Where does the energy come from?

[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?

 

[mfb]

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

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.

 

[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.

 

[mfb]

I can place the latter in the region with higher potential.

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

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.

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.

 

[Jano L.]

Energy is conserved in quantum mechanics.

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?

 

[mfb]

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?

I answered this in my previous posts.

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

It is a direct result of quantum mechanics via the Hamilton formalism.

 

[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.

 

[Jano L.]

It is a direct result of quantum mechanics via the Hamilton formalism.

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 ?

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.

The state is not localized, it is rather a wave packet close to a plane wave of definite momentum, traveling 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.

 

[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.

 

[Jano L.]

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.

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).

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.

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

 

[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:
https://www.physicsforums.com/showthread.php?t=658128

 

[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?

 

[Jano L.]

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 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.

 

[mfb]

I am not sure what do you mean - I do not remember any energy conservation theorem based on Schroedinger's equation.

Well, use other descriptions of the same thing?

Also, quantum theory is not only Hamilton's formalism.

But Hamilton is sufficient.

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 is an interaction as well, and can be described with QM.

This detection process is not very well understood,

Depends on the interpretation.

and the original Hamiltonian cannot be applied.

With the proper Hamiltonian, it can.

I do not think it is practicable to include the accelerator into the wave function description, as it contains too many particles.

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.

 

[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.

 

[mfb]

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.

 

[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...)

 

[mfb]

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.

 

[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.

 

[mfb]

Yes, but that implies already that the energy of the particle is higher than E with some probability.

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.

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.

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

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.

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.

 

[Jano L.]

OK, thanks to all for their answers.

 

[Khashishi]

You can think of the particles as having negative kinetic energy in the evanescent region. The total kinetic+potential energy is still conserved.

 

[Jano L.]

I do not see how that could be. Kinetic energy is given by $\frac{1}{2}mv^2$, so it is always positive or zero.

 

[mfb]

Classic kinetic energy is given by 1/2mv^2. The effect we discuss does not even exist in classical mechanics, you cannot use those formulas here.

In (nonrelativistic) quantum mechanics, kinetic energy is $\frac{p^2}{2m}$ and the momentum p can be a complex number.

 

[Jano L.]

Classic kinetic energy is given by 1/2mv^2. The effect we discuss does not even exist in classical mechanics, you cannot use those formulas here.

I do not see a problem, unless there is some contradiction or a reason to avoid using pre-quantum notions. If we talk about "kinetic energy", we have to agree onsome meaning of the term. The standard meaning I assumed is given by the classical formula I wrote above.

In (nonrelativistic) quantum mechanics, kinetic energy is p22m and the momentum p can be a complex number.

Can you give an example? As far as I know, expected momentum for configuration q is given by the formula

$$
\mathrm{Re}~ \frac{\psi^*(q) \hat{\mathbf p} - \frac{q}{c}\mathbf A \psi(q)}{\psi^* \psi},
$$

which can be inferred from the equation of continuity for the probability of q.

which is always real. Do you assume some other definition of momentum?

EDIT: I've corrected the above formula.

 

[Khashishi]

I meant the statement somewhat informally, but it does make sense. The kinetic energy in nonrelativistic quantum mechanics is p^2/2m, where p is the momentum operator. If you compute the value of p^2/2m for a given wavefunction, you will see that the value is negative for the region inside a potential barrier.

 

[Nugatory]

From the basic interpretation of $|\psi|^2$ as probability density, the particles should access these forbidden regions spontaneously, without the help of the detector

You're interpreting $|\psi|^2$ as a probability density, which is fine. But probability of what?

It is not the density function for the probability of the particle being at a particular location; instead it is the density function for the probability that a detector at that location will detect the particle.

 

[Jano L.]

That is interesting! I assume you mean the fact that in the time-independent Schr. equation

$$
\left( \hat T + V(\mathbf r) \right) \phi = \epsilon \psi
$$

the expression

$$
\hat T \phi(\mathbf r)
$$

gives product

$$
T\phi(\mathbf r)
$$

with T < 0 whenever $V(\mathbf r) >\epsilon$. Is that right?

That is quite an interesting feature of the Schr. equation. I agree that it may be possible to define such concept as negative energy. However, any average value of the operator $\hat T$

$$
\int \psi^* \hat T \psi d\tau
$$

is always positive for normalizable functions; it is not possible to have potential in which this would be negative. So I think that the above property of the Schr. equation should not be taken as a sufficient reason for the adoption of the concept of "negative energy of particle"; it should be possible to get it more naturally, in measurements or as average value, but this does not seem to be possible.

I think that if one wants to understand this and other mysteries in physics, one should distinguish two things: the system and its description. The latter gives interesting info about the former (for example, average values), but not every feature has to have direct counterpart in the real system. I think this negative kinetic energy is such feature of the formalism that does not have counterpart in the system itself, but I could be wrong. Perhaps there are some other arguments for/against such a concept.

 

[Jano L.]

You're interpreting |ψ|2 as a probability density, which is fine. But probability of what?

It is not the density function for the probability of the particle being at a particular location; instead it is the density function for the probability that a detector at that location will detect the particle.

I think that what you say is quite common viewpoint. However, I can not support it, because that would require that one can put detector inside the hydrogen atom or molecules. This never happened, yet the calculations based on the idea of probability/charge density seem to work well. Also such interpretation both requires presence of detector and assumes that the resulting probability density does not depend on the nature of this detector in any way; that is very strange. If the detector was crucial for the actual frequency of arrivals, then it should enter the calculation of the frequency in a non-trivial manner.

The picture of fluctuating electron position can be supported indirectly by the example of wave-mechanical calculation of the van der Waals forces between molecules. There one calculates dipole-dipole forces between two molecules from the idea that the probability distribution $|\psi|^2$ gives the actual frequency of position of electron, and hence of its instantaneous dipole moment. The average dipole moments are zero, but the fluctuations are present, influence each other and give the vdW forces.

This makes me believe that the probability density at $\mathbf r$ is an estimate of the actual frequency the electron spends at $\mathbf r$. The quantity $|\psi|^2$ is gauge invariant, so it seems to have more reality than $\psi$ or expansion coefficients, which depend on the chosen basis.

 

[Nugatory]

I think that what you say is quite common viewpoint. However, I can not support it, because that would require that one can put detector inside the hydrogen atom or molecules... The picture of fluctuating electron position can be supported indirectly by the example of wave-mechanical calculation of the van der Waals forces between molecules.

I'm not suggesting that the detector has to be at the point at which electron is detected; indirect means of localizing the electron, such as measuring van der Waals forces, seems to me a quite satisfactory detector. So although the picture of fluctuating electron position is supported, the detector is still present (and I don't see how you could eliminate it and still obtain any sort of measurement to support that or any other picture) and the magnitude of ψ² can be understood as the probability that the electron will be detected at a location, not that is actually there.

Interpretations are a matter of personal taste, so if you cannot support mine we can agree to disagree on that matter while still agreeing on the mathematical formalism and the results it predicts. Nonetheless, I cannot resist pointing out that energy is conserved across the source, the detector, and the electron even when the electron is detected in the classically forbidden region. Thus we wouldn't even have to ask the question that started this thread ("where does the energy come from?") if we interpret the wave function as giving the probability that the electron will be detected in the classically forbidden zone, rather than as the probability that the electron is violating conservation of energy by being there.

 

[Jano L.]

Thus we wouldn't even have to ask the question that started this thread ("where does the energy come from?") if we interpret the wave function as giving the probability that the electron will be detected in the classically forbidden zone, rather than as the probability that the electron is violating conservation of energy by being there.

Why do you think so - do you think that when the electron is detected, the description of such process including the detector will be such that the total energy is conserved?

 

[Khashishi]

If you include the detector in the description then the total energy is conserved.

For example, if you have an electron gun at 0V firing 10keV electrons at a -11kV wall, then they're supposed to be reflected. Now if you put a probe in the forbidden region near the wall, you should be able to get some current in the probe. The probe is at some voltage > -10kV, so the electrons are allowed to propagate in the probe. You can measure the energy of the electrons in the probe by looking at the IV characteristics.

You can't directly measure a negative energy electron, but once the electron appears on the other side of the barrier (and has positive k. energy again), you can check that the energy was indeed conserved.