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The finite well and tunneling

Exponential decay into forbidden regions, penetration depths, the bound states of a finite well, and tunneling through barriers with $T \approx e^{-2\kappa L}$.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to find decay constants and penetration depths, count the bound states of a finite well, and estimate tunneling probabilities through barriers.

2. What you already have

You solved the infinite square well, whose wave functions vanish at rigid walls, and you know the free particle's oscillating solutions $e^{\pm ikx}$. You also saw that the harmonic oscillator's ground state reaches beyond the classical turning points. This lesson makes the walls finite and follows the wave function into the region where, classically, the particle could never be.

3. Words for this lesson

4. Wave functions leak through finite walls

Where $V = V_0$ is larger than $E$, the Schrödinger equation becomes

$$\psi'' = \kappa^2\psi, \qquad \kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar},$$

whose solutions are $e^{\pm\kappa x}$, real exponentials instead of oscillations. In a wall that extends to infinity only the decaying one is allowed, so the wave function reaches into the wall and fades over a penetration depth $1/\kappa$. For an electron a few electron volts below the top, that depth is about $0.1$ nm.

In a finite square well, matching the oscillating interior to the decaying exterior (both $\psi$ and $\psi'$ continuous at each wall) allows only certain energies. The wave function now spills past the walls, so it fits more easily and every level lies a little below its infinite-well value. And there are only finitely many bound states: a well of width $2a$ has $N = \lceil z_0/(\pi/2)\rceil$ of them, with $z_0 = (a/\hbar)\sqrt{2mV_0}$. A one-dimensional well always has at least one.

If the wall is a barrier of finite width $L$, the decaying wave function reaches the far side with amplitude reduced by about $e^{-\kappa L}$ and continues as a traveling wave. The probability of getting through, for a thick barrier with $\kappa L \gg 1$, is

$$T \approx \frac{16E(V_0 - E)}{V_0^2}\,e^{-2\kappa L} \sim e^{-2\kappa L}.$$

The exponential dominates: this is tunneling, and its sensitivity to $L$ and to $V_0 - E$ is what makes it both useful and dangerous in technology.

5. The method, step by step, and how to check it

1. The gap. Find $V_0 - E$, the height of the barrier above the particle's energy. For electrons leaving a metal, the relevant gap is the work function, about $4$ to $5$ eV.
2. The decay constant. $\kappa = \sqrt{2m(V_0 - E)}/\hbar$. For an electron, $\hbar^2/2m = 0.0381$ eV nm², so $\kappa = \sqrt{(V_0 - E)/0.0381}$ nm⁻¹. For a proton or an alpha particle use its own mass: heavier particles have much larger $\kappa$ and tunnel far less.
3. The exponent. $2\kappa L$. If it is below about $1$, the thick-barrier formula fails and the exact result is needed; if it is large, $T \approx e^{-2\kappa L}$ is good to within a factor of a few, which the prefactor fixes.
4. Ratios. When asked how the transmission changes with width or height, work with ratios, where the uncertain prefactor cancels.

Checks. A higher or wider barrier must give less transmission. $\kappa$ grows only as the square root of the gap, but the transmission depends on it exponentially. And bound-state energies in a finite well must lie below their infinite-well values and below $V_0$.

6. Why the exponent has a factor of two

Inside the barrier the wave function is mostly $Ae^{-\kappa x}$, so across a width $L$ its amplitude falls by $e^{-\kappa L}$. Probability is the square of the amplitude, so the probability falls by $e^{-2\kappa L}$. Forgetting the square is the most common slip in tunneling estimates; it changes the answer by a factor that can be millions.

The prefactor $16E(V_0 - E)/V_0^2$ comes from matching the waves at the two edges: some of the incoming wave reflects at the first surface before it ever begins to decay. It is of order one, between $0$ and $4$, so for rough estimates the exponential alone is enough, and in ratios it cancels exactly.

The exact transmission for a rectangular barrier, $T^{-1} = 1 + V_0^2\sinh^2(\kappa L)/[4E(V_0 - E)]$, reduces to the thick-barrier form when $\sinh\kappa L \approx \tfrac{1}{2}e^{\kappa L}$, and it shows that transmission is never exactly zero for a finite barrier, nor exactly one.

7. Why tunneling is so sensitive

The exponential makes tunneling extraordinarily sensitive to the barrier. For an electron and a $1$ eV gap, $\kappa = 5.1$ nm⁻¹: a barrier $0.5$ nm wide transmits about $0.6$ percent, one $1$ nm wide about $0.004$ percent, and one $2$ nm wide about $10^{-9}$. Each extra tenth of a nanometer cuts the transmission by a factor of about $2.8$.

Mass matters as much. A proton is $1836$ times heavier than an electron, so its $\kappa$ is $\sqrt{1836} = 43$ times larger at the same gap, and its exponent for the same barrier is $43$ times larger. That is why electrons tunnel freely through nanometer gaps in electronics while protons tunnel appreciably only across distances comparable to a chemical bond, as they do in some enzyme reactions and in the fusion inside the Sun.

This sensitivity is a gift for measurement, as in the scanning tunneling microscope, and a nuisance for engineering, where it sets the minimum thickness of insulating layers in transistors.

8. Bound states of a finite well

For a finite well of width $2a$ and depth $V_0$, set $z = la$ with $l = \sqrt{2mE_{\text{kin}}}/\hbar$ inside, and $z_0 = (a/\hbar)\sqrt{2mV_0}$. Matching the even solutions at the wall gives the transcendental equation $\tan z = \sqrt{(z_0/z)^2 - 1}$, solved graphically: the right side falls from infinity to zero as $z$ goes from $0$ to $z_0$, and each branch of $\tan z$ crosses it once. The odd solutions satisfy $-\cot z = \sqrt{(z_0/z)^2 - 1}$.

Two limits check the result. A wide, deep well has $z_0$ large, the crossings sit just below $z = n\pi/2$, and the energies approach the infinite well's. A shallow, narrow well has $z_0 < \pi/2$ and only one bound state, the even ground state, which in one dimension always exists, however weak the well. The number of bound states, $\lceil 2z_0/\pi\rceil$, grows with both the width and the square root of the depth.

Semiconductor quantum wells, a few nanometers of gallium arsenide between layers of aluminum gallium arsenide, are finite wells a few hundred milli-electron volts deep, and their handful of bound states set the colors of laser diodes.

9. Tunneling in nature

Tunneling is not a curiosity. Alpha decay is an alpha particle tunneling out of the nucleus through the Coulomb barrier; George Gamow explained it this way in 1928, and the exponential sensitivity explains why half-lives range from microseconds to billions of years for alpha energies that differ by only a factor of two. The Sun shines because protons tunnel through their mutual Coulomb repulsion: at the core's temperature they have only about a thousandth of the energy needed to get over the barrier classically.

In chemistry, hydrogen atoms tunnel in some reactions, which shows up as a large difference in rate between ordinary hydrogen and deuterium, twice as heavy. In electronics, tunnel diodes, flash memory and the Josephson junctions of superconducting quantum computers all depend on it. The WKB method, later in this course, generalizes $e^{-2\kappa L}$ to barriers of any shape by replacing $\kappa L$ with $\int\kappa(x)\,dx$.

10. In the world: the scanning tunneling microscope

In 1981 Gerd Binnig and Heinrich Rohrer at IBM Zurich built a microscope that images individual atoms by tunneling, winning the 1986 Nobel Prize. A metal tip, sharpened to a single atom at its apex, is held a few tenths of a nanometer above a conducting surface with a small voltage between them. Electrons tunnel across the vacuum gap, whose barrier height is about the work function, $4$ to $5$ eV.

With $\phi = 4.5$ eV, $\kappa = 5.123\sqrt{4.5} = 10.9$ nm⁻¹, so moving the tip $0.1$ nm closer multiplies the current by $e^{2 \times 10.9 \times 0.1} = 8.8$. A feedback loop raises and lowers the tip to hold the current fixed as it scans, and the tip's height traces the surface with a vertical resolution of a few picometers, a hundredth of an atom's diameter. Because almost all the current flows through the single atom closest to the surface, the lateral resolution is atomic too. IBM researchers later used the same instrument to push individual xenon atoms into place, spelling out the company's initials in 35 atoms.

11. In the world: flash memory

Every USB drive, solid-state disk and smartphone stores data in flash memory, where each bit is a small amount of charge trapped on a floating gate surrounded by insulating oxide. To write, a high voltage tilts the oxide's barrier into a triangle thin enough for electrons to tunnel through it (Fowler–Nordheim tunneling); to hold the data, the voltage is removed and the full barrier, about $3$ eV high and several nanometers thick, keeps the electrons in place.

The design balances the exponential in both directions. Through a $7$ nm oxide with a $3.2$ eV barrier, $2\kappa L = 2 \times 5.123\sqrt{3.2} \times 7 = 128$, so $T \sim e^{-128} \approx 10^{-56}$ per attempt, and the charge stays put for years. Thinning the oxide to squeeze more cells onto a chip raises the leakage exponentially, one reason flash cells stopped shrinking and manufacturers began stacking them in three dimensions, more than two hundred layers high, instead.

12. A particle does not need to go over a barrier to get past it

Classically, a particle with energy below a barrier's top is always reflected. Quantum mechanically its wave function extends into the barrier, decaying exponentially, and if the barrier is thin enough some of it emerges on the far side: the particle tunnels through with probability about $e^{-2\kappa L}$. It does not borrow energy or go over the top; its energy is the same on both sides.

Two related errors are to think the transmission falls in proportion to the width, when it falls exponentially, and to think a higher barrier matters as much as a wider one. Width enters the exponent directly, while the height enters only through $\sqrt{V_0 - E}$: doubling the width squares the attenuation, while doubling the gap raises it to the power $\sqrt{2} = 1.41$.

13. How far an electron leaks into a wall

1. An electron sits $2.0$ eV below the top of a well's walls. Write the equation outside.

   $\psi'' = \kappa^2\psi$

   Where $V > E$ the curvature has the same sign as $\psi$.

2. Write the allowed solution in the wall.

   $\psi = Ce^{-\kappa x}$

   The growing exponential cannot be normalized.

3. Evaluate the decay constant.

   $\kappa = \sqrt{\dfrac{2.0}{0.0381}} = 7.25\ \text{nm}^{-1}$

   With $\hbar^2/2m = 0.0381$ eV nm².

4. Find the penetration depth.

   $\delta = \dfrac{1}{\kappa} = 0.138\ \text{nm}$

   The amplitude falls by $e$ over this distance.

5. Find the probability density one depth into the wall.

   $\dfrac{|\psi(\delta)|^2}{|\psi(0)|^2} = e^{-2} = 0.135$

   The density falls twice as fast as the amplitude.

14. Tunneling through a thin barrier

1. An electron of energy $1.0$ eV meets a barrier of height $3.0$ eV and width $0.40$ nm. Find the gap.

   $V_0 - E = 3.0 - 1.0 = 2.0\ \text{eV}$

   What the barrier rises above the electron's energy.

2. Find the decay constant.

   $\kappa = \sqrt{\dfrac{2.0}{0.0381}} = 7.25\ \text{nm}^{-1}$

   From the gap.

3. Find the exponent.

   $2\kappa L = 2 \times 7.25 \times 0.40 = 5.80$

   Well above one, so the thick-barrier formula applies.

4. Find the prefactor.

   $\dfrac{16E(V_0 - E)}{V_0^2} = \dfrac{16 \times 1.0 \times 2.0}{9.0} = 3.56$

   From matching at the edges.

5. Evaluate the transmission.

   $T \approx 3.56 \times e^{-5.80} = 3.56 \times 0.00303 = 0.0108$

   About one electron in a hundred gets through.

6. Compare with the rough estimate.

   $e^{-5.80} = 0.0030$

   The exponential alone is within a factor of four.

15. Counting bound states in a finite well

1. An electron is in a well $1.0$ nm wide and $1.5$ eV deep. Write the well's strength.

   $z_0 = \dfrac{a}{\hbar}\sqrt{2mV_0} = a\sqrt{\dfrac{V_0}{\hbar^2/2m}}$

   Here $a$ is the half-width, $0.50$ nm.

2. Evaluate the square root.

   $\sqrt{\dfrac{1.5}{0.0381}} = 6.27\ \text{nm}^{-1}$

   The decay constant a particle at the bottom would have.

3. Multiply by the half-width.

   $z_0 = 0.50 \times 6.27 = 3.14$

   Almost exactly $\pi$.

4. Count the bound states.

   $N = \left\lceil \dfrac{z_0}{\pi/2} \right\rceil = \lceil 2.00 \rceil = 2$

   One new state appears each time $z_0$ passes a multiple of $\pi/2$.

5. Find the infinite-well ground energy for comparison.

   $E_1^{\infty} = \dfrac{0.376}{1.0^2} = 0.376\ \text{eV}$

   With $E_1 = 0.376$ eV nm²$/a^2$ for a well of full width $1.0$ nm.

6. Estimate the finite-well ground energy.

   $z \approx 1.18: \quad E_1 = \dfrac{\hbar^2}{2m}\left(\dfrac{z}{a}\right)^2 = 0.0381 \times \left(\dfrac{1.18}{0.50}\right)^2 = 0.214\ \text{eV}$

   The root of $\tan z = \sqrt{(3.14/z)^2 - 1}$, found numerically.

7. Compare the two ground energies.

   $0.214\ \text{eV} < 0.376\ \text{eV}$

   Leaking into the walls gives the wave more room, lowering its energy.

8. Check that the level is bound.

   $0.214\ \text{eV} < V_0 = 1.5\ \text{eV}$

   Measured from the bottom of the well, the state lies below the rim.

16. Your turn: an electron sits $4.0$ eV below the top of a barrier. How far does its wave function penetrate?

1. Write the decay constant.

   $\kappa = \sqrt{(V_0 - E)/0.0381}$

   In nm⁻¹ with the gap in eV.

2. Substitute the values.

   $\kappa = \sqrt{4.0/0.0381} = 10.2\ \text{nm}^{-1}$

   Four times the $1$ eV gap doubles $\kappa$.

3. Your turn: work this step out. Its working is at the end of the packet.

   Invert for the depth.

17. Guided practice

An electron's wave function penetrates $0.18$ nm into a wall that rises some height above its energy. If the wall rose four times as far above the electron's energy, how far would the wave function penetrate?

• $0.09$ nm, half as far
• $0.045$ nm, a quarter as far
• $0.36$ nm, twice as far
• zero, because the electron cannot enter the wall

18. Guided practice

Complete the worked solution: a barrier has decay constant $\kappa = 7$ nm⁻¹. Compare the transmission through a barrier $2$ nm wide with one three times as wide.

1. Find the exponent for the narrow barrier.

   $2\kappa L = 2 \times 7 \times 2 =$ _a

   The transmission is about $e$ to minus this number.

2. Find the exponent for the wide barrier.

   $2\kappa(3L) =$ _b

   Tripling the width triples the exponent.

3. Subtract the two exponents.

   $\dfrac{T_{\text{narrow}}}{T_{\text{wide}}} = e^{\,b - a} = e^{\,\text{gap}}$ with gap $=$ _c

   Each extra decay length cuts the transmission by a constant factor.

19. Guided practice

A particle of energy $E$ meets a region where $V = V_0 > E$. Match each quantity to its expression.

20. Practice

For an electron, $\kappa = 5.123\sqrt{V_0 - E}$ nm⁻¹ with $V_0 - E$ in eV. Fill in $\kappa$, in nm⁻¹, for each gap.

21. Practice

An electron in a finite well has an energy $2$ eV below the top of the well's walls. How far into the walls does its wave function penetrate, in nm? Use $\hbar^2/2m = 0.0381$ eV nm².

Answer: nm

22. Practice

An electron meets a rectangular barrier that rises $0.5$ eV above its energy and is $1$ nm wide. Estimate the probability that it tunnels through, in percent, using $T \approx e^{-2\kappa L}$ and $\hbar^2/2m = 0.0381$ eV nm².

Answer: %

23. Somewhere new

A scanning tunneling microscope holds a sharp tip above a metal whose work function is $4.5$ eV; electrons tunnel across the vacuum gap. By what factor does the tunneling current rise when the tip moves $0.1$ nm closer?

Answer:

24. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

25. Test question

For an electron, $\kappa = 5.123\sqrt{V_0 - E}$ nm⁻¹ with $V_0 - E$ in eV. Fill in $\kappa$, in nm⁻¹, for each gap.

26. What you can do now

You can estimate how a wave function leaks through a barrier. Explain to someone how a scanning tunneling microscope sees individual atoms.

Working for the steps left to you

16. Your turn: an electron sits $4.0$ eV below the top of a barrier. How far does its wave function penetrate?, step 3