Inside a well, WKB gives an oscillating solution starting from each turning point. The two must be the same function, which is possible only when the phase accumulated across the well is a half-integer multiple of $\pi$. Written as the area of the classical orbit in the plane of position and momentum, the condition says: each state occupies an area $2\pi\hbar$ of phase space, the first one half of it. For the harmonic oscillator this is exact; for other wells it is best for high levels.

Standing waves on a rope fit a whole number of half-waves between its ends. A particle in a well fits a whole number of half-waves between its turning points too — plus a quarter at each soft end.

Classical orbits of the harmonic oscillator in the plane of $x$ and $p$, in units of $x_{0}$ and $p_{0}$, at the energies Bohr–Sommerfeld allows. Each encloses an area , and consecutive orbits differ by exactly : each quantum state takes one cell of that size.

For a potential well with two smooth turning points $a<b$, WKB with the connection formulas gives the condition below. It is exact for the harmonic oscillator and asymptotically exact for large $n$ in general.

• $\oint p\,dx$ is the area enclosed by the classical orbit in phase space.
• Each smooth turning point contributes $\pi/4$; a hard wall contributes $\pi/2$. A box has $\int_{0}^{L}p\,dx=(n+1)\pi\hbar$, exact.

Carry the decaying solution across each turning point into the well. From the left one gets a cosine of the phase measured from $a$, minus $\pi/4$; from the right, a cosine of the phase measured to $b$, minus $\pi/4$. Their phases add to the whole integral minus $\pi/2$, and the two cosines are the same function up to a sign only if that is a whole multiple of .

From the turning point $a$.

From the turning point $b$, whose phase $\frac1\hbar\int_{x}^{b}p\,dx'$ is .

The cosine is even.

Two cosines of the same variable agree up to sign only if their phases differ by a multiple of $\pi$.

Rearrange.

Each quantum state occupies an area $2\pi\hbar$ of the classical phase space; the lowest occupies half of that.

Each smooth turning point contributes $\pi/4$ of phase; the two together add $\pi/2$, which is the half in $n+\tfrac12$.

For the harmonic oscillator the Bohr–Sommerfeld condition gives the exact levels, $\hbar\omega(n+\tfrac12)$; for other potentials it is best for large $n$.

The number of states below an energy is roughly the phase-space area of the orbit at that energy divided by $2\pi\hbar$.

Great work! You now know how to:

• derive the quantisation condition from two connection formulas
• read it as one phase-space cell $2\pi\hbar$ per state
• apply it to the oscillator, a box and a linear well