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.

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 $E$.

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.

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. Evaluate

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

4. Locate sign changes or bracketed roots, while also checking for missed roots near tangent poles.
5. Normalize the resulting piecewise wavefunction and verify continuity of $\psi$ and $\psi'$ at both interfaces.

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.

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.

Where This Is Used

• Finite Square Well is the symmetric benchmark.
• Boundary Conditions gives the continuity rules at finite jumps.
• Qualitative Features of One-Dimensional Bound States explains node ordering, tails, and localization costs.
• Eigenvalue Problems gives the boundary-value-problem framework.
• Matrix Diagonalization and Sparse Eigensolvers are practical tools for asymmetric wells without closed-form parity reductions.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• E. Merzbacher, Quantum Mechanics, 3rd ed., Wiley, 1998.

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. Why does the symmetric finite-well parity shortcut fail when $V_L\ne V_R$?

Solution

Parity about the midpoint requires the potential to be invariant under $x\mapsto L-x$. Unequal barriers violate that symmetry. The Hamiltonian no longer commutes with the midpoint-reflection operator, so energy eigenstates need not be even or odd. Both interfaces must be matched explicitly.