# Conservation of energy and wavefunctions

I'd appreciate a bit of clarification on how conservation of energy works in QM.

The infinite square well has a set of stationary states, each corresponding to one of the discrete energy levels of the well. A particle in a particular state can be represented as a sum of these stationary states. The expectation value of the Hamiltonian, < H > , of this particle does not change with time. The state however (by virtue of being a sum of multiple stationary states) has a non-zero energy-uncertainty.

Measuring the energy of the particle collapeses its wavefunction into one of the stationary states with a corresponding energy value.

This bothers me. While < H > is constant, the "actual" energy of the particle (upon measurement) could vary greatly. This seems like a violation of energy conservation. [And while I've posited this question in terms of the infinite well, any wavefunction composed of non-degenerate stationary states seems like a violation to me]

So, is the particle entangled with some other particle (through whatever process put it in the well in the first place) such that the net energy in some larger system is constant? Or is energy only conserved is larger systems (which are compromised of so many particles as to approach < H > )? Or what am I misunderstanding?

• please explain why you feel like it is a violation of conservation energy? Unless you clarify that, I can't answer your question. (by the way, there actually is a scope of violation of conservation of energy by uncertainty principal. But I don't think you are stuck at that point.) Please elaborate your problem. Sep 21 '17 at 7:31
• I had always intrepreted conservation of energy to be a very exact statement. What seems strange to me is that I can have a particle with energy expectation of (let's say) 5, but if I measure it, I could collapse the particle into a stationary state with energy (let' say) 3 Where did the extra 2 energy go? It would seem on a very small scale energy is not truely conserved. Sep 21 '17 at 12:41
• What do you mean by "could vary greatly"? Is this during normal unitary evolution, or during a projective measurement of energy? Sep 21 '17 at 13:27
• ... and, more importantly, what do you mean by the "actual" energy of the particle? Sep 21 '17 at 13:28
• I mean we have a number of different energy values the particle might collapse into when measured. 'Actual' energy as in the value the particle assumes upong measurement (there is a reason I put it in quotes) Sep 21 '17 at 13:36

That's a great question! Indeed, in quantum mechanics, energy is only defined in expectation, and conservation of energy refers solely to expectation values. It's straightforward to show from the Schrodinger equation that this expectation value is constant in time.

However, measurement does not break conservation of energy as long as we keep track of the energy of the measurement apparatus. This has to be true because we can always consider the system plus the measurement apparatus as one joint quantum system, whose energy we know is conserved.

As a specific example, suppose we measure the position of a low-energy particle by firing a photon at it. If we measure the position very accurately, then the wavefunction collapses to a sharp spike, which has a very high energy which wasn't present in the original particle. This energy must have been absorbed from the incoming photon -- and indeed, by the de Broglie formula, $$\lambda = h/p$$, a high-precision position measurement can only be done using a high energy photon.

Some subtleties arise if the measurement device is macroscopic. For example, let's say that the photon in the previous example has a chance of missing, so the quantum state is a superposition $$|\text{low energy particle and high energy photon} \rangle + |\text{high energy particle and low energy photon} \rangle.$$ The (expectation value) of the energy is the same as before at this point. Now suppose we use a high-energy photon detector and read off the result in the lab, giving a superposition of $$|\text{low energy particle, you see '1' on detector} \rangle + |\text{high energy particle, you see '0' on detector} \rangle.$$ At this point energy is still perfectly conserved if you account for the energy of everything in each branch, but if you subscribe to the Copenhagen interpretation, you might say this state is unacceptable because macroscopic observers can't be in superpositions. In other words, you say the state must be $$|\text{low energy particle, you see '1' on detector} \rangle \text{ or } |\text{high energy particle, you see '0' on detector} \rangle$$ rather than a superposition.

This is also fine. The problem is that one is then tempted to remove oneself and the detector from the picture entirely, arriving at $$|\text{low energy particle} \rangle \text{ or } |\text{high energy particle} \rangle$$ which are two states with different energy. This thinking led the pioneers of quantum mechanics to propose that energy was conserved only on average (e.g. in BKS theory, see here).

This is how it's usually presented in introductory textbooks, but that is only done for simplicity. When you think about it, it's a profoundly unphysical picture -- it basically treats measurements as occurring magically, without any physical cause. Modern-day users of the Copenhagen interpretation don't use it in this unphysical way, and we have known for almost a century that energy non-conserving theories like BKS are completely wrong. Energy is exactly conserved in all genuine interpretations of quantum mechanics.

• @ValterMoretti I just mean in terms of the expectation value $\langle H \rangle$ for the system plus measurement apparatus, I don't assume everything is in an eigenstate. Sep 21 '17 at 13:16
• OK, sorry I misunderstood your answer, I cancel my comment. Sep 21 '17 at 14:08
• Is it not the case that "Low energy particle & high energy photon" cancel each other out so that energy is conserved even when the wavefunction collapses (Since it's conserved in each state)? I thought that's what you were explaining with your first paragraph. Dec 27 '18 at 2:23
• @NicholasPipitone Belated response, but you're right, I forgot the last part of the explanation. Sep 18 '19 at 7:50