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Accepted

### Understanding derivation Klein-Gordon equation

Because there seems to be some confusion about "squaring the wave function", you need to do no such thing, because you're not "squaring Schrödingers equation". Like hft's answer ...

### Landau QM Angular Momentum Eigenvalues Derivation

The argument is that (single particle) wavefunctions should satisfy $\psi(\phi)=\psi(\phi+2\pi)$ since rotating a distribution by $2\pi$ about the $z$ axis should leave the distribution invariant. ...

### Understanding derivation Klein-Gordon equation

The Klein-Gordon Equation is given as follows, using natural units ($\hbar \to1, c \to 1$): $-\frac{\partial^2 \Psi}{{\partial t}^2} = -\nabla^2 \Psi + m^2 \Psi$ Yes, this is the Klein-Gordon ...
1 vote

### Why are all purely real solutions to the time-independent schrodinger equation non-degenerate?

Since the time-independent Schrödinger equation is real, is it such a surprise that the solutions should be real? The reverse of your claim is definitely not true (at least not in general): it is ...
1 vote
Accepted

### Expectation value of momentum in a particular eigenstate

Yes, it is true. An eigenstate of the Hamiltonian $H={p^2\over 2m}+V(x)$ being of the form $$\psi_n(x,t)=\phi_n(x)e^{-iE_nt/\hbar}$$ the average position \eqalign{ \langle x(t)\rangle &=\int \...
1 vote

### Justification of discrete spectrum for $V(x)$ unbounded at $\pm \infty$ in Pauling and Wilson

Yes, there is actually a rigorous proof, for detail's read Simon and Rid, methods of modern mathematical physics, it's the theorem 13.16. (13 roman). It holds for bounded from below hamiltonians (...

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