4
votes
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 ...
- 709
3
votes
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.
...
- 45.1k
2
votes
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 ...
- 17.6k
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 ...
- 45.1k
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 \...
- 3,298
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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