# Where does the energy come from?

1. Apr 22, 2013

### Jano L.

Consider the following experiment. We accelerate electrons in an electrostatic accelerator with potential drop U and shoot them towards a sensitive detector. Between the accelerator and detector we set up a repulsive potential barrier V with help of an electrical condenser. In the plates there are holes along the path, so there is possibility that electron passes the barrier of the condenser and arrives at the detector on the other side.

What will happen to the electrons and the detector if the potential barrier is higher than the accelerating potential?

According to quantum theory, there will be particles arriving at the detector even if V > U (tunnel effect).

1) Does not this contradict conservation of energy? From the description it is clear that the potential energy of the electron at the detector is higher than potential energy at the source, so it seems like the detector receives more energy than the source loses by accelerating the electrons.

2) If not, where does the extra energy come from?

2. Apr 22, 2013

### Staff: Mentor

Only if you detect them in a region where the potential is lower again, or have some mechanism to give them energy in the detection process.
This is similar to the orbital wave functions of electrons in atoms, you don't need accelerators.

3. Apr 22, 2013

### Jano L.

No, I really mean macroscopic setup with accelerator, tunable potential barrier and clear detection. With atom there is little possibility to detect where the electron is.

The potential does not need to be lower at the detector. I can place the latter in the region with higher potential. The $\psi$ function will come out non-zero in that region despite that it is at higher potential, so according to the probabilistic meaning of $|\psi|^2$, there should be some particles coming.

4. Apr 22, 2013

### Staff: Mentor

Sure, but then you'll need some way to provide energy to the electrons. Energy is conserved in quantum mechanics.

A realistic macroscopic setup will give you a detection probability so close to 0 that you won't see the effect. Okay, does not matter for a hypothetical experiment.
It is possible to detect where an electron is in an atom.

5. Apr 22, 2013

### Jano L.

But the scenario described above suggests the opposite conclusion. So how is the energy conserved when particle penetrates to the region with too high potential energy?

More generally, is there some general justification for the conservation of energy in quantum theory?

6. Apr 22, 2013

### Staff: Mentor

I answered this in my previous posts.
It is a direct result of quantum mechanics via the Hamilton formalism.

7. Apr 22, 2013

### Sonderval

How would you create an electron state with a definite energy that is sufficienty localised in the beginning anyway? Since energy eigenstates are time-independent, if it's an energy eigenstate, the electron cannot travel anywhere. If it is not an energy eigenstate, then you create you electron in a superpositon of initial states - and then you have to think about the entanglement with whatever you used to create the electron state in the first place.

8. Apr 23, 2013

### Jano L.

I am not sure what do you mean - I do not remember any energy conservation theorem based on Schroedinger's equation. Also, quantum theory is not only Hamilton's formalism. There is also the part with measurement. When we detect particle in the high potential region, we measure its position and thus its potential energy. This detection process is not very well understood, and the original Hamiltonian cannot be applied. Perhaps some other can, but I do not know. Can you explain your position a little bit ?

The state is not localized, it is rather a wave packet close to a plane wave of definite momentum, travelling towards the detector. I do not think it is practicable to include the accelerator into the wave function description, as it contains too many particles.

9. Apr 23, 2013

### Sonderval

But if your initial state is a wave packet, it already does not have a definite energy - there is an amplitude for any possible energy value. So it may be conceptually simpler to forget about your measurement apparatus and just imagine that you measure the energy of the wave packet, where any energy value is now possible. But since an isolated wave packet is not in an energy eigenstate, the notion of energy conservation does not really apply to it anyway - for that you need to include whatever machine you used to create the wave packet and consider that the energy of the machine is entangled to the energy of the wave packet.

10. Apr 23, 2013

### Jano L.

Assume that the wave function is such that variance in kinetic energy was lowered down to negligible value.

Also imagine that its energy spectrum is cut-off below V, so there is no component with energy enough to get to the region with potential (the Fourier expansion of the wave has no components above some max frequency).

Are you suggesting I should describe the accelerator with wave function too?

11. Apr 23, 2013

### Sonderval

Even with the cut-off, there is still the problem of making the measurement: If you actuzally measure the electron in a classically forbidden region (for example by doing a position measurement), your system is not in an energy eigenstate.

You can use the simpler situation of an electron in a finite potential well in the ground state. If you make a position measurement, there is some prob. to find the electron in the wall of the well - but this localised state is not an energy eigenstate. To perform the measurement, you have to use some apparatus, though.

I just found the old thread where we discussed exactly this kind of problem in some detail:

12. Apr 23, 2013

### derek101

I don't understand,why if 10% of electrons were to pass through a barrier,must I assume that the electron has a 10% chance of tunnelling it's way through.
Why can't I assume that the barrier only has a 90% chance of stopping the electrons?

13. Apr 23, 2013

### Jano L.

I do not understand you. We measure position, we infer the potential energy. The shape of the wave function after the measurement will be some sharp peak with great variance in energy, but the kinetic energy, however we are uncertain about it, cannot be negative. In a result, we just have violation of energy conservation.

14. Apr 23, 2013

### Staff: Mentor

Well, use other descriptions of the same thing?
But Hamilton is sufficient.
This detection is an interaction as well, and can be described with QM.
Depends on the interpretation.
With the proper Hamiltonian, it can.
It is not necessary, but it would be possible.

The key point is: You want to perform a position measurement. The measurement results are not energy eigenstates - you have to modify the energy of the particle in the measurement process.

15. Apr 23, 2013

### Jano L.

Do you mean that the additional energy comes from the measurement device? But what if I do not put the detector there? The particles are still coming - think of tunnel effect - there is no detector in the potential wall.

16. Apr 23, 2013

### Staff: Mentor

If you don't detect a particle (if you don't interact with it), there is no problem with a wavefunction with an energy below the potential.

17. Apr 23, 2013

### Jano L.

Of course, there is no problem with the wave function being non-zero in the forbidden region. That is a common result of Schroedinger's equation. The problem I am trying to understand is where is the additional energy of the _particle_ coming from, when it is in the region where V > E, E being the energy associated with wave function.

Your arguments seem to suggest that the detector is the necessary prerequisite to get the particles going into the region with V > E.

But there is no such requirement for the detector in the calculation of the wave function - the detector is always neglected. From the basic interpretation of $|\psi|^2$ as probability density, the particles should access these forbidden regions spontaneously, without the help of the detector (think of the tunnel effect...)

18. Apr 23, 2013

### Staff: Mentor

It does not need to have additional energy to have a wavefunction there. The interpretation of that wavefunction is up to you. If you think it is real and its squared modulus is a probability density, it is in a "forbidden" region with non-zero probability. Note that "forbidden" comes from a classical view. In QM, this is not forbidden any more.

19. Apr 23, 2013

### Jano L.

Yes, but that implies already that the energy of the particle is higher than E with some probability. I do not see how to square this with energy conservation. The only reasonable explanation I see is that there is no energy conservation in this theory - the probabilistic meaning of the wave function seems to make it impossible.

You obviously think otherwise, but from you posts I was unable to understand your reasons. Is your reason that average value of the Hamiltonian is time-independent for closed system? But that has only statistical meaning; the average value of

$$(H - \langle H\rangle)^2$$

is non-zero as well, which means that there are fluctuations in the actual energy going on.

20. Apr 24, 2013

### Staff: Mentor

No, it just implies that the classical description is wrong. The particle has an energy lower than the potential, and there is nothing wrong with that.

Energy is conserved, both in probabilistic and non-probabilistic interpretations.

If you start with a well-defined energy of the system (include the detector, if you like), this energy will be conserved exactly. If you start with a superposition of different energies, do not include the environment, and introduce "collapses", only the expectation value is conserved I think.