# Energy conservation in the classically forbidden region

1. Jun 12, 2012

### QuasiParticle

Consider a particle approaching a finite potential step or inside a potential well. As we know, the particle has a finite probability to penetrate into the region where E < V. The probability remains finite arbitrarily deep into the classically forbidden region. Suppose we set a detector far from the step and wait for a very long time. When we finally detect a particle, what saves the law of conservation of energy?

Unless we have an absolutely crazy potential, the Lagrangian is time-translational invariant, so energy should be conserved. What am I not getting?

The issue was discussed, e.g., here:

physics.stackexchange.com/questions/11188/can-a-particle-be-physically-observed-inside-a-quantum-barrier

I do not think the explanation about the penetration depth is sufficient, since we can set the detector arbitrarily far from the step.

Last edited: Jun 13, 2012
2. Jun 14, 2012

### Bill_K

As you point out, E is conserved. In the classically forbidden region V > E, implying the kinetic energy p2/2m is negative, p is imaginary and the particle does not propagate, it's exponentially damped.

In order to be detected, the particle has to cause a transition in the detector. The detector also conserves energy, and so if the particle lacks sufficient energy to cause the transition, it will not be detected.

3. Jun 14, 2012

### Demystifier

First of all, there is no region where E<V, just as there is no region where E>V. That's because E is NOT a function of position x. Instead, E is a functional of the whole wave function at ALL points at once.

Second, if you decide to detect the particle inside the classically forbidden region, nobody can stop you to try to do that, and there is a finite probability that you will be successfull. But in the case of successfull detection, the act of detection will effectively collapse the wave function into a new wave function localized within the classically forbidden region only. The collapse will be such that energy after the collapse will be equal to energy before the collapse.

4. Jun 14, 2012

### Staff: Mentor

E is constant, but V is a function which depends on the position (and maybe some other variables). Just like there are regions where 2<f(x) with f(x)=x^2, there can be regions with E<V(x) and E>V(x).

5. Jun 14, 2012

### Demystifier

Yes, you are right. What I really wanted to say is that, given E and psi(x), in standard QM E cannot be interpreted as energy at position x. That's because, in standard QM, a particle with energy E and wave function psi(x) does not have a well-defined position x.

Attributing energy to a position x makes sense in the non-standard Bohmian interpretation, but then (instead of introducing negative kinetic energy) one should think of energy as a sum of positive kinetic energy p^2/2m, classical potential energy V(x), and quantum potential energy Q(x).

6. Jun 14, 2012

### Bill_K

Very good, Demystifier, in your effort to avoid non-standard interpretations you have invented a "quantum potential" Q(x) which does not appear in textbooks. You have also decided that the position variable x should not be called position. Well call it anything you like. E is constant and does not depend on it anyway. I am mystified.

7. Jun 14, 2012

### Ken G

The question seems related to one that I have had-- if I prepare a particle in a spin state "up", then I send it through a Stern-Gerlach instrument at 90 degrees to that spin, I have a 50% chance of deflecting the particle left or right, associated with a state of left or right spin instead of "up." Let's say it deflects so as to have "left" spin. Does that not imply that the apparatus itself has picked up the angular momentum change? Or is it so impossible to measure the angular momentum change in the apparatus that we need an interpretation to tell us what the discrepancy is in the first place? (Like, in the Bohm approach, could we have started with a particle in a superposition of left and right spin that is actually one or the other but we had no way to determine that from the details of its preparation, so if which way it goes is predetermined, then there might not have been any angular momentum discrepancy in the first place?)

8. Jun 14, 2012

### TrickyDicky

He was actually addressing the non standard Bohmian QM when referring to that Q(x).

9. Jun 14, 2012

### Demystifier

Thanks, but it was invented by David Bohm, who was much more clever and famous than me.

Remarkably, it appears in one of the most famous and most standard textbooks of physics - Feynman lectures on physics. See Eq. (21.38) in vol. 3 of that textbook.

10. Jun 14, 2012

### Demystifier

That's good because, according to Bohr, those who are not mystified by quantum mechanics do not yet understand it.