Double Delta Potential

The double delta potential is the simplest exactly solvable model of two coupled attractive wells. It has enough structure to show tunneling splitting, even and odd parity states, and the bonding-antibonding pattern familiar from molecules, while remaining algebraically much simpler than a smooth double well.

Take two identical attractive delta wells separated by distance $2a$:

$$V(x) =-g\left[ \delta(x-a)+\delta(x+a) \right], \qquad g\gt0.$$

The parameter $a$ is half the separation, and $g$ is the integrated strength of each well. Define

$$\kappa_0=\frac{mg}{\hbar^2}.$$

For a single attractive delta well of strength $g$, the bound-state decay constant is $\kappa_0$. In the double-well problem, the two delta wells communicate through the evanescent tails between them. That communication splits the two one-well states into a lower even state and, when the separation and strength are large enough, an upper odd state.

Hamiltonian And Jump Conditions

The stationary Schrödinger equation is

$$-\frac{\hbar^2}{2m}\psi''(x) -g\left[ \delta(x-a)+\delta(x+a) \right]\psi(x) =E\psi(x).$$

Away from $x=\pm a$, the particle is free. A bound state has $E\lt0$, so write

$$E=-\frac{\hbar^2\kappa^2}{2m}, \qquad \kappa\gt0.$$

At each delta well the wavefunction is continuous, while the derivative jumps. For a well at $x=x_0$,

$$\psi(x_0^-)=\psi(x_0^+)=\psi(x_0),$$

and

$$\psi'(x_0^+)-\psi'(x_0^-) =-2\kappa_0\psi(x_0).$$

These matching conditions are inherited directly from the Delta-Function Potential. The only new feature is that the two singular points must be matched simultaneously.

Parity Sectors

The potential is symmetric:

$$V(x)=V(-x).$$

Therefore bound states can be chosen with definite parity. The even state is lower because it has no node between the wells. The odd state, when it exists, has a node at the origin and is higher in energy.

This parity reduction is not just a convenience. It turns one coupled matching problem into two scalar equations:

$$\kappa_{\mathrm e} =\kappa_0\left(1+e^{-2\kappa_{\mathrm e}a}\right),$$

for the even state, and

$$\kappa_{\mathrm o} =\kappa_0\left(1-e^{-2\kappa_{\mathrm o}a}\right),$$

for the odd state. The rest of the page derives and interprets these equations.

Even Bound State

For an even bound state, choose the form

$$\psi_{\mathrm e}(x) = \begin{cases} B e^{-\kappa(x-a)}, & x\gt a,\\ A\cosh(\kappa x), & -a\lt x\lt a,\\ B e^{\kappa(x+a)}, & x\lt-a. \end{cases}$$

Continuity at $x=a$ gives

$$B=A\cosh(\kappa a).$$

The derivative just outside the right well is

$$\psi_{\mathrm e}'(a^+)=-\kappa B,$$

while the derivative just inside is

$$\psi_{\mathrm e}'(a^-)=A\kappa\sinh(\kappa a).$$

The jump condition at $x=a$ gives

$$-\kappa B-A\kappa\sinh(\kappa a) =-2\kappa_0B.$$

Using $B=A\cosh(\kappa a)$,

$$\kappa\left[ 1+\tanh(\kappa a) \right] =2\kappa_0.$$

Equivalently,

$$\kappa =\kappa_0\left(1+e^{-2\kappa a}\right).$$

This equation has one positive solution for every $g\gt0$ and $a\gt0$. The even bound state therefore always exists.

The energy is

$$E_{\mathrm e} =-\frac{\hbar^2\kappa_{\mathrm e}^2}{2m}.$$

Because $\kappa_{\mathrm e}\gt\kappa_0$, the even state is more deeply bound than the bound state of a single isolated delta well of strength $g$.

Odd Bound State

For an odd bound state, use

$$\psi_{\mathrm o}(x) = \begin{cases} B e^{-\kappa(x-a)}, & x\gt a,\\ A\sinh(\kappa x), & -a\lt x\lt a,\\ -B e^{\kappa(x+a)}, & x\lt-a. \end{cases}$$

Continuity at $x=a$ gives

$$B=A\sinh(\kappa a).$$

The derivative jump at $x=a$ is

$$-\kappa B-A\kappa\cosh(\kappa a) =-2\kappa_0B.$$

After dividing by $B$,

$$\kappa\left[ 1+\coth(\kappa a) \right] =2\kappa_0.$$

Equivalently,

$$\kappa =\kappa_0\left(1-e^{-2\kappa a}\right).$$

This equation does not always have a positive solution. Near $\kappa=0$, the right-hand side behaves as

$$\kappa_0\left(1-e^{-2\kappa a}\right) =2\kappa_0a\,\kappa+O(\kappa^2).$$

A nonzero positive intersection requires

$$2\kappa_0a\gt1.$$

Thus the odd bound state exists only when the wells are sufficiently strong or sufficiently separated:

$$\frac{2mga}{\hbar^2}\gt1.$$

At the threshold $2\kappa_0a=1$, the odd solution reaches $\kappa=0$ and is not a normalizable bound state. Below threshold, the model has only the even bound state.

The odd-state energy, when it exists, is

$$E_{\mathrm o} =-\frac{\hbar^2\kappa_{\mathrm o}^2}{2m}.$$

Since $\kappa_{\mathrm e}\gt\kappa_{\mathrm o}$, the odd state lies above the even state:

$$E_{\mathrm o}\gt E_{\mathrm e}.$$

Tunneling Splitting

When both states exist, the splitting is

$$\Delta E =E_{\mathrm o}-E_{\mathrm e} =\frac{\hbar^2}{2m} \left( \kappa_{\mathrm e}^2-\kappa_{\mathrm o}^2 \right).$$

For large separation, $a\kappa_0\gg1$, the two wells are almost independent. The exponential overlap is small, and the two transcendental equations give

$$\kappa_{\mathrm e} \approx \kappa_0\left(1+e^{-2\kappa_0a}\right), \qquad \kappa_{\mathrm o} \approx \kappa_0\left(1-e^{-2\kappa_0a}\right).$$

Therefore

$$\Delta E \approx \frac{2\hbar^2\kappa_0^2}{m} e^{-2\kappa_0a}.$$

This exponential dependence is the exact-model version of tunneling splitting. The two isolated wells would give degenerate states. Finite overlap through the middle region produces a lower even combination and a higher odd combination.

Bonding And Antibonding Interpretation

Let $\phi_L$ and $\phi_R$ denote approximate bound states localized near the left and right delta wells when $a$ is large. The exact low-energy states are approximately

$$\psi_{\mathrm e} \approx \frac{1}{\sqrt2} \left( \phi_L+\phi_R \right),$$

and

$$\psi_{\mathrm o} \approx \frac{1}{\sqrt2} \left( \phi_L-\phi_R \right).$$

The even state has constructive amplitude between the wells and no node at the center. It is the bonding state in the elementary molecular-orbital analogy. The odd state has destructive amplitude at the center and one node. It is the antibonding state.

The same structure is captured abstractly by a two-state Hamiltonian

$$H_{\mathrm{eff}} = \begin{pmatrix} E_0 & -K \\ -K & E_0 \end{pmatrix}, \qquad K\gt0,$$

whose eigenstates are the symmetric and antisymmetric combinations. The exact double-delta solution shows where this two-state picture comes from and when the splitting is exponentially small. The finite-dimensional version is developed in Two-State Hamiltonians.

Small-Separation Limit

When the two wells approach each other, the potential tends to a single delta well of twice the strength:

$$-g\left[ \delta(x-a)+\delta(x+a) \right] \longrightarrow -2g\delta(x).$$

The even equation becomes

$$\kappa_{\mathrm e}\to 2\kappa_0,$$

which is exactly the single-delta result for strength $2g$. The energy approaches

$$E_{\mathrm e} \to -\frac{\hbar^2(2\kappa_0)^2}{2m} =-\frac{2mg^2}{\hbar^2}.$$

The odd state disappears in this limit. Physically, an odd wavefunction vanishes at the origin, so it cannot benefit from a delta attraction concentrated at the origin after the two wells merge.

Relation To Smooth Double Wells

The double delta model is not meant to be a realistic molecular potential. Its value is conceptual:

• it shows exactly how parity splits a pair of nearly degenerate localized states;
• it gives an explicit exponential splitting at large separation;
• it displays a threshold for the first odd bound state;
• it separates matching-condition algebra from WKB approximation;
• it provides a clean bridge from one-dimensional bound states to two-level systems.

The smoother Double-Well Potential page keeps the same physical ideas but drops exact solvability. Detailed WKB estimates for smooth barriers belong to Barrier Penetration and Tunneling.

Common Mistakes

• Requiring $\psi'$ to be continuous at $x=\pm a$.
• Forgetting that the wavefunction itself remains continuous at each delta well.
• Assuming the odd bound state always exists.
• Confusing the half-separation $a$ with the full well separation $2a$.
• Calling the localized left and right states exact energy eigenstates when the potential is symmetric.
• Missing the sign difference between the even equation and the odd equation.
• Treating the large-separation exponential estimate as valid near the odd-state threshold.

Where This Is Used

• Delta-Function Potential supplies the single-well matching condition and bound-state scale.
• Bound-State Counting explains threshold entry of additional one-dimensional bound states.
• Boundary Conditions explains why delta interactions are encoded by derivative jumps.
• Qualitative Features of One-Dimensional Bound States gives the node and parity reasoning behind the even-lower, odd-higher ordering.
• Double-Well Potential generalizes the same splitting language to smooth wells.
• Two-State Hamiltonians abstracts the bonding-antibonding pair into a finite-dimensional model.

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.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• S. Flügge, Practical Quantum Mechanics, Springer, 1999.

Exercises

1. Derive the even-state equation for the double delta potential.

Solution

Use

$$\psi_{\mathrm e}(x) = \begin{cases} B e^{-\kappa(x-a)}, & x\gt a,\\ A\cosh(\kappa x), & -a\lt x\lt a,\\ B e^{\kappa(x+a)}, & x\lt-a. \end{cases}$$

Continuity at $x=a$ gives

$$B=A\cosh(\kappa a).$$

The derivative jump is

$$-\kappa B-A\kappa\sinh(\kappa a) =-2\kappa_0B.$$

Divide by $B$ and use $A\sinh(\kappa a)/B=\tanh(\kappa a)$:

$$\kappa\left[ 1+\tanh(\kappa a) \right] =2\kappa_0.$$

Since

$$1+\tanh z =\frac{2}{1+e^{-2z}},$$

the equation becomes

$$\kappa =\kappa_0\left(1+e^{-2\kappa a}\right).$$

2. Show that the odd bound state requires $2\kappa_0a\gt1$.

Solution

The odd-state equation is

$$\kappa =\kappa_0\left(1-e^{-2\kappa a}\right).$$

For small $\kappa$,

$$1-e^{-2\kappa a} =2\kappa a+O(\kappa^2).$$

Thus the right-hand side initially has slope $2\kappa_0a$ as a function of $\kappa$, while the left-hand side has slope $1$. A positive nonzero intersection can emerge from threshold only if

$$2\kappa_0a\gt1.$$

At equality the solution is at $\kappa=0$, which is not a normalizable bound state.

3. Find the small-separation limit of the even bound-state energy.

Solution

As $a\to0$, the double delta potential tends to

$$V(x)\to -2g\delta(x).$$

Equivalently, the even equation gives

$$\kappa_{\mathrm e} =\kappa_0\left(1+e^{-2\kappa_{\mathrm e}a}\right) \to 2\kappa_0.$$

Therefore

$$E_{\mathrm e} \to -\frac{\hbar^2(2\kappa_0)^2}{2m} =-\frac{2mg^2}{\hbar^2}.$$

4. Estimate the large-separation splitting.

Solution

For $a\kappa_0\gg1$, insert $\kappa\approx\kappa_0$ into the small exponential terms:

$$\kappa_{\mathrm e} \approx \kappa_0\left(1+e^{-2\kappa_0a}\right), \qquad \kappa_{\mathrm o} \approx \kappa_0\left(1-e^{-2\kappa_0a}\right).$$

Then

$$\Delta E = \frac{\hbar^2}{2m} \left( \kappa_{\mathrm e}^2-\kappa_{\mathrm o}^2 \right) \approx \frac{2\hbar^2\kappa_0^2}{m} e^{-2\kappa_0a}.$$

The splitting is exponentially small in the separation measured in units of the single-well localization length.