $$ T_a|x\rangle = |x + a\rangle $$ …

This is not a peculiar physicist notation oddly enough. The notation allows one to interpret $1/x$ as a distribution (which makes sense since it's being added to the delta distribution on the right h …

The Hilbert dimension of the Hilbert space of a free particle is countable. To see this, note that The Hilbert space of a free particle in three dimensions is $L^2(\mathbb{R}^3)$. An orthonormal ba …

A "kosher" way to do this employs test functions. Consider a test function $\phi:\mathbb R^4\to \mathbb R$. Notice that \begin{align} \int_{\mathbb R^4} d^4 x\, \partial_\mu j^\mu(x) \phi(x) &= ec …

One needs to be careful about what one mean by the "size" of a vector space. A theorem of functional analysis tells us that any two Hilbert bases for a Hilbert space must have the same cardinality. T …

No. The expectation value of the square of the momentum operator cannot be negative. The other answers address your particular problem on an integration level, but also notice that this can be easily …

1) Notice that by inserting a complete set of position states we can write $$ \hat p \psi(x) = \langle x|\hat p|\psi\rangle = \int dx'\langle x|\hat p|x'\rangle\langle x'|\psi\rangle =\int dx'\langl …