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Because tunneling can happen at all accessible energies, the spectrum will be purely continuous, i.e., there will be no stationary states in the Hilbert space of square integrable wave functions. The solutions of $H\psi=E\psi$ corresponding to the continuous spectrum have real $E$ and are concentrated near $x=0$, with sinusoidal tails outside $[-x_0,x_0]$. ...

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I'm not sure if I understand your question, but let me give it a shot. Many times in physics, I often encounter a lot of trial and error functions and the rule of thumb my professors say is if it fits the boundary conditions, it is correct. An example I can give is solutions to laplace's equation where more often than not, after using separation of ...

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In$$x=A\cos(\omega t+\phi)$$ $\phi$ is not the left shift by $\phi$ radians, it actually is the left shift by $\dfrac{\phi}{\omega}$ radians. You can see it by factoring $\omega$: $$x=A\cos(\omega(t+\dfrac{\phi}{\omega}))$$ $T$ is the time taken for $1$ revolution, so there will be $\dfrac{1}{T}$ revolutions in $1$ second. Since there are $2\pi$ radians in $... 1 How to qualitatively draw the stationary states of this problem (of course, without solving it in detail) Start with the obvious facts: The potential is symmetric:$V(-x)=V(x)$. The potential is bounded and is a potential barrier of finite width. The solution for$|x|>x_0$is a plane wave. The first fact lets you classify the solutions in two kinds: ... 1 We can numerically determine the first n eigenfunctions in a box$-L\le x\le L$with such a potential. We put $$H=-\frac{1}{2}\nabla ^2+\frac{1}{2}x^2$$ for$-4\le x\le 4$, and $$H=-\frac{1}{2}\nabla ^2$$ for$|x|>4\$. Figure 1 shows the modulus of the wave function for n = 30 and L = 20. Eigenvalues are shown above. It can be seen that there are ...

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