Asymmetric Square Well

The asymmetric square well is the finite-well problem without parity symmetry. The same matching principles used in the symmetric finite well still apply, but one can no longer split the calculation into even and odd sectors. The eigenstates are not centered mirror images, and their probability density generally leans toward the side with the lower or wider barrier tail.

Use the piecewise-constant potential

$$V(x)= \begin{cases} V_L, & x\lt0,\\ 0, & 0\lt x\lt L,\\ V_R, & x\gt L, \end{cases} \qquad V_L\gt0,\quad V_R\gt0.$$

Bound states in this convention have

$$0\lt E\lt\min(V_L,V_R).$$

The wave oscillates inside the well and decays outside both sides. If $V_L=V_R$, the problem is symmetric about the midpoint after shifting the origin; otherwise parity is absent.

Helpful background. Finite Square Well to match oscillating interiors to decaying tails.

Piecewise Solutions

Define

$$k=\frac{\sqrt{2mE}}{\hbar}, \qquad \kappa_L=\frac{\sqrt{2m(V_L-E)}}{\hbar}, \qquad \kappa_R=\frac{\sqrt{2m(V_R-E)}}{\hbar}.$$

Square integrability selects the decaying exterior forms:

$$\psi(x)= \begin{cases} A e^{\kappa_L x}, & x\lt0,\\ B\sin(kx)+C\cos(kx), & 0\lt x\lt L,\\ D e^{-\kappa_R(x-L)}, & x\gt L. \end{cases}$$

The left exterior decays as $x\to-\infty$. The right exterior decays as $x\to+\infty$.

Matching Conditions

Because the potential has finite jumps but no delta-function singularity, both $\psi$ and $\psi'$ are continuous at $x=0$ and $x=L$.

At $x=0$,

$$C=A, \qquad kB=\kappa_L A.$$

Thus the interior solution can be written in a form that already satisfies the left matching condition:

$$\psi_{\mathrm{in}}(x) = C \left[ \cos(kx) +\frac{\kappa_L}{k}\sin(kx) \right].$$

At $x=L$, the right matching condition is equivalently a logarithmic-derivative condition:

$$\frac{\psi_{\mathrm{in}}'(L)} {\psi_{\mathrm{in}}(L)} = -\kappa_R.$$

Substituting the interior solution gives the eigenvalue equation

$$(\kappa_L+\kappa_R)\cos(kL) + \left( \frac{\kappa_L\kappa_R}{k}-k \right) \sin(kL) =0.$$

Equivalently, away from poles and zeros where division would obscure roots,

$$\tan(kL) = \frac{k(\kappa_L+\kappa_R)} {k^2-\kappa_L\kappa_R}.$$

This is a transcendental equation for $E$, because $k$, $\kappa_L$, and $\kappa_R$ all depend on $E$.

Boundary Phases and the Number of Levels

A phase form makes it possible to count every level before finding its energy. Write the interior as $A\sin(kx+\delta_L)$, where left matching fixes

$$\delta_L(E)=\arctan\frac{k}{\kappa_L} =\arcsin\sqrt{\frac{E}{V_L}}, \qquad 0<\delta_L<\frac{\pi}{2}.$$

Define $\delta_R(E)=\arcsin\sqrt{E/V_R}$ in the same way. Right matching requires $k\cot(kL+\delta_L)=-\kappa_R$, hence

$$\Phi(E)\equiv kL+\delta_L(E)+\delta_R(E)=n\pi, \qquad n=1,2,\ldots.$$

These boundary phases are exact for the piecewise-constant model; no WKB approximation has been made. An equivalent arccosine form is derived in Kaplunovsky’s problem-set solution, Problem 1(b,d).

Each term in $\Phi$ increases strictly with $E$. Starting at $\Phi(0)=0$, it approaches

$$\Phi_* =\frac{L\sqrt{2mV_<}}{\hbar} +\frac{\pi}{2} +\arcsin\sqrt{\frac{V_<}{V_>}}, \qquad V_<=\min(V_L,V_R),\quad V_>=\max(V_L,V_R),$$

as $E\to V_<^-$, including the symmetric case $V_<=V_>$. There is exactly one bound level for each positive integer satisfying $n\pi<\Phi_*$. Thus

$$N_{\mathrm{bound}}=\left\lceil\frac{\Phi_*}{\pi}\right\rceil-1.$$

The inequality is strict: $n\pi=\Phi_*$ gives a nondecaying threshold tail, not another bound state. In particular, unequal exterior barriers can give $\Phi_*\le\pi$ and no bound state. For equal barriers, $\Phi_*=L\sqrt{2mV_L}/\hbar+\pi>\pi$, so at least one level exists for every positive width and depth.

No Even Or Odd Classification

For the symmetric finite well, parity lets one solve even and odd sectors separately. Here, if $V_L\ne V_R$,

$$V(x)\ne V(L-x),$$

so reflection about the midpoint is not a symmetry. A bound state is not generally even or odd about the center of the well.

This changes the workflow:

• use both matching interfaces, not just one edge plus a parity condition;
• expect $\langle x\rangle$ to differ from $L/2$;
• expect the evanescent tails to have different lengths;
• classify states by energy ordering and node count, not by parity.

The node ordering from one-dimensional bound-state theory still applies under ordinary conditions: the ground state has no interior node, the first excited state has one, and so on.

Probability Density Shift

The exterior decay lengths are

$$\ell_L=\frac{1}{\kappa_L}, \qquad \ell_R=\frac{1}{\kappa_R}.$$

If $V_R\lt V_L$ for a fixed bound-state energy, then $\kappa_R\lt\kappa_L$, so the right tail is longer. The probability density extends farther into the lower barrier side. This does not mean the particle has enough energy to classically enter that region; it means the classically forbidden decay is weaker there.

Asymmetry can also shift the interior density. Low-lying states tend to lean toward the side where leakage is easier or the effective confinement is weaker. The precise shift depends on the full matching problem, not only on the barrier heights.

Worked Example: Unequal Exterior Barriers

Let $E_L=\hbar^2/(2mL^2)$, $V_L=9E_L$, and $V_R=4E_L$. For $\epsilon=E/E_L$, the phase equation is

$$\sqrt\epsilon +\arcsin\sqrt{\epsilon/9} +\arcsin\sqrt{\epsilon/4} =n\pi.$$

Here $\Phi_*=2+\pi/2+\arcsin(2/3)\simeq4.30052$, between $\pi$ and $2\pi$, so there is exactly one bound state. Bisection gives

$$\epsilon\simeq2.63449, \qquad kL\simeq1.62311, \qquad\kappa_LL\simeq2.52300, \qquad\kappa_RL\simeq1.16855.$$

To normalize without discarding either tail, use interior amplitude $A\sin(kx+\delta_L)$. Then

$$|A|^{-2} =\int_0^L\sin^2(kx+\delta_L)\,dx +\frac{\sin^2\delta_L}{2\kappa_L} +\frac{\sin^2(kL+\delta_L)}{2\kappa_R}.$$

The three terms are the interior, left-tail, and right-tail norm contributions. They give probabilities $0.69822$, $0.05152$, and $0.25026$, respectively, and $\langle x\rangle\simeq0.72970L$. The longer right tail carries almost five times the probability of the left tail. This is stationary penetration into the lower barrier: the bound eigenfunction is real and has zero current.

Numerical Solution

A robust numerical workflow is:

1. Choose $V_L$, $V_R$, $L$, and units.
2. Search the interval $0\lt E\lt\min(V_L,V_R)$.
3. Use $\Phi_*$ to enumerate the integers $n\pi<\Phi_*$. For each one, solve the strictly increasing equation $\Phi(E)-n\pi=0$ by bisection or Brent’s method. Its endpoint limits at $0$ and $\min(V_L,V_R)$ have opposite signs, so each permitted integer brackets precisely one level.
4. Independently check the original, undivided matching function

$$F(E) = (\kappa_L+\kappa_R)\cos(kL) + \left( \frac{\kappa_L\kappa_R}{k}-k \right) \sin(kL).$$

The identity $\sin\Phi=kF/\sqrt{(k^2+\kappa_L^2)(k^2+\kappa_R^2)}$ explains the agreement for $E>0$. The undivided equation has no tangent poles. A coarse sign scan of $F$ alone can miss pairs of roots; the phase count supplies the missing completeness check.

5. Normalize the resulting piecewise wavefunction, including both tails, and verify continuity of $\psi$ and $\psi'$ at both interfaces. Close to threshold, also check that the positive binding energy $\min(V_L,V_R)-E$ is numerically resolved.

Finite-difference or basis diagonalization avoids explicit transcendental root finding and is often the better route for more complicated asymmetric wells. Use Finite Difference Methods, Matrix Diagonalization, and Sparse Eigensolvers for the numerical toolkit.

Limiting Cases

Several limits are useful checks.

If $V_L,V_R\to\infty$, then $\kappa_L,\kappa_R\to\infty$ and the wavefunction is forced toward

$$\psi(0)=0, \qquad \psi(L)=0.$$

The spectrum approaches the infinite square well:

$$E_n\to \frac{n^2\pi^2\hbar^2}{2mL^2}.$$

If $V_L=V_R$, the problem is symmetric about $x=L/2$. The eigenstates can be recast as even and odd functions about the midpoint, and the single matching equation above contains both parity families.

If one barrier becomes infinite while the other remains finite, the problem becomes a half-hard-wall finite well. One side has a hard Dirichlet boundary, while the other side has an evanescent tail. Unlike a symmetric full-line attractive well, unequal exterior thresholds can leave no bound state. For $V_L>V_R$, put $k_*=\sqrt{2mV_R}/\hbar$ and $\kappa_*=\sqrt{2m(V_L-V_R)}/\hbar$. The first state becomes bound only when

$$k_*L>\arctan(\kappa_*/k_*).$$

At equality the right tail reaches threshold and is not normalizable. This follows by writing the interior as $\sin(kx+\delta_L)$, with $\delta_L=\arctan(k/\kappa_L)$, and matching $kL+\delta_L+\arctan(k/\kappa_R)=\pi$ for the nodeless state.

If $E$ approaches the lower of $V_L$ and $V_R$ from below, the corresponding tail length diverges. At threshold, the state can cease to be a normalizable bound state and merge with the continuum on that side.

Common Mistakes

• Using even and odd finite-well equations when $V_L\ne V_R$.
• Setting $\psi=0$ at a finite barrier.
• Forgetting that the two decay constants are different.
• Solving only one interface and assuming the other is automatic.
• Dividing by $\cos(kL)$ or $\sin(kL)$ and accidentally losing roots.
• Expecting $\langle x\rangle=L/2$ without symmetry.
• Calling a long tail classically allowed just because the probability density is nonzero.

Exercises

1. Derive the left-matched interior solution.

Solution

Inside the well,

$$\psi_{\mathrm{in}}(x)=B\sin(kx)+C\cos(kx).$$

At $x=0$, continuity gives $C=A$. Derivative continuity gives

$$kB=\kappa_L A=\kappa_L C.$$

Thus

$$B=\frac{\kappa_L}{k}C,$$

and therefore

$$\psi_{\mathrm{in}}(x) = C \left[ \cos(kx) +\frac{\kappa_L}{k}\sin(kx) \right].$$

2. Starting from the left-matched interior solution, derive the eigenvalue equation at $x=L$.

Solution

The interior solution is

$$\psi_{\mathrm{in}}(x) = C \left[ \cos(kx) +\frac{\kappa_L}{k}\sin(kx) \right].$$

Its derivative is

$$\psi_{\mathrm{in}}'(x) = C \left[ -k\sin(kx) +\kappa_L\cos(kx) \right].$$

Matching to the right decaying exponential gives

$$\psi_{\mathrm{in}}'(L) = -\kappa_R\psi_{\mathrm{in}}(L).$$

Substitution yields

$$(\kappa_L+\kappa_R)\cos(kL) + \left( \frac{\kappa_L\kappa_R}{k}-k \right) \sin(kL) =0.$$

3. If $V_R\lt V_L$, which tail is longer for the same bound-state energy?

Solution

The decay constants are

$$\kappa_L=\frac{\sqrt{2m(V_L-E)}}{\hbar}, \qquad \kappa_R=\frac{\sqrt{2m(V_R-E)}}{\hbar}.$$

If $V_R\lt V_L$, then $\kappa_R\lt\kappa_L$. The right decay length $1/\kappa_R$ is therefore larger, so the right tail is longer.

4. Keep the physical barrier heights of the worked example fixed, but reduce its width from $L$ to $cL$. How many bound states remain for $c=1/2$ and $c=1/4$? Find the critical width and explain why equality does not count as a bound state.

Solution

The ratio $V_R/V_L=4/9$ stays fixed, while the propagation phase at threshold scales with the width:

$$\Phi_*(c)=2c+\frac{\pi}{2}+\arcsin\frac23.$$

At $c=1/2$, this is about $3.30052$, just above $\pi$, so one bound state remains. At $c=1/4$, it is about $2.80052$, below $\pi$, so no bound states remain. The first level is bound precisely when

$$c>c_{\mathrm{crit}} =\frac12\arccos\frac23 \simeq0.42053.$$

At equality, $E=V_R$ and the exterior solution on the right is constant or linear instead of decaying. No nonzero such tail is square integrable. Setting it identically to zero would force both the value and derivative at the interface to vanish, giving the zero solution everywhere. Thus the threshold candidate must be excluded.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• V. Kaplunovsky, “Solutions for problem set #6,” graduate Quantum Mechanics, University of Texas at Austin, Fall 2023, Problem 1(b,d), course solution.
• E. Merzbacher, Quantum Mechanics, 3rd ed., Wiley, 1998.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.