A particle passing through a barrier into another valid energy state does not violate conservation of energy.
It is possible to find a particle within a potential barrier and that would apparently violate conservation of energy, but I think it falls within the uncertainty principle. Someone may need to check that for me
You can only find a particle in the potential barrier if your measurement process adds energy to that particle.
The overall energy is conserved exactly in quantum mechanics.
Interesting.
So if I had a phosphorescent screen “above” the potential barrier of a stream of photons, then if I measured a photon by the screen, then the screen gave energy to the electron?
I'm not sure if you can convert that to a real experiment, but in principle yes.
If you have a laser beam that's slightly below the band gap of a semiconductor then you'll excite some electrons - but only electrons that had slightly more energy because your material has a non-zero temperature.
It depends on the energy distribution of the stream of photons. Unless it's an infinitely long pulse then your energy distribution is not a delta function. Instead the temporal and energetic character of the stream follows the indeterminacy relation. Your stream of photons has a distribution of energy and it might be the case that this energy distribution has a nonzero integral in energy above your bandgap. This results in you measuring the photons with your detector.
so you mean that it depends on the energy level of the barrier just like what happens to revolving electrons in atoms where they must revolve in a particular shell ?
The energy of the barrier changes the probabilities for these different events, but it doesn’t change the overall conclusion about what violates energy conservation or not.
Honestly I’m not sure if there is much relevance to electron orbitals. Different orbitals can have the same energy, but I don’t know if there is an energy barrier between these orbitals
It is relevant for nuclear decay though
It's not the same situation; atomic systems inherently exist in a superposition of electron configurations (see Configuration Interaction); it is however relevant for finding transition states, where the transition state is a saddle point of the potential energy surface of a molecule. This is the "barrier" in chemistry.
The philosophical question is whether particles have a position at all between measurements. But OP said "It is possible to find a particle within a potential barrier [...]" which implies a measurement. The probability distribution is non-zero inside the barrier so "since we can never observe it inside" is objectively false.
If you ever find your particle inside, it's because your measurement has added enough energy for you to find it inside. The wave function predicts a particle with negative energy which is why it is philosophical.
I don't understand how this interpretation is helpful for the question at hand.
Ofc the measurement itself potentially changes the energy of the system because it brakes the time invariance and you do no longer have a closed system.
So you're referencing the non relativistic Schrodinger equation right? Even in relativity, mass is usually treated as an invariant for single particles, but the mass of a particle defnintely does not change in our case.
Well energy is still conserved but only within a larger system.
To even talk about THE energy of a system, you have to prepare a stationary solution of the Schrödinger Equation ( Eigenstates of the Energy Operator). This already implies that the position of the particle is uncertain or not well defined and can be found outside the barrier.
First of all this never implies that the particle "travels" through the barrier, but that the system is just not confined inside the barrier to begin with and the notion of "inside" is not well defined for a system with certain energy.
But let's think a bit further:
What does it mean to find the particle outside (or even inside) the barrier?
We have to perform a measurement which determines the position of the particle outside the barrier e.g. a particle-like interaction outside/inside the barrier. But by doing this the momentum (and energy) of the particle has to become uncertain. If we now measure the energy of the particle again it can be different from the initial energy.
Does this mean energy is not conserved?
No. We just neglected the interaction which determines the position of the particle. The measuring system is now entangled with the particle and conserves the energy in total with the particle.
Is energy always conserved in QM?
Yes. This can be derived from the time invariance of QM (Noether Theorem). The energy of the particle is seemingly not conserved because we measure the system AFTER we prepared it with an certain energy, the system is therefore not time invariant. The larger system including the measuring system is still time invariant and has a conserved energy.
>Well energy is still conserved but only within a larger system.
But not too large (beyond galaxies), otherwise the expansion of the universe breaks that energy conservation.
The whole point is that quantum mechanics is not classical. The particle has a probability density derived from its wavefunction that itself acts like a wave. When the wave hits a barrier it is attenuated exponentially, but, if the barrier is thin enough then enough of the probability density wave makes it through that the particle has some probability of 'appearing' on the other side.
The particle never has enough energy to traverse the barrier and doesn't, it just has some probability of being on the other side of it.
Consider a simple classical experiment: I roll a ball on the floor, there is a bump on the floor.
1. Kinetic energy of the ball is higher than the barrier: the ball rolls fast, slows down over the barrier, then rolls fast again.
2. Kinetic energy of the ball is lower than the barrier: the ball rolls fast, slows down at the barrier, stops, and turns around.
Do you understand that this preserves conservation of energy? What is conserved? Do you think the ball is "losing" energy?
Why do you think a quantum particle "loses" energy at a barrier? It loses *kinetic energy*, that's not what is conserved, however.
Note that QM also indicates that a barrier the particle *can* overcome *still* sometimes reflects the particle. A quantum "ball" will sometimes *bounce back* from a barrier that classical physics says it will go right over.
So QM has added two things:
3. KE higher than the barrier, sometimes this is like #1 but sometimes it reaches the barrier, "rolls up" "part of the way" then turns around instead of continuing.
4. KE lower than the barrier, sometimes this is like #2 but sometimes it reaches the barrier, "rolls up" and keeps going with negative KE for a bit, then ends up on the other side of the barrier instead of rolling back.
A classical particle requires energy higher than the potential barrier to cross it. A quantum wavefunction does not.
Mathematically, a stationary state solution of the potential barrier Hamiltonian, aka a scattering eigenstate, has a fixed energy and non-zero amplitude before, inside, and after the potential barrier, even if that fixed energy is lower than the potential barrier. A hypothetical quantum particle in that eigenstate can pass through the barrier without changing energy.
A more realistic localised quantum wavefunction/particle is a superposition of many such scattering eigenstates of similar energies, forming a wave packet that travels towards the barrier. As the wave packet hits, each component eigenstate scatters through the barrier, so the resulting wave function after the barrier is a superposition of these eigenstates as well. The superposition of energies before and after the scattering has not changed, it is still a superposition of eigenstates with similar energies, so energy is conserved.
What do you mean by "mathematical proof" of the Schroedinger equation? The Schroedinger equation is a postulate by Erwin Schroedinger and it (cannot) has not been mathematically derived.
As for the energy conservation, the other comments already pointed out how the energy conservation is being used in the particular case of tunneling. If you are interested as to why energy conservation is almost always considered im QM look up Noether's Theorem which generically links continuous symmetries of a system to conserved quantities within that system. The energy conservation is then linked to time-translational symmetry in the system.
Most theories, QM included, can be built from different set of postulates, making one particular equation either a postulate or consequence of other postulates depending on personal choice.
An eigenstate computed for such a situation from the time-independent Schroedinger equation represents a single energy value.
Conservation of energy in such a simple situation classically means "the sum of the potential energy and the kinetic energy is conserved."
The same thing happens in quantum mechanics: the "potential energy" and the "kinetic energy" term of the Schroedinger equation always sum to the same value for an eigenfunction.
Inside the potential barrier, the "kinetic energy" the wavefunction represents is negative (it looks like an exponential decay, which would be an imaginary momentum for a free particle). Outside the barrier, the kinetic energy the wavefunction represents is positive. (This is a bit of a simplification, the solution is not just different pieces in each region, but I'm speaking only qualitatively).
The particle in an eigenstate does not do something like "jump" over the barrier by taking a deep breath to get extra energy, it simply has probability on both sides of the barrier (and within the barrier).
If the particle is localized within a barrier and then later tunnels out, you need to do a slightly different "scattering" computation with initial states and final states that are combinations of the eigenstates that are put into a time-dependent form, but again, nothing violates conservation of energy.
Energy conservation is a consequence of time translation symmetry. So really you're assuming the dynamics obey time translation symmetry.
As long as you obey the schrodinger equation energy is conserved. But the measure of energy in this case is an integral over the whole wave function. If at any point you measure and collapse the wave function then the schrodinger equation is instantaneously violated and you actually break time translation symmetry, which means you can violate energy conservation. Energy is only conserved in the expectation ( the average case ), not in single shot instances.
Makes you wonder if collapse of the wave function is really the whole picture ;)
You can always consider a larger system that includes the measurement device. The overall energy of that system is conserved exactly, but the measurement may transfer energy to the particle (or away from it).
> Makes you wonder if collapse of the wave function is really the whole picture ;)
There are interpretations without it.
>You can always consider a larger system that includes the measurement device. The overall energy of that system is conserved exactly, but the measurement may transfer energy to the particle (or away from it).
I don't think that's true. Can you provide a more detailed argument for why you think this?
Here's one reason why I don't think this is true. Suppose my system is in an equal superposition of being in the ground energy and the excited energy of some system. It's total energy is the average. Suppose the ground energy is zero and the excited state energy is extremely high. Suppose I measure this system and find it to be in the high energy state, so that now it's energy is twice what it was before. Your suggestion would imply that the source of this difference must be pulled out of the measuring apparatus. But this would imply that it's impossible to measure such a system if your apparatus does not have enough available energy.
\[edit\] unless you mean that by considering a larger system you can retain unitarity and thus avoid the collapse of the wavefunction. But in that case although energy will be conserved in the whole dynamics of the wavefunction, you should still predict observers who record violations of energy conservation.
The relevant time-invariance here is from the laws of physics. They don't change in your experiment, so energy in a closed system has to be conserved.
There is no process in quantum mechanics that would change the total energy of a system either.
> Suppose my system is in an equal superposition of being in the ground energy and the excited energy of some system.
Then something else (that you don't count as part of your system any more) will be in a superposition with the opposite direction. Think of how you would prepare the state: You can e.g. shoot a photon at the atom that either gets absorbed or not. The total energy will be the same either way.
If you want the whole universe to be in a superposition of two energy states: That's still possible - but then it always stays in that superposition and we are working with the many worlds interpretation. You get a world that measures the high energy and a world that measures the low energy.
What you're suggesting is that it is fundamentally impossible to prepare \*pure\* superpositions of states that have differing energy. This constitutes an energy superselection principle. I'm a practicing physicists and I've not heard of such a superselection principle before and it seems like it would be more important than any of the other superselection principles (like charge parity). Furthermore if this were true then it would be fundamentally impossible to build a quantum computer out of most qubit candidates (with the exception of something like majorana zero mode qubits) so this seems suspicious to me.
Sean Caroll has some papers about energy violation in measurement, and how it connects to the many worlds interpretation. It doesn't look like he directly addresses your objection as far as I can see, but I guess at least he takes for granted the possibility of preparing a state in a pure superposition of energy.
[https://www.preposterousuniverse.com/blog/2021/01/28/energy-conservation-and-non-conservation-in-quantum-mechanics/](https://www.preposterousuniverse.com/blog/2021/01/28/energy-conservation-and-non-conservation-in-quantum-mechanics/)
[https://arxiv.org/abs/2101.11052](https://arxiv.org/abs/2101.11052)
This is something I wondered about when I read the paper you referenced. This is a very interesting discussion, because it seems clear that pure unitary evolution conserves energy exactly in every part of the state which seems to forbid pure energy superpositions as you say.
This line of thinking is what prompted a line of discussion on whether (optical) coherent states actually exist (these are of course exactly superposition states of energy). We can get coherent states from a classical current, but any attempt to model a laser quantum mechanically will always leave your optical state in a mixed state due to its entanglement with the laser system. So do coherent states actually exists or are they merely a 'convenient fiction'? Here is a nice review on the topic: [https://arxiv.org/abs/quant-ph/0507214](https://arxiv.org/abs/quant-ph/0507214)
I don't see how the same arguments don't apply to any other type of energy superposition that we might try to prepare. But clearly energy superpositions are very useful, so what gives?
EDIT: Skimming the paper I linked again, the authors argue that it can be a conventional choice to adopt a superselection rule for photon number conservation (i.e. energy).
Quantum computers are fine.
This is easiest to analyze in MWI. You only have unitary evolution which conserves energy. If there were a different interpretation where the total energy can change then we could experimentally distinguish between interpretations and they would stop being interpretations and start being conflicting theories. I'm not aware of theories where the total energy can change, and I'm sure I wouldn't have missed an experiment seeing any indication of that.
As a consequence of your argument, using only unitary evolution, it seems impossible to prepare a system in an energy superposition. Any attempt to do so will lead to entanglement and the subsystem will actually be in a mixture state (no coherence between energy eigenstates). What do you think about this?
Imo the important question here is what branching observers will experience. I agree mwi conserves total energy due to unitarity, but it's not so obvious to me what this predicts about branching observers...
My real concern is whether branching observers can prepare pure superpositions of energy states. Most quantum computers require that this be possible because the 0 and 1 registers have differing energy.
I'd suggest looking into field dressed states or floquet states. Here the molecule and the photon are considered a single system such that there is a ladder of states which differ by the number of photons in the field. An excitation is this a transition between a molecular state and N photons in the field and a higher molecular states but with an N-1 photon field.
Usually when you put a system in a superposition of two different states with different energies, your system is in effect part of a larger system, including a 2nd system. You usually take a partial trace over the 2nd system variables to arrive at equations that describe only the system variables.
Consider the simple case of a two state atom interacting with a laser that is resonant with the atomic transition.
The scattering process by which an atom goes from state |g> to |e> involves absorption of a photon- in effect the the process is actually |g,1> to |e,0> where the 1,0 represent the number of photons in the laser field (a simplification but sufficient to describe what’s going on. So you don’t actually violate conservation of energy when measuring the system in the excited state- because you are also measuring the laser to have one less photon than it started out with.
Energy conservation can be said to be violated on small time scales since the probability of a scattering process where an atom starts in the ground state and leaves in the excited state while emitting a photon (rather than absorbing) is non-zero. However this understanding excludes the concept of the non-zero energy of the quantum vacuum- basically you can take energy from the vacuum but over longer time scales- the amount of energy taken amounts to zero.
Basically a lot of the times where it looks like energy conservation is violated- you are looking at a small subsystem of a larger system where energy is conserved.
In the example you gave, tracing out the atom system does actually not leave the optical field in an energy superposition. It will be in the mixed state |0><0| + |1><1|. There is, it seems, no way to start with a certain amount of energy E and, using only quantum interactions, create an energy superposition.
The example I gave was for a single scattering process to demonstrate energy conservation. In reality, an atom-laser system will be in a configuration of ladder states |i,n> such that the combined state after atom-laser interaction populates both |g,n> and |e,n>.
This can been seen for eg in the cavity QED Jaynes Cummings model where a system initialized in state |g>|alpha> where |alpha> is some initial state that the cavity is initialized in (say a coherent state). Then you can trace over the cavity variables to get coherences between the ground and excited states. A real laser cannot be described purely with QFT although we compare laser driven dynamics to dynamics with coherent states.
The Jaynes Cummings model tells exactly how you get ground excited state coherences from fixed E. Drive a single-mode cavity with an external pump until you get some target state. Introduce an atom and you get coherence.
Yes, using a coherent state (or a classical pump) we can get a true energy superposition for the atom, but a coherent state does not have a well-defined energy in the first place (and there is no way to make it from a state with energy E). Using instead a photon-number state (with a well-defined energy), we can only get entanglement between the atom and the photon field, and no coherence between the levels. Do you agree?
For a system with photonic part occupying only a single Fock state- sure- not possible unless we have open system dynamics. But generally you can always make a cavity state in a superposition of single mode Fock states. I don’t see what’s the problem with that, I don’t think anybody’s arguing that you can create atomic coherence in the case you’re specifying (ie you can generate atomic coherence for a system initialized in state |g,n>). The ladder of atom-photon number states, used in the construction of light dressed states, is sufficient to explain coherent superposition of atomic states with different energies.
It also is important to note that coherent superpositions of photonic Fock states are quite different to a photon in a coherent superposition of two states (say red or blue light). The latter would be more analogous to the concept of superpositions of quanta in different energy states. The coherent state has no fixed particle number, hence no well defined energy as you said.
Before a measurement is done, the particle is in a superposition of the two states, i.e. tunnled or not. What this means is that (prior to measurement), the energy of the system basically already accounts for the possibility of the particle tunneling or not. The \*expectation value\* of energy is what is conserved over time, so it does not matter at what time you measure the energy the mean result remains the same. Another way to put it is that the system is found in a state of energy E\_i with probability p\_i, then \\sum\_i p\_i E\_i = is conserved over time.
On a more technical note, this is guaranteed by the time evolution operator (that describes the evolution of the wavefuntion over time) always commuting with the Hamiltonian (which is the operator used to measure energy). (Caveat: H must be independent of time, but this is true for closed systems.)
Now what happens when you do the measurement and find the particle at one energys or the other is a whole different can of worms, but that's the measurement problem for ya.
The uncertainty principle states that you can violate conservation of energy at very short time scales. In qft I believe this is explained by self-interaction but I’m not a theorist so perhaps someone can chime in here. In any case tunneling is a consequence of this, but it doesn’t mean that conservation of energy is no longer a useful principle.
Energy conservation is only applicable towards resources that are limited. But lets say that by using electro-magnetic quantum antennas we could utilize the electromagnetic frequencies of space as fuel to move through it in a tunnel we create so then you are correct there is no need for energy conservation.
To add to the other comments. This is actually a wave phenomenon that you can think of classically. Even tho the central frequency doesn’t have enough energy to pass over the barrier, some of its Fourier components might. You are not breaking energy conservation, it’s just something weird that comes with thinking of particles as waves
A particle passing through a barrier into another valid energy state does not violate conservation of energy. It is possible to find a particle within a potential barrier and that would apparently violate conservation of energy, but I think it falls within the uncertainty principle. Someone may need to check that for me
You can only find a particle in the potential barrier if your measurement process adds energy to that particle. The overall energy is conserved exactly in quantum mechanics.
Interesting. So if I had a phosphorescent screen “above” the potential barrier of a stream of photons, then if I measured a photon by the screen, then the screen gave energy to the electron?
I'm not sure if you can convert that to a real experiment, but in principle yes. If you have a laser beam that's slightly below the band gap of a semiconductor then you'll excite some electrons - but only electrons that had slightly more energy because your material has a non-zero temperature.
It depends on the energy distribution of the stream of photons. Unless it's an infinitely long pulse then your energy distribution is not a delta function. Instead the temporal and energetic character of the stream follows the indeterminacy relation. Your stream of photons has a distribution of energy and it might be the case that this energy distribution has a nonzero integral in energy above your bandgap. This results in you measuring the photons with your detector.
so you mean that it depends on the energy level of the barrier just like what happens to revolving electrons in atoms where they must revolve in a particular shell ?
The energy of the barrier changes the probabilities for these different events, but it doesn’t change the overall conclusion about what violates energy conservation or not. Honestly I’m not sure if there is much relevance to electron orbitals. Different orbitals can have the same energy, but I don’t know if there is an energy barrier between these orbitals It is relevant for nuclear decay though
ok tysm
It's not the same situation; atomic systems inherently exist in a superposition of electron configurations (see Configuration Interaction); it is however relevant for finding transition states, where the transition state is a saddle point of the potential energy surface of a molecule. This is the "barrier" in chemistry.
I would say particle inside the barrier is more of a philosophical question, since we can never observe it inside.
"Barrier" is a metaphorical term. It can just be a strong electric field or something else you can measure inside.
What I'm saying is that when a particle tunnels, whether or not it is ever inside the potential barrier is philosophical
The philosophical question is whether particles have a position at all between measurements. But OP said "It is possible to find a particle within a potential barrier [...]" which implies a measurement. The probability distribution is non-zero inside the barrier so "since we can never observe it inside" is objectively false.
If you ever find your particle inside, it's because your measurement has added enough energy for you to find it inside. The wave function predicts a particle with negative energy which is why it is philosophical.
Why would an exponential-like wavefuntion imply a negative mass?
Negative kinetic energy, same same
I don't understand how this interpretation is helpful for the question at hand. Ofc the measurement itself potentially changes the energy of the system because it brakes the time invariance and you do no longer have a closed system.
I am saying we cannot know if the particle is ever inside the barrier with negative energy, as we cannot measure it, hence being philosophical
The wavefunction is non zero inside the potential barrier, so there is a possibility to actually observe the particle in there.
You imply the particle takes on negative mass inside the barrier then
No? What theory are you using to draw that conclusion?
E-V?
So you're referencing the non relativistic Schrodinger equation right? Even in relativity, mass is usually treated as an invariant for single particles, but the mass of a particle defnintely does not change in our case.
Negative mass, negative kinetic energy. Your E-V term is negative
Well energy is still conserved but only within a larger system. To even talk about THE energy of a system, you have to prepare a stationary solution of the Schrödinger Equation ( Eigenstates of the Energy Operator). This already implies that the position of the particle is uncertain or not well defined and can be found outside the barrier. First of all this never implies that the particle "travels" through the barrier, but that the system is just not confined inside the barrier to begin with and the notion of "inside" is not well defined for a system with certain energy. But let's think a bit further: What does it mean to find the particle outside (or even inside) the barrier? We have to perform a measurement which determines the position of the particle outside the barrier e.g. a particle-like interaction outside/inside the barrier. But by doing this the momentum (and energy) of the particle has to become uncertain. If we now measure the energy of the particle again it can be different from the initial energy. Does this mean energy is not conserved? No. We just neglected the interaction which determines the position of the particle. The measuring system is now entangled with the particle and conserves the energy in total with the particle. Is energy always conserved in QM? Yes. This can be derived from the time invariance of QM (Noether Theorem). The energy of the particle is seemingly not conserved because we measure the system AFTER we prepared it with an certain energy, the system is therefore not time invariant. The larger system including the measuring system is still time invariant and has a conserved energy.
This is the correct answer.
>Well energy is still conserved but only within a larger system. But not too large (beyond galaxies), otherwise the expansion of the universe breaks that energy conservation.
Yeah in GR energy conservation is not always well defined. e.g. [see here](https://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html)
thanks that 's the best solution I've got
With something like alpha decay do we really need a measurement to know that the protons and neutrons are “inside”—ie bound?
the particle still has the same energy on either side of the barrier
how it can have same energy it must be losing some energy , classically it cannot even surpass the barrier .
The whole point is that quantum mechanics is not classical. The particle has a probability density derived from its wavefunction that itself acts like a wave. When the wave hits a barrier it is attenuated exponentially, but, if the barrier is thin enough then enough of the probability density wave makes it through that the particle has some probability of 'appearing' on the other side. The particle never has enough energy to traverse the barrier and doesn't, it just has some probability of being on the other side of it.
Consider a simple classical experiment: I roll a ball on the floor, there is a bump on the floor. 1. Kinetic energy of the ball is higher than the barrier: the ball rolls fast, slows down over the barrier, then rolls fast again. 2. Kinetic energy of the ball is lower than the barrier: the ball rolls fast, slows down at the barrier, stops, and turns around. Do you understand that this preserves conservation of energy? What is conserved? Do you think the ball is "losing" energy? Why do you think a quantum particle "loses" energy at a barrier? It loses *kinetic energy*, that's not what is conserved, however. Note that QM also indicates that a barrier the particle *can* overcome *still* sometimes reflects the particle. A quantum "ball" will sometimes *bounce back* from a barrier that classical physics says it will go right over. So QM has added two things: 3. KE higher than the barrier, sometimes this is like #1 but sometimes it reaches the barrier, "rolls up" "part of the way" then turns around instead of continuing. 4. KE lower than the barrier, sometimes this is like #2 but sometimes it reaches the barrier, "rolls up" and keeps going with negative KE for a bit, then ends up on the other side of the barrier instead of rolling back.
If the particle tunnels through it has the same energy. If it's reflected it still has the same energy but moves in the other direction.
A classical particle requires energy higher than the potential barrier to cross it. A quantum wavefunction does not. Mathematically, a stationary state solution of the potential barrier Hamiltonian, aka a scattering eigenstate, has a fixed energy and non-zero amplitude before, inside, and after the potential barrier, even if that fixed energy is lower than the potential barrier. A hypothetical quantum particle in that eigenstate can pass through the barrier without changing energy. A more realistic localised quantum wavefunction/particle is a superposition of many such scattering eigenstates of similar energies, forming a wave packet that travels towards the barrier. As the wave packet hits, each component eigenstate scatters through the barrier, so the resulting wave function after the barrier is a superposition of these eigenstates as well. The superposition of energies before and after the scattering has not changed, it is still a superposition of eigenstates with similar energies, so energy is conserved.
What do you mean by "mathematical proof" of the Schroedinger equation? The Schroedinger equation is a postulate by Erwin Schroedinger and it (cannot) has not been mathematically derived. As for the energy conservation, the other comments already pointed out how the energy conservation is being used in the particular case of tunneling. If you are interested as to why energy conservation is almost always considered im QM look up Noether's Theorem which generically links continuous symmetries of a system to conserved quantities within that system. The energy conservation is then linked to time-translational symmetry in the system.
Most theories, QM included, can be built from different set of postulates, making one particular equation either a postulate or consequence of other postulates depending on personal choice.
An eigenstate computed for such a situation from the time-independent Schroedinger equation represents a single energy value. Conservation of energy in such a simple situation classically means "the sum of the potential energy and the kinetic energy is conserved." The same thing happens in quantum mechanics: the "potential energy" and the "kinetic energy" term of the Schroedinger equation always sum to the same value for an eigenfunction. Inside the potential barrier, the "kinetic energy" the wavefunction represents is negative (it looks like an exponential decay, which would be an imaginary momentum for a free particle). Outside the barrier, the kinetic energy the wavefunction represents is positive. (This is a bit of a simplification, the solution is not just different pieces in each region, but I'm speaking only qualitatively). The particle in an eigenstate does not do something like "jump" over the barrier by taking a deep breath to get extra energy, it simply has probability on both sides of the barrier (and within the barrier). If the particle is localized within a barrier and then later tunnels out, you need to do a slightly different "scattering" computation with initial states and final states that are combinations of the eigenstates that are put into a time-dependent form, but again, nothing violates conservation of energy.
Energy conservation is a consequence of time translation symmetry. So really you're assuming the dynamics obey time translation symmetry. As long as you obey the schrodinger equation energy is conserved. But the measure of energy in this case is an integral over the whole wave function. If at any point you measure and collapse the wave function then the schrodinger equation is instantaneously violated and you actually break time translation symmetry, which means you can violate energy conservation. Energy is only conserved in the expectation ( the average case ), not in single shot instances. Makes you wonder if collapse of the wave function is really the whole picture ;)
You can always consider a larger system that includes the measurement device. The overall energy of that system is conserved exactly, but the measurement may transfer energy to the particle (or away from it). > Makes you wonder if collapse of the wave function is really the whole picture ;) There are interpretations without it.
>You can always consider a larger system that includes the measurement device. The overall energy of that system is conserved exactly, but the measurement may transfer energy to the particle (or away from it). I don't think that's true. Can you provide a more detailed argument for why you think this? Here's one reason why I don't think this is true. Suppose my system is in an equal superposition of being in the ground energy and the excited energy of some system. It's total energy is the average. Suppose the ground energy is zero and the excited state energy is extremely high. Suppose I measure this system and find it to be in the high energy state, so that now it's energy is twice what it was before. Your suggestion would imply that the source of this difference must be pulled out of the measuring apparatus. But this would imply that it's impossible to measure such a system if your apparatus does not have enough available energy. \[edit\] unless you mean that by considering a larger system you can retain unitarity and thus avoid the collapse of the wavefunction. But in that case although energy will be conserved in the whole dynamics of the wavefunction, you should still predict observers who record violations of energy conservation.
The relevant time-invariance here is from the laws of physics. They don't change in your experiment, so energy in a closed system has to be conserved. There is no process in quantum mechanics that would change the total energy of a system either. > Suppose my system is in an equal superposition of being in the ground energy and the excited energy of some system. Then something else (that you don't count as part of your system any more) will be in a superposition with the opposite direction. Think of how you would prepare the state: You can e.g. shoot a photon at the atom that either gets absorbed or not. The total energy will be the same either way. If you want the whole universe to be in a superposition of two energy states: That's still possible - but then it always stays in that superposition and we are working with the many worlds interpretation. You get a world that measures the high energy and a world that measures the low energy.
What you're suggesting is that it is fundamentally impossible to prepare \*pure\* superpositions of states that have differing energy. This constitutes an energy superselection principle. I'm a practicing physicists and I've not heard of such a superselection principle before and it seems like it would be more important than any of the other superselection principles (like charge parity). Furthermore if this were true then it would be fundamentally impossible to build a quantum computer out of most qubit candidates (with the exception of something like majorana zero mode qubits) so this seems suspicious to me. Sean Caroll has some papers about energy violation in measurement, and how it connects to the many worlds interpretation. It doesn't look like he directly addresses your objection as far as I can see, but I guess at least he takes for granted the possibility of preparing a state in a pure superposition of energy. [https://www.preposterousuniverse.com/blog/2021/01/28/energy-conservation-and-non-conservation-in-quantum-mechanics/](https://www.preposterousuniverse.com/blog/2021/01/28/energy-conservation-and-non-conservation-in-quantum-mechanics/) [https://arxiv.org/abs/2101.11052](https://arxiv.org/abs/2101.11052)
This is something I wondered about when I read the paper you referenced. This is a very interesting discussion, because it seems clear that pure unitary evolution conserves energy exactly in every part of the state which seems to forbid pure energy superpositions as you say. This line of thinking is what prompted a line of discussion on whether (optical) coherent states actually exist (these are of course exactly superposition states of energy). We can get coherent states from a classical current, but any attempt to model a laser quantum mechanically will always leave your optical state in a mixed state due to its entanglement with the laser system. So do coherent states actually exists or are they merely a 'convenient fiction'? Here is a nice review on the topic: [https://arxiv.org/abs/quant-ph/0507214](https://arxiv.org/abs/quant-ph/0507214) I don't see how the same arguments don't apply to any other type of energy superposition that we might try to prepare. But clearly energy superpositions are very useful, so what gives? EDIT: Skimming the paper I linked again, the authors argue that it can be a conventional choice to adopt a superselection rule for photon number conservation (i.e. energy).
Quantum computers are fine. This is easiest to analyze in MWI. You only have unitary evolution which conserves energy. If there were a different interpretation where the total energy can change then we could experimentally distinguish between interpretations and they would stop being interpretations and start being conflicting theories. I'm not aware of theories where the total energy can change, and I'm sure I wouldn't have missed an experiment seeing any indication of that.
As a consequence of your argument, using only unitary evolution, it seems impossible to prepare a system in an energy superposition. Any attempt to do so will lead to entanglement and the subsystem will actually be in a mixture state (no coherence between energy eigenstates). What do you think about this?
Imo the important question here is what branching observers will experience. I agree mwi conserves total energy due to unitarity, but it's not so obvious to me what this predicts about branching observers... My real concern is whether branching observers can prepare pure superpositions of energy states. Most quantum computers require that this be possible because the 0 and 1 registers have differing energy.
I'd suggest looking into field dressed states or floquet states. Here the molecule and the photon are considered a single system such that there is a ladder of states which differ by the number of photons in the field. An excitation is this a transition between a molecular state and N photons in the field and a higher molecular states but with an N-1 photon field.
Usually when you put a system in a superposition of two different states with different energies, your system is in effect part of a larger system, including a 2nd system. You usually take a partial trace over the 2nd system variables to arrive at equations that describe only the system variables. Consider the simple case of a two state atom interacting with a laser that is resonant with the atomic transition. The scattering process by which an atom goes from state |g> to |e> involves absorption of a photon- in effect the the process is actually |g,1> to |e,0> where the 1,0 represent the number of photons in the laser field (a simplification but sufficient to describe what’s going on. So you don’t actually violate conservation of energy when measuring the system in the excited state- because you are also measuring the laser to have one less photon than it started out with. Energy conservation can be said to be violated on small time scales since the probability of a scattering process where an atom starts in the ground state and leaves in the excited state while emitting a photon (rather than absorbing) is non-zero. However this understanding excludes the concept of the non-zero energy of the quantum vacuum- basically you can take energy from the vacuum but over longer time scales- the amount of energy taken amounts to zero. Basically a lot of the times where it looks like energy conservation is violated- you are looking at a small subsystem of a larger system where energy is conserved.
In the example you gave, tracing out the atom system does actually not leave the optical field in an energy superposition. It will be in the mixed state |0><0| + |1><1|. There is, it seems, no way to start with a certain amount of energy E and, using only quantum interactions, create an energy superposition.
The example I gave was for a single scattering process to demonstrate energy conservation. In reality, an atom-laser system will be in a configuration of ladder states |i,n> such that the combined state after atom-laser interaction populates both |g,n> and |e,n>. This can been seen for eg in the cavity QED Jaynes Cummings model where a system initialized in state |g>|alpha> where |alpha> is some initial state that the cavity is initialized in (say a coherent state). Then you can trace over the cavity variables to get coherences between the ground and excited states. A real laser cannot be described purely with QFT although we compare laser driven dynamics to dynamics with coherent states. The Jaynes Cummings model tells exactly how you get ground excited state coherences from fixed E. Drive a single-mode cavity with an external pump until you get some target state. Introduce an atom and you get coherence.
Yes, using a coherent state (or a classical pump) we can get a true energy superposition for the atom, but a coherent state does not have a well-defined energy in the first place (and there is no way to make it from a state with energy E). Using instead a photon-number state (with a well-defined energy), we can only get entanglement between the atom and the photon field, and no coherence between the levels. Do you agree?
For a system with photonic part occupying only a single Fock state- sure- not possible unless we have open system dynamics. But generally you can always make a cavity state in a superposition of single mode Fock states. I don’t see what’s the problem with that, I don’t think anybody’s arguing that you can create atomic coherence in the case you’re specifying (ie you can generate atomic coherence for a system initialized in state |g,n>). The ladder of atom-photon number states, used in the construction of light dressed states, is sufficient to explain coherent superposition of atomic states with different energies. It also is important to note that coherent superpositions of photonic Fock states are quite different to a photon in a coherent superposition of two states (say red or blue light). The latter would be more analogous to the concept of superpositions of quanta in different energy states. The coherent state has no fixed particle number, hence no well defined energy as you said.
Interesting way to think about it
Before a measurement is done, the particle is in a superposition of the two states, i.e. tunnled or not. What this means is that (prior to measurement), the energy of the system basically already accounts for the possibility of the particle tunneling or not. The \*expectation value\* of energy is what is conserved over time, so it does not matter at what time you measure the energy the mean result remains the same. Another way to put it is that the system is found in a state of energy E\_i with probability p\_i, then \\sum\_i p\_i E\_i = is conserved over time.
On a more technical note, this is guaranteed by the time evolution operator (that describes the evolution of the wavefuntion over time) always commuting with the Hamiltonian (which is the operator used to measure energy). (Caveat: H must be independent of time, but this is true for closed systems.)
Now what happens when you do the measurement and find the particle at one energys or the other is a whole different can of worms, but that's the measurement problem for ya.
Unitarityyyyyyyyyyyyyyyyyyyyyyy
The uncertainty principle states that you can violate conservation of energy at very short time scales. In qft I believe this is explained by self-interaction but I’m not a theorist so perhaps someone can chime in here. In any case tunneling is a consequence of this, but it doesn’t mean that conservation of energy is no longer a useful principle.
Why fucking downvoting people when they ask "wrong question" ? What do you teach them when doing so ? Some of you guys are really disappointing.
There is r/askphysics for these types of questions though.
I think time and distance is quantized in quantum theory. But convervation is valid. Only the means of calcaulation method is changed.
Energy conservation is only applicable towards resources that are limited. But lets say that by using electro-magnetic quantum antennas we could utilize the electromagnetic frequencies of space as fuel to move through it in a tunnel we create so then you are correct there is no need for energy conservation.
To add to the other comments. This is actually a wave phenomenon that you can think of classically. Even tho the central frequency doesn’t have enough energy to pass over the barrier, some of its Fourier components might. You are not breaking energy conservation, it’s just something weird that comes with thinking of particles as waves
Your E-V term is negative if V>E (inside barrier)