Wavefunction Collapse and Energy Conservation (and Stationary Tunneling)

Question

Let's say I have a simple isolated quantum system of one particle, 1-D, without spin or relativistic effects. The system was prepared in a way that it begins in an eigentstate of the Hamiltonian, with Total Energy = E1.

Then, I measure the position of the particle. The wavefunction collapses, and since the Hamiltonian and position operators don't commute (in most cases at least), that means the system is no longer in an eigenstate of Total Energy = E1, but in a linear combination of various eigenstates, E1, E2, E3, etc. So the Total Energy changed.

For example, a finite potential well. Let's say the potential outside the well is 100 J, inside it's 0 J, and that the initial Total Energy is 20 J. Upon measuring, I find that the particle is outside the well. Its potential energy is now 100 J. The kinetic energy is not -80 J (that is impossible, right?), rather, there is now a superposition of kinetic energy eigenstates with various values.

At first I would think that Energy Conservation was violated (was it?).

But perhaps what happened is that the interaction with the measuring device meant the system wasn't truly isolated. The device 'gave' or 'took' the extra energy. Then that makes me wonder:

Do the classicaly forbidden sections of the wavefunction (like outside the well in the example above) 'rely' on someone measuring the particle in order to be probable? Could you propose that, as long as nobody measures it, the particle will never actually be in that section (since it would have to have negative kinetic energy)?

Answer 1

Collapse doesn't happen. The way to understand quantum mechanics is to apply it universally to both the measured system and the measuring instrument:

https://arxiv.org/abs/1212.3245.

A real measuring instrument will be able to detect some set of states that are something like Gaussians since those states are selected by decoherence:

https://arxiv.org/abs/quant-ph/0306072

Those Gaussians will be peaked around some particular energy and will fall off as you move away from those energies. If all of the states the detector can detect are peaked far above those of the particle being detected then the detection probability will be low and the detector won't work well.

You ask whether the particle could never be in that section. The only existing explanation of the particle's behaviour is quantum mechanics. Quantum mechanics claims that the particle wave function goes into the barrier and that its phase and amplitude outside the barrier is changed by what happens to the wave function inside the barrier. So there is something happening in the barrier involving the particle. So it makes no sense to claim that the particle was never in the barrier regardless of where and how you measure it.

Answer 2

Do the classicaly forbidden sections of the wavefunction (like outside the well in the example above) 'rely' on someone measuring the particle in order to be probable?

No.

Could you propose that, as long as nobody measures it, the particle will never actually be in that section (since it would have to have negative kinetic energy)?

No, we are talking quantum mechanics here, which does not work with classical potentials intuition. This link shows what happens with tunneling:

Note that the energy level does not change inside or outside the barrier. That is one of the things quantum mechanics is about, probabilities. The amplitude of $Ψ$ changes and this means that $Ψ^*Ψ$, where $Ψ^*$ is the complex conjugate of $Ψ$, the probability is reduced . The total wavefunction is continuous, the solution of the quantum mechanical barrier set up.

When you measure the energy of the particle that has tunneled, the wavefunction changes, due to the interactions necessary for the experimenter to define the energy. It is a different quantum mechanical boundary problem.

If the particle is not detected it will go off to infinity fulfilling the probability distribution which predicts its tunneling. To test the model, one has to have a large number of similar setups, and build up the probability distribution. Radioactive decay of nuclei is studied with tunneling .

Answer 3

The system is initially in the energy eigenstate $E_1$. After the position measurement, according to orthodox quantum mechanics, the wave function has collapsed to the position eigenfunction at say $x_0$ which is the delta function $\delta(x-x_0)$. This says nothing about the energy of the particle, even if $x_0$ is outside the potential well. This delta wave function can be expressed with complex coefficients $a_i$ as a linear combination of the energy eigenfunctions $\psi_i$ corresponding to the energy eigenvalues $E_i$: $$\psi=\sum_i a_i\psi_i$$ When you subsequently measure the energy of the system, you will find it with probability $|a_i|^2$ in the energy eigenstate $E_i$ which can be in the potential well or above the potential barriers. Then you know nothing about the position of the particle.

When you consider the particle in eigenstate $E_1$ and make a position measurement, you have a certain probability also to find the particle outside the well in an interval $dx$ at $x$ according to $|\psi_1(x)|^2dx$ This doesn't violate the energy conservation because the energy is unknown after the position measuerement.

Answer 4

Projection measurement is not an Unitary Evolution, thus if you only consider the system itself, the average energy during one specific realization of measurement does not conserve indeed. For example, The initial state of the system is a superposition of two energy eigenstates (the eigenstates of the Hamiltonian), $$|\psi \rangle=a_0|0\rangle+a_1|1 \rangle$$ and one can calculate the average energy of this state, which is $$\langle H \rangle=|a_0|^2E_0+|a_1|^2E_1$$ If one realize a single projection measurement, say we get the first eigenstate, which leads the average energy of the system to be simply $E_0$. But if you conduct many measurements and calculate the average energy of the ensembles after the measurements, it is the same as $\langle H \rangle$, thus the energy does conserve. Actually, this calculated quantity can only be perceived as the average value of the measurements. The initial state does not have a definite energy value, so there is no meaning talk about the energy conservation.