Let's first do an example without any interference at all - imagine an electron trapped in a box with some total energy $E$. At some point it releases a photon and we capture that new particle in a box.

By conservation of energy wavefunction of the two particles looks like $\sum\limits_{E_i+E_n = E} a_{in} e^{-iEt} | E_i \rangle | E_n \rangle$

If we measure box with the photon and find an energy of $E_n$, then by the usual copenhagen interpretation the electron immediately collapses into the eigenstate $|E_i \rangle$. Now one has to be careful, because quantum mechanics doesn't really deal with the creation of particles - but we can analogously think of the situation where we dont measure any energy in the photon box - we would immediately know that the electron has energy $E$. The fact that there is no photon in the box does not violate the conservation of energy, it simply tells us that no photon was released in the first place.

This should tell us that even if perfect destructive interference were possible, energy would still be conserved. We can only definitely say that energy has been transferred once we do a measurement - if there is no light to measure then there is no energy that has been transferred.

Now, is perfect destructive interference actually possible to have? Let's look at the wavefunction of two particles $\Psi(x_1,x_2,t)$. Generally we force the norm of $\Psi$ to equal 1 for normalization, but lets relax that condition and take $\Psi(x_1,x_2,t_1)=0$ We then have that $\Psi(x_1,x_2,t_2) = U(t_2,t_1) \Psi(x_1,x_2,t_1) = 0$ And so the wavefunction is 0 for all time - the wavefunction can't be 0 unless it was always 0, assuming unitarity. Now, if we think of a system where destructive interference happens, at time $t$ our wavefunction is non-zero, and at time $t + dt$ our wavefunction is exactly 0. Since time-evolution is given by the flow of the hamiltonian, this means that our hamiltonian has a delta-function spike at time $t$, which we can think of as a change in the energy of our system. This is really an indicator that we are not looking at a large enough system - that part of the energy we thought was in our system was really coming from somewhere else. Note that this argument actually works for any system whose time-evolution is described by hamiltonian flow.

This also partially explains why the answers to this question are so varied. Destructive interference is common but perfect destructive interference is the same as nothing happening. We can say that perfect destructive interference does not happen and that would be correct. We could also say that two waves perfectly destructively interfering give their energy back to the source that made them and that would also be correct. If two waves cancel out in a forest and nobody sees them, were they ever there?

(Note that destructively interfering waves can also lose their energy to another system, such as heat. It is not always the case that the energy lies within the system that created the waves. For example, one could absorb a sound wave to gain energy, which we could model by destructive interference - but also another way to destructively interfere a sound wave is to turn it into heat.)

Let's first do an example without any interference at all - imagine an electron trapped in a box with some total energy $E$. At some point it releases a photon and we capture that new particle in a box.

By conservation of energy wavefunction of the two particles looks like $\sum\limits_{E_i+E_n = E} a_{in} e^{-iEt} | E_i \rangle | E_n \rangle$

If we measure box with the photon and find an energy of $E_n$, then by the usual copenhagen interpretation the electron immediately collapses into the eigenstate $|E_i \rangle$. Now one has to be careful, because quantum mechanics doesn't really deal with the creation of particles - but we can analogously think of the situation where we dont measure any energy in the photon box - we would immediately know that the electron has energy $E$. The fact that there is no photon in the box does not violate the conservation of energy, it simply tells us that no photon was released in the first place.

This should tell us that even if perfect destructive interference were possible, energy would still be conserved. We can only definitely say that energy has been transferred once we do a measurement - if there is no light to measure then there is no energy that has been transferred.

Now, is perfect destructive interference actually possible to have? Let's look at the wavefunction of two particles $\Psi(x_1,x_2,t)$. Generally we force the norm of $\Psi$ to equal 1 for normalization, but lets relax that condition and take $\Psi(x_1,x_2,t_1)=0$ We then have that $\Psi(x_1,x_2,t_2) = U(t_2,t_1) \Psi(x_1,x_2,t_1) = 0$ And so the wavefunction is 0 for all time - the wavefunction can't be 0 unless it was always 0, assuming unitarity. Now, if we think of a system where destructive interference happens, at time $t$ our wavefunction is non-zero, and at time $t + dt$ our wavefunction is exactly 0. Since time-evolution is given by the flow of the hamiltonian, this means that our hamiltonian has a delta-function spike at time $t$, which we can think of as a change in the energy of our system. This is really an indicator that we are not looking at a large enough system - that part of the energy we thought was in our system was really coming from somewhere else. Note that this argument actually works for any system whose time-evolution is described by hamiltonian flow.

This also partially explains why the answers to this question are so varied. Destructive interference is common but perfect destructive interference is the same as nothing happening. We can say that perfect destructive interference does not happen and that would be correct. We could also say that two waves perfectly destructively interfering never described a system with energy and that would also be correct. If two waves cancel out in a forest and nobody sees them, were they ever there?

(Note that destructively interfering waves can also lose their energy to another system, such as heat. It is not always the case that the energy lies within the system that created the waves. For example, one could absorb a sound wave to gain energy, which we could model by destructive interference - but also another way to destructively interfere a sound wave is to turn it into heat.)

Let's first do an example without any interference at all - imagine an electron trapped in a box with some total energy $E$. At some point it releases a photon and we capture that new particle in a box.

By conservation of energy wavefunction of the two particles looks like $\sum\limits_{E_i+E_n = E} a_{in} e^{-iEt} | E_i \rangle | E_n \rangle$

If we measure box with the photon and find an energy of $E_n$, then by the usual copenhagen interpretation the electron immediately collapses into the eigenstate $|E_i \rangle$. Now one has to be careful, because quantum mechanics doesn't really deal with the creation of particles - but we can analogously think of the situation where we dont measure any energy in the photon box - we would immediately know that the electron has energy $E$. The fact that there is no photon in the box does not violate the conservation of energy, it simply tells us that no photon was released in the first place.

This should tell us that even if perfect destructive interference were possible, energy would still be conserved. We can only definitely say that energy has been transferred once we do a measurement - if there is no light to measure then there is no energy that has been transferred.

Now, is perfect destructive interference actually possible to have? Let's look at the wavefunction of two particles $\Psi(x_1,x_2,t)$. Generally we force the norm of $\Psi$ to equal 1 for normalization, but lets relax that condition and take $\Psi(x_1,x_2,t_1)=0$ We then have that $\Psi(x_1,x_2,t_2) = U(t_2,t_1) \Psi(x_1,x_2,t_1) = 0$ And so the wavefunction is 0 for all time - the wavefunction can't be 0 unless it was always 0, assuming unitarity. Now, if we think of a system where destructive interference happens, at time $t$ our wavefunction is non-zero, and at time $t + dt$ our wavefunction is exactly 0. Since time-evolution is given by the flow of the hamiltonian, this means that our hamiltonian has a delta-function spike at time $t$, which we can think of as a change in the energy of our system. This is really an indicator that we are not looking at a large enough system - that part of the energy we thought was in our system was really coming from somewhere else. Note that this argument actually works for any system whose time-evolution is described by hamiltonian flow.

This also partially explains why the answers to this question are so varied. Destructive interference is common but perfect destructive interference is the same as nothing happening. We can say that perfect destructive interference does not happen and that would be correct. We could also say that two waves perfectly destructively interfering give their energy back to the source that made them and that would also be correct. If two waves cancel out in a forest and nobody sees them, were they ever there?

Let's first do an example without any interference at all - imagine an electron trapped in a box with some total energy $E$. At some point it releases a photon and we capture that new particle in a box.

By conservation of energy wavefunction of the two particles looks like $\sum\limits_{E_i+E_n = E} a_{in} e^{-iEt} | E_i \rangle | E_n \rangle$

If we measure box with the photon and find an energy of $E_n$, then by the usual copenhagen interpretation the electron immediately collapses into the eigenstate $|E_i \rangle$. Now one has to be careful, because quantum mechanics doesn't really deal with the creation of particles - but we can analogously think of the situation where we dont measure any energy in the photon box - we would immediately know that the electron has energy $E$. The fact that there is no photon in the box does not violate the conservation of energy, it simply tells us that no photon was released in the first place.

This should tell us that even if perfect destructive interference were possible, energy would still be conserved. We can only definitely say that energy has been transferred once we do a measurement - if there is no light to measure then there is no energy that has been transferred.

Now, is perfect destructive interference actually possible to have? Let's look at the wavefunction of two particles $\Psi(x_1,x_2,t)$. Generally we force the norm of $\Psi$ to equal 1 for normalization, but lets relax that condition and take $\Psi(x_1,x_2,t_1)=0$ We then have that $\Psi(x_1,x_2,t_2) = U(t_2,t_1) \Psi(x_1,x_2,t_1) = 0$ And so the wavefunction is 0 for all time - the wavefunction can't be 0 unless it was always 0, assuming unitarity. Now, if we think of a system where destructive interference happens, at time $t$ our wavefunction is non-zero, and at time $t + dt$ our wavefunction is exactly 0. Since time-evolution is given by the flow of the hamiltonian, this means that our hamiltonian has a delta-function spike at time $t$, which we can think of as a change in the energy of our system. This is really an indicator that we are not looking at a large enough system - that part of the energy we thought was in our system was really coming from somewhere else. Note that this argument actually works for any system whose time-evolution is described by hamiltonian flow.

This also partially explains why the answers to this question are so varied. Destructive interference is common but perfect destructive interference is the same as nothing happening. We can say that perfect destructive interference does not happen and that would be correct. We could also say that two waves perfectly destructively interfering give their energy back to the source that made them and that would also be correct. If two waves cancel out in a forest and nobody sees them, were they ever there?

(Note that destructively interfering waves can also lose their energy to another system, such as heat. It is not always the case that the energy lies within the system that created the waves. For example, one could absorb a sound wave to gain energy, which we could model by destructive interference - but also another way to destructively interfere a sound wave is to turn it into heat.)