# Is there a time delay during tunnelling?

A particle hitting a square potential barrier can tunnel through it to get to the other side and carry on. Is there a time delay in this process?

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– user7757 Jan 23 '14 at 13:19

There is a roughly constant time delay during tunnelling that is roughly proportional to $\hbar$ over the barrier height.

Consider for simplicity a monoenergetic particle beam of energy $E=\hbar^2k^2/2m$, trying to tunnel through a barrier of length $l$ and height $V_0$. This means that the wavefunction coming up to the barrier will go as $e^{ikx}$, and some portion of it will tunnel through and make a tunnelled wavefunction right of the barrier going as $A_\textrm{tunnel}e^{ikx}$. This is the standard treatment as given in UG texts on QM (I'm working off of Cohen-Tannoudji) and it can be expanded to accommodate a time-dependent, moving wavepacket by superposing different energies in the usual way.

The tunnelling time is given by the phase difference between the incoming and the tunnelled components - that is, in the phase of $A_\textrm{tunnel}$. This phase counts the number of wave peaks and troughs that the outgoing wavefunction has been delayed by, which essentially gives the time delay you want. For a time-dependent wavepacket, the different phases between different-energy components make up the position where the wavepacket is, so it should be no surprise. In such a situation, dividing the accumulated phase by the central energy will give the time.

(In a related note, it's important to realize that unless the pulse is very narrow in energy and thus quite broad in time then the barrier will be highly dispersive and can quite radically alter the pulse's shape. This is because the higher-energy components will tunnel more easily than the longer-wavelength ones.)

The details of the square-barrier tunnelling are given in Cohen-Tannoudji, for example (vol I, 1977 English edition, p. 72), from which one can get the tunnelling amplitude as $$A_\textrm{tunnel}=\frac{e^{-\kappa l}}{\cosh(\kappa l)- i\frac{k^2-\kappa^2}{k\kappa}\sinh(\kappa l)}$$ where $\kappa=\sqrt{2m(V_0-E)}/\hbar$. The first important feature is that this is indeed exponentially decreasing with $l$, going as $A_\textrm{tunnel}\propto e^{-2\kappa l}$ for largish $l\gtrsim 1/\kappa$. The second important feature is that it does have a nontrivial phase. For simplicity consider the case of a pretty large barrier, for which $$A_\textrm{tunnel}\approx\frac{e^{-2\kappa l}}{1- i\frac{k^2-\kappa^2}{k\kappa}} =\frac{ik\kappa e^{-2\kappa l}}{k^2-\kappa^2+ik\kappa} =\frac{ik\kappa e^{-2\kappa l}}{(k+i\kappa)^2} = e^{-2\kappa l}\frac{k\kappa}{|k+i\kappa|^2}e^{i\frac{\pi}{2}-2i\arctan(\kappa/k)} .$$ For $E=V_0/2$, you get $k=\kappa$ and the phase vanishes. For $E<V_0$ the phase approaches $e^{i\pi/2}$ as $E\rightarrow0$, and similarly it approaches $e^{i\pi/2}$ as $E\rightarrow V_0$ from the left. That is, the wavefunction gets delayed/advanced by at most 1/4 of a wavelength.

For a real pulse the situation is a bit more complicated because you should distinguish between a shift in the global phase, which is irrelevant, and a $k$-dependent phase that will shift the pulse. Thus it's the derivative $\partial \varphi/$ that matters (where $A_\textrm{tunnel}=|A_\textrm{tunnel}|e^{i\varphi}$). Since $\varphi$ is more or less a straight line from $E=0$, $\varphi=\pi/2$ to $E=V_0$, $\varphi=-\pi/2$, the upshot is that the final, tunnelled pulse is displaced to the left in space by about $$\Delta x=\frac{\partial \varphi}{\partial k}(k_0)=-\frac{\pi\hbar}{V_0}\frac{\hbar k_0}{m}$$ and this corresponds to a delay in time by $h/2V_0$. (Note that the weird scaling at "low" $V_0$ is offset by the requirement that $\kappa l\leq \sqrt{V_0}l$ be large in natural units, so the limit doesn't really make sense.)

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if the delay is independent from the width of the tunnel, can't this in principle lead to FTL signal propagation? – lurscher Dec 3 '12 at 20:01
I don't think it can, since it's a further delay on the slower-than-light, nonrelativistic Schrödinger propagation of a free particle. – Emilio Pisanty Dec 4 '12 at 2:27
On the other hand, though, to be quite honest I think the maths is right but would be glad for someone else to go over it with a fine comb and to correct where necessary. – Emilio Pisanty Dec 4 '12 at 2:28

For a wave packet, it should be the barrier width divided by the packet group velocity, I think.

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Well, what if the group velocity is imaginary? – Emilio Pisanty Dec 3 '12 at 17:28