Semicircular potential well

Semicircular potential well

In quantum mechanics, the case of a particle in a one-dimensional ring is similar to the particle in a box. The particle follows the path of a semicircle from 0 to pi where it cannot escape, because the potential from pi to 2 pi is infinite. Instead there is total reflection, meaning the particle bounces back and forth between 0 to pi . The Schrödinger equation for a free particle which is restricted to a semicircle (technically, whose configuration space is the circle S^1) is

-frac{hbar^2}{2m}nabla^2 psi = Epsi quad (1)

Wave function

Using cylindrical coordinates on the 1 dimensional semicircle, the wave function depends only on the angular coordinate, and so

nabla^2 = frac{1}{s^2} frac{partial^2}{partial phi^2} quad (2)

Substituting the Laplacian in cylindrical coordinates, the wave function is therefore expressed as

-frac{hbar^2}{2m s^2} frac{d^2psi}{dphi^2} = Epsi quad (3)

The moment of inertia for a semicircle, best expressed in cylindrical coordinates, is I stackrel{mathrm{def}}{=} iiint_V r^2 ,rho(r,phi,z),r dr,dphi,dz !. Solving the integral, one finds that the moment of inertia of a semicircle is I=m s^2 , exactly the same for a hoop of the same radius. The wave function can now be expressed as -frac{hbar^2}{2I} psi = Epsi , which is easily solvable.

Since the particle cannot escape the region from 0 to pi , the general solution to this differential equation is

psi (phi) = A cos(m phi) + B sin (m phi) quad (4)

Defining m=sqrt {frac{2 I E}{hbar^2}} , we can calculate the energy as E= frac{m^2 hbar ^2}{2I} . We then apply the boundary conditions, where psi and frac{dpsi}{dphi} are continuous and the wave function is normalizable:

int_{0}^{pi} left| psi (phi ) right|^2 , dphi = 1 quad (5) .

Like the infinite square well, the first boundary condition demands that the wave function equals 0 at both phi = 0 and phi = pi . Basically

psi (0) = psi (pi) = 0 quad (6) .

Since the wave function psi(0) = 0 , the coefficient A must equal 0 because cos (0) = 1 . The wave function also equals 0 at phi= pi so we must apply this boundary condition. Discarding the trivial solution where B=0, the wave function psi (pi) = 0 = B sin (m pi) only when m is an integer since sin (n pi) = 0 . This boundary condition quantizes the energy where the energy equals E= frac{m^2 hbar ^2}{2I} where m is any integer. The condition m=0 is ruled out because psi = 0 everywhere, meaning that the particle is not in the potential at all. Negative integers are also ruled out.

We then normalize the wave function, yielding a result where B= sqrt{frac{2}{pi}} . The normalized wave function is

psi (phi) = sqrt{frac{2}{pi}} sin (m phi) quad (7) .

The ground state energy of the system is E= frac{hbar ^2}{2I} . Like the particle in a box, there exists nodes in the excited states of the system where both psi (phi) and psi (phi) ^2 are both 0, which means that the probability of finding the particle at these nodes are 0.


Since the wave function is only dependent on the azimuthal angle phi , the measurable quantities of the system are the angular position and angular momentum, expressed with the operators phi and L_z respectively.

Using cylindrical coordinates, the operators phi and L_z are expressed as phi and -i hbar frac{d}{dphi} respectively, where these observables play a role similar to position and momentum for the particle in a box. The commutation and uncertainty relations for angular position and angular momentum are given as follows:

[phi, L_z] = i hbar psi(phi) quad (8)

(Delta phi) (Delta L_z) geq frac{hbar}{2} where Delta_{psi} phi = sqrt{langle {phi}^2rangle_psi - langle {phi}rangle_psi ^2} and Delta_{psi} L_z = sqrt{langle {L_z}^2rangle_psi - langle {L_z}rangle_psi ^2} quad (9)

Boundary conditions

As with all quantum mechanics problems, if the boundary conditions are changed so does the wave function. If a particle is confined to the motion of an entire ring ranging from 0 to 2 pi , the particle is subject only to a periodic boundary condition (see particle in a ring). If a particle is confined to the motion of frac{- pi}{2} to frac{pi}{2} , the issue of even and odd parity becomes important.

The wave equation for such a potential is given as:

psi_o (phi) = sqrt{frac{2}{pi}} cos (m phi) quad (10)
psi_e (phi) = sqrt{frac{2}{pi}} sin (m phi) quad (11)

where psi_o (phi) and psi_e (phi) are for odd and even m respectively.

Similarly, if the semicircular potential well is a finite well, the solution will resemble that of the finite potential well where the angular operators phi and L_z replace the linear operators x and p.

See also

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