# Negative kinetic energy in tunneling

1. Oct 31, 2006

### gulsen

Say I have a particle with a kinetic energy of 5 (in some units). And I have a potential barrier of 10 between 0<x<1 (again, in ome unit system), and 0 elsewhere. According to quantum theory, the partcile may be found between 0 and 1. And in this region, if the energy is conserved (5 = T + 10), shouldn't the kinetic energy be -5?!? So that $$\hat T \psi_2 = -5 \psi_2$$.

What's going on here??

2. Oct 31, 2006

### ZapperZ

Staff Emeritus
How do you know that your $\psi$ inside the barrier is an eigenfunction of your T operator in the first place? Go ahead and solve for the wavefunction inside the barrier, and see if it is an eigenfuntion of your operator.

Furthermore, why should KE be conserved? Shouldn't you be more concerned with H?

Zz.

3. Oct 31, 2006

### gulsen

I didn't say kinetic energy should be conserved. Note that I'm conserving energy and not kinetic energy, by saying: 5 = T + 10 (initial energy = energy inside bump)
$$\psi_2$$ should be eigenfunction of T because $$H = T + V_0$$ in this case. Also, the solution for 2nd region is
$$(T + V_0)\psi_2 = E\psi_2$$
$$T\psi_2 = -5\psi_2$$

4. Nov 5, 2006

### wavemaster

There's indeed such a problem. In regions where energy of particle is smaller than the minumum of particle, the solutions are exponential rather than oscialating, and kinetic energy operator becomes negative. For instance, (in nuclear physics) in a spherical potential well, the solution in the sphere (r<R) is $$F \sinh(qr)$$ where $$q^2 > 0$$.

$$<T> = <\psi | T | \psi> = <F \sinh(qr) | -\frac{\hbar^2}{2m} \frac{\partial ^2}{\partial r^2} | F \sinh(qr)>$$

$$<T> = -\frac{\hbar^2}{2m} q^2 <F \sinh(qr) | F \sinh(qr)>$$
Which is a negative number since $$<F \sinh(qr) | F \sinh(qr)>$$ is the probability of finding the particle within the sphere. I don't know whether such a problem arises in the field theory, so I expect some educated ones shed light on the topic.

Last edited: Nov 5, 2006
5. Nov 5, 2006

### lalbatros

This all goes together:

- evanescent wave
- imaginary wavevector
- negative "kinetic energy"

why would that be a problem ?

The important thing is not the "kinetic energy" would be positive.
The important thing is the Noether theorem: a symmetry implies a conserved quantity, time invariance implies energy conservation.

Dealing with (quantum) waves brings the possibility of evanescent wave and tunelling and this translates in negative kinetic energy "during tunneling".

6. Nov 19, 2006

### wavemaster

Because eigenvalues of an observable, namely momentum, becomes imaginary then.