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How is quantum tunneling possible?

According to quantum mechanics, each particle is represented by a probability density function. This function must be continuous, and therefore when we look at a particle near a potential barrier, we deduce that there is a finite probability for finding the particle inside the barrier (and as a result, beyond the barrier). If the particle can be found inside the barrier, his energy will be negative. This state sounds impossible. Where does the extra energy come from?

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The instantaneously computed value for the kinetic energy would appear to be negative, as you propose. However, the reason for quantum tunneling has to do with Heisenberg uncertainty relations. You may find a particle in a classically forbidden region, but you are much less likely to find it here. The length of time of any departure from this energy conservation condition is limited by the energy departure, so that \begin{equation} \Delta E \Delta t \ge \hbar/2. \end{equation} The larger the apparent energy violation, the more fleeting the event. The uncertainty principle essentially allows the system to momentarily have enough energy for the particle to be in the forbidden region, with the proviso that it may not do so for very long. (This weirdness is also responsible for the Casimir effect.)

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    $\begingroup$ I can't comment, yet. I'd just like to add to KDN's answer that this is called Quantum Fluctuation. $\endgroup$ – seb Feb 17 '13 at 19:11
  • $\begingroup$ Thanks! this uncertainty relation is quite unique, because there is no time operator in quantum mechanics. I googled it and found this proof. it seems that delta t means evolution time of some arbitrary operator rather than simple clock. I'm not really sure what is the meaning of this relation. can energy preservation principle be broken for a short time? $\endgroup$ – Helios Feb 17 '13 at 19:53
  • $\begingroup$ related: physics.stackexchange.com/questions/53802/… $\endgroup$ – joshphysics Feb 17 '13 at 20:15
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    $\begingroup$ It's not so much can energy conservation be broken, It's more that defining the energy of a state takes time. The larger interval you take to measure the energy of a system, the more certain you are about it's energy. If you measure over a very short time and happen to find the particle tunneling, you have only a rough idea of the energy of the system. Measuring the same system over a long time (so that $\Delta t$ is large) yields a much smaller uncertainty in the energy. However, measuring the particle over a long time, you will only rarely find it tunneling. $\endgroup$ – KDN Feb 18 '13 at 1:19
  • $\begingroup$ A quotation from Griffiths' Introduction to Quantum Mechanics, at the end of par. 3.5.3: "It is often said that the uncertainty principle means energy is not strictly conserved in quantum mechanics [...]; the greater the violation, the briefer the period over which it can occur. Now, there are many legitimate readings of the energy-time uncertainty principle, but this is not one of them. Nowhere does quantum mechanics license violation of energy conservation [...]". $\endgroup$ – HicHaecHoc Jul 23 '20 at 7:37
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In quantum mechanics you have to be careful what you mean by 'the energy'. In QM the energy is calculated using the Hamiltonian operator $\hat H=-\frac{\hbar^2}{2m}\nabla^2+U(x)$. Naively you would expect you can just apply it to wavefunctions $\hat H\psi(x)$ and wherever this gives you a larger number the energy is high. But the only way to calculate the energy that makes sense is to calculate the expectation value of the Hamiltonian $\langle \hat H\rangle=\int \text{d}x\ \psi^*(x)\hat H\psi(x)$. For this expectation value you can't just look at the energy at a point you have to consider the entire space.

What you are trying to calculate is the energy at a specific point $x$. This only makes sense when you are talking about energy eigenstates. Or states $\psi(x)$ that satisfy $$-\frac{\hbar^2}{2m}\nabla^2\psi(x)+U(x)\psi(x)=E\psi(x)$$ This means you can talk about 'the energy' without confusion: the energy is given by $E$ and likewise $\langle\hat H\rangle$=E. If you take a look at this last equation you see that there are two regions: $E>U(x)$, the classically allowed region, and $E<U(x)$, the classicaly forbidden region. If you work it out you'll find that in the allowed region the wavefunction curves towards the zero axis like a sine-wave and in the forbidden region the wavefunction curves away from the zero axis like an exponential function.

These two regions just tell you in which way the wavefunction curves. They both have the same energy since they are part of the same energy eigenstate. The best you can do is say that the kinetic energy $\hat T\psi(x)=-\frac{\hbar^2}{2m}\nabla^2\psi(x)$ is negative in the forbidden region but again this is not really a useful quantity because ultimately you are interested in expectation values $\langle\hat T\rangle$

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Think about single eigenstate, energy just how the state change with time goes (well by factor of $i\hbar$) but it never say negative energy is not allowed . In fact negative energy mean growing and decay for single energy eigenstate and positive energy mean wave oscillation. Inside of the wall, it will grow or decay because energy is negative.

Energy always describe how things evolve (no matter it's QM or classical) but QM say it is evolve of wave function and it could be negative just like it could be positive. There is no bias in QM. Then you just sum up all the eigenstates and get the final function.

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The argument in the question is as follows:

  • There is a finite probability of finding the particle inside of the barrier
  • If the particle is inside the barrier its energy is negative
  • States with negative (kinetic) energy are impossible

The seeming paradox here is due to mixing classical and quantum concepts. Let us first note that there is no state inside the barrier. The state of a particle in this case a scattered wave, extending to infinity. The probability of measurement finding a particle inside the barrier is indeed non-zero. On the other hand in classical physics the state is identified with a specific location and momentum, i.e. the measured position of the particle is its state. Once we localized a quantum particle at a specific location, it is no more in the scattering state, and has undetermined momentum and energy.

Secondly, let us consider what the conservation of energy means in quantum mechanics.

  • We could view it as a fact that a particle in an energy eigenstate remains forever in this eigenstate. But this is not relevant here, since we perform a measurement, collapsing the wave function to a position eigenstate.
  • It can be viewed from the point of view of the uncertainty relation, as discussed in the answer by KDN.
  • Finally, it can be viewed from the point of view of the correspondence principle and the Ehrenfest theorem, where it will be conservation of the energy average in the quantum mechanical sense: $$\langle\frac{p^2}{2m}\rangle + \langle U(x)\rangle \neq \frac{\langle p\rangle^2}{2m} + U(\langle x\rangle),$$ where on the left we have the average energy in QM sense, whereas on the right is its classical estimate implied by the argument in the question. These are not the same.

To summarize: quantum mechanics appear paradoxial only as long as one relies on the intuition, grounded in classical phsyics.

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Apparently, extra energy can be extracted out of the universe for a very short period of time, corresponding to the uncertainty relation given by $$\Delta E\times \Delta t\ge\frac{h}{4\pi}$$ This is what quantum electrodynamics (the quantum theory explaining electrostatic force between charges) is based on. Consider this A electron A is present in space, another electron B is placed in the first electrons electric field. Now this electron A would release a photon(apparently borrowing the energy for it from the universe), the other electron B would absorb it, and send another photon back to electron A and hence experience the electron A's presence there (force) through the photons(the first electron would also experience electron B's presence similarly). This energy borrowed by electron A for its photon release must be returned within an interval $\Delta t$,(by the returning photon it absorbs from electron B). This is how electrons sense each other's presence. These photons are called virtual ones because we can't detect them. This might not be literally physically true, but the theory does involve violating the energy conservation law through the Heisenberg uncertainty principle.

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Here is what Erwin Schrödinger did:

    Imagine you put a cat in a box, and leave it in there for a day. Now imagine 
you are going to take the cat out, but before you open the box, you don't know if 
it is dead or alive. So in that moment, the cat exists in two states at once, both
dead *and* alive. Therefore the cat is now in two states, in two places, and in two
forms of reality. This is known as quantum tunneling, as reality diverges into two 
types: one where the cat is dead and one where the cat is alive. This can apply to
particles and quantum fields, and allow for a gap to open in spacetime.

Does this make sense now?

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