I have read in Griffiths' Quantum Mechanics that there is a phenomenon called tunneling, where a particle has some nonzero probability of passing through a potential even if $E < V(x)_{max}$.

What I don't understand about this is how to conceptualize how this can happen. I have read on Wikipedia that tunneling means that objects can "in a sense, borrow energy from their surroundings to cross the wall". How can the object "know" that across the wall there's going to be a lower energy and, thus, the borrowed energy will be restored and not depleted.


3 Answers 3


You're just beginning your study of quantum mechanics, so I would advise you to be careful not to try to interpret quantum mechanics through the lens of classical mechanics. It's a very reasonable thing to imagine quantum tunneling as a little ball which magically pops through a barrier and emerges on the other side, but that is an outstanding way to develop bad intuition which you'll need to fix down the line. Quantum mechanics is fundamentally different from classical mechanics, and it is the latter which should be understood as a limiting case of the former, not the other way around. In that sense, the real question should be not why quantum particles can tunnel, but why classical particles (whatever that means) apparently cannot.

With that being said, the rough idea is the following. We can gain some useful intuition by studying the simpler case of what happens when a particle encounters a potential step of the form $$V(x) = \begin{cases} 0 & x<0 \\V_0 & x\geq 0\end{cases}$$ and then extend this to a potential barrier of width $L$, because the latter is just a step up followed by a step down.

The (generalized) eigenstate corresponding to a particle incident on the barrier from the left with energy $E=\hbar^2k^2/2m<V_0$ takes the form $$\psi_k(x) = \begin{cases} e^{ikx} + r_k e^{-ikx} & x < 0 \\ t_k e^{-q_k x} & x \geq 0\end{cases}$$ where $$\matrix{q_k \equiv \sqrt{\frac{2m(V_0 - E)}{\hbar^2}} \\ r_k \equiv \frac{2iq_k}{k-iq_k}\\ t_k \equiv 1+r_k = \frac{k+iq_k}{k-iq_k}}$$

enter image description here

Based on this picture, we might imagine (correctly) that there is a nonzero probability of measuring a particle with $E<V_0$ within the potential step. However, we need to be a bit careful - this is a non-normalizable (and hence unphysical) state, after all, so if we want to understand what happens dynamically, we should construct a real, physical state. Such states take the form of wavepackets, which may be written

$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int \mathrm dk \ A(k) \psi_k(x) e^{-iE_kt/\hbar}$$ for some square-integrable function $A(k)$ (where $E_k \equiv \hbar^2 k^2/2m$). In essence, $A(k)$ tells us how much of the state with energy $E_k$ is present in the wavepacket. The take-away is that real states consist of an integral superposition of energy eigenstates, not specific energies, and if we want to understand what happens dynamically when a particle encounters a potential step, we need to consider what happens to one of these wavepackets.

The specifics of this are actually rarely covered in detail because while the process is conceptually fairly simple, the calculations are tedious and need to be performed numerically. The qualitative picture goes like this:

  • The components of the wavepacket with energy $E>V_0$ are partially reflected and partially transmitted. The transmitted parts propagate forever in the $+x$ direction.
  • The components of the wavepacket with energy $E<V_0$ are all reflected eventually; however, they penetrate into the barrier by an exponentially small distance ($\psi_k\sim e^{-x/\ell_k}$, where $\ell_k=1/q_k$) and are delayed by a correspondingly small amount of time before being reflected.

In particular, if all of the components of the wavepacket have energy less than $V_0$, then the wavepacket will be perfectly reflected - however, it will be distorted because the different components penetrate different depths into the step before being reflected, and during the reflection there will be a nonzero (but exponentially small) chance of measuring the particle to be physically located at some $x>0$.

We can now turn our attention to your main question of what happens when we have a potential barrier of width $L$, and a wavepacket whose components all have energy less than $V_0$. From a qualitative and dynamic perspective, everything proceeds exactly as it did with the potential step. As the wavepacket approaches the barrier, its components penetrate into the classically forbidden region by an exponentially small distance before being reflected. However, because the barrier has a finite width $L$, a fraction $\sim e^{-L/\ell_k}\equiv e^{-q_k L}$ of the components of the wavepacket will make it all the way through the barrier and escape to the other side$^\dagger$.

You can find an animation of such a process here. Note that the mean energy of the wavepacket in this simulation is much lower than $V_0$, and so essentially none of the wavepacket is able to reach the far end of the barrier. However, observe the exponentially-suppressed penetration of the wavepacket into the front side of the barrier, and then imagine what would happen if the barrier were significantly thinner so the wave amplitude at the back edge was not effectively zero.

How can the object "know" that across the wall there's going to be a lower energy and, thus, the borrowed energy will be restored and not depleted.

I think the "borrowing energy" metaphor is not really a good way to think about it, for essentially the reason you mention. The particle doesn't need to know that the barrier has finite width; the penetration of the wavepacket into the barrier proceeds the same way in both cases, but if the barrier is not infinitely long then an exponentially small fraction of the wavepacket will reach the back edge and escape.

$^\dagger$In fact, this is an oversimplification. In reality, the components of the wavepacket which reach the back edge of the potential are not perfectly transmitted - some of them reflect backward into the barrier, so the precise expression for the tunneling amplitude is a bit more subtle than simply calculating $e^{-q_k L}$ (though that does provide the right order of magnitude).

Remark on Localization

(My initial reading of the question was sloppy, and I thought OP was asking about a potential step rather than a potential barrier. As a result, this is no longer particularly relevant, but it is mildly interesting, so I elected to include it as an afterthought.)

As an interesting side note, it turns out that a particle which is initially localized to some compact interval $[x_1,x_2]$ to the left of the barrier (by which I mean, $\psi_0(x)=0$ for all $x\notin[x_1,x_2]$), then the wavepacket must contain components with energy $E>V_0$. This is related to a well-known theorem about Fourier transforms which says that a function and its Fourier transform cannot both be compactly-supported; in this context, the interpretation is that the better-localized you want your initial particle to be, the more high-energy components you will need to include in the wavepacket.

As a result, a wavepacket with average energy $E<V_0$ which is initially localized to a compact interval $[x_1,x_2]$ will always be partially transmitted, even through an infinitely long potential step, because it will contain some high-energy components which exceed the barrier height. Of course, even more of such a wavepacket would be transmitted through a potential barrier of width $L$, because the high-energy components would be partially transmitted and an exponentially small fraction of the low-energy components would be able to tunnel.

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    $\begingroup$ this is an awesome response thanks! and thanks for that suggestion on reformulating my way of thought - I will for sure keep this in mind. so regarding why classical particles don't exhibit tunneling - i am trying to think about this, but it's unclear to me what the classical limit actually is. i read online that it is when hbar goes to zero, but I am not sure what this physically means. my thought was to see what the limit is of and to see if maybe in this limit something is gone that leads to no tunneling, but I am stuck at the hbar -> 0 concept. should I just keep studying? $\endgroup$ Jul 11, 2022 at 20:51
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    $\begingroup$ @Relativisticcucumber Keep studying, but this might help for now. The quantum leeway in momentum, which translates into a leeway in energy in J. Murray's argument, is something like $\hbar/\epsilon$ by dimensional analysis. As $\hbar\to0$, this leeway $\to0$. But as you study more, you'll see that the classical loss of tunneling opportunities is also tied to particle count (due to e.g. law of large numbers) & overall mass. $\endgroup$
    – J.G.
    Jul 11, 2022 at 21:36
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    $\begingroup$ "Indeed, if all of the energies present in the initial superposition are below the barrier height, then the probability that the particle will tunnel is precisely zero" -- this is incorrect. This would imply (in 1D, which is implicit in your statement) that a sufficiently high potential barrier, no matter how narrow, will result in complete reflection (and no transmission) of an incoming plane wave. ... $\endgroup$
    – nanoman
    Jul 12, 2022 at 7:56
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    $\begingroup$ ... But this isn't true -- e.g., as the barrier becomes higher and narrower, it approaches a delta function potential, which has a nonzero transmission amplitude. More generally, any barrier of finite height and width has some nonzero (perhaps very small) probability of being tunneled through by a particle of any energy. $\endgroup$
    – nanoman
    Jul 12, 2022 at 8:10
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    $\begingroup$ @Relativisticcucumber IMO, J. Murray's answer attempts to give an intuitive explanation but is not technically accurate. Superposition of energies is not the controlling factor because even a particle with a definite energy behaves nonclassically (is not bounded to the classically allowed region). $\endgroup$
    – nanoman
    Jul 12, 2022 at 18:46

From the viewpoint of the Schrödinger equation, the key relation is $$\hbar k = \sqrt{2m(E - V)},$$ where $k$ is the wave number ($\psi \propto e^{ikx}$). This holds locally in a spatially varying potential $V$ under the semiclassical WKB approximation. In the classically allowed region, $E > V$ and $k$ is real, i.e., the wave function is oscillatory. As $V$ increases and approaches $E$ (classical turning point), the kinetic energy and momentum go to zero and the oscillations halt. Then when $E < V$ (tunneling region), $k$ is imaginary ($k = i\kappa$) and the wave function decays rather than oscillating ($\psi \propto e^{-\kappa x}$). This spatial decay occurs at a finite rate $\hbar\kappa = \sqrt{2m(V - E)}$, and so some of the wave function reaches the other side as long as the barrier is finite in height and width.

How can the object "know" that across the wall there's going to be a lower energy and, thus, the borrowed energy will be restored and not depleted.

It doesn't need to know. If the barrier extends for infinite distance (e.g., if $V$ continues to grow without bound), the wave function will still decay at the prescribed finite rate, and there will be some nonzero probability of finding the particle far into the classically forbidden region. This probability merely continues to rapidly decrease rather than ever "breaking free".


You're treating Quantum objects as balls. This is misleading. When working with Quantum Mechanics, picture the object as the entire wave. So tunneling happens because parts of the wave pass through the potential. If there isn't a lower energy across the wall, the wave won't pass.

I like to imagine it as a water wave and it might leak if there is nothing behind the wall.


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