Consider a particle in a potential well. Let’s assume it’s a simple harmonic oscillator potential and the particle is in its ground state with energy E0 = (1/2) ℏω0. We measure its position (measurement-1) with a high degree of accuracy which localises the particle, corresponding to a superposition of momentum (and therefore energy) states.

Now we measure the particle’s energy (measurement-2) and happen to find that it’s E10 = (21/2) ℏω0. Where did the extra energy come from?

In the textbooks it’s claimed that the extra energy comes from the act of observation but I wonder how that could work. Measurement-1 which probed the position of the particle can’t have delivered to it a precise amount of energy, while measurement-2 might just have been passive. No doubt there is entanglement here between the particle state and the measuring device but where, and which measurement?


For a high degree of accuracy you would have to probe the particle with a high energy (short wavelength) photon so there is plenty of energy that can go into vibrational excitation. After such hard hit the particle will be smeared across a wide range of states $$\Psi=a_0*\Psi_0+a_1*\Psi_1+...$$This is not an entanglement but a simple superposition of eigenstates. The expectation value of the particle's energy $\bar{E}=\sum_{i}a_i^2E_i$ should not be equal to the energy of any particular state and the second measurement will yield $E_i$ with $a_i^2$ probability.

So, the extra energy comes from an interaction with a probe particle and it doesn't have to be precisely equal to the energy of a certain vibrational state.

  • $\begingroup$ Excellent. I think that if we take this answer and then add what Lubos said to the end of it we have the complete answer. Thanks. +1. $\endgroup$ – Nigel Seel Jan 28 '11 at 20:49

Dear Nigel, if you measure the particle's position and find it in a small region, you also change its state.

As you correctly wrote, a localized wave packet (I don't talk about a delta-function state vector whose average kinetic energy would be infinite) can be rewritten as a linear superposition of energy eigenstates.

It means that before you measured the energy, there was a nonzero probability for the energy to be $(21/2)\hbar\omega$, and this particular outcome was ultimately realized. There is no violation of the energy conservation law here.

What you may be annoyed by is that the final energy of the electron usually isn't equal to the expectation value of energy in the state before the measurement. It's surely true. But there is no reason why it should be. The expectation value isn't any "objective value" of energy. It's just a statistical average of many possibilities, and only some of them will be realized, as dictated by the probabilities predicted by quantum mechanics.

If you wish, the process of measurement violates the "conservation of the expectation value of energy".

Needless to say, this "violation" can't be used to obtain any sharp contradiction with the conservation law for the "actual" energy. You may only interpret the measured energy in a "classical way" after decoherence, which means after the measuring apparatus has interacted with the environmental degrees of freedom (that are needed for decoherence). If you prepare your harmonic oscillator plus the apparatus in a state whose total energy is sufficiently sharply well-defined to reveal the "violation of the conservation law for the expectation value", the environmental degrees of freedom will inevitably spoil this accuracy.

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    $\begingroup$ Nigel considers a particle in $E_0$ state which means that $\Psi=\Psi_0$. It means that the probability of other outcomes was zero before the energy measurement. It is the act of measurement itself that introduced a perturbation and mixed states. $\endgroup$ – gigacyan Jan 28 '11 at 15:00
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    $\begingroup$ Right, well, in that case, the first measurement of the position decomposed the wave function according to positions. The fact that we measured the energy again doesn't mean that the particle could return to the very same state. Such a return can't occur because $[X,H]\neq 0$, the measurements don't commute, and similarly the projectors for $X$ and $H$ don't commute. $\endgroup$ – Luboš Motl Jan 28 '11 at 15:28
  • $\begingroup$ Lubos, please read the question carefully. You will find that the actual question is in the final sentence. Thanks. $\endgroup$ – Nigel Seel Jan 28 '11 at 16:13
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    $\begingroup$ Dear Nigel, well, every interaction - including every measurement - creates entanglement. But the very point of measurement is that the measured particle interacts with a "classical" system that quickly decoheres. So one shouldn't use an entangled pure state but a density matrix for which the notion of entanglement loses its usual strength. $\endgroup$ – Luboš Motl Jan 28 '11 at 21:15

What is interesting about this question is that the answers may display its author's Interpretation of QM. Let us first review the order of events here:

  1. Particle is prepared in Ground State $\psi_0$ (an Energy Eigenstate)
  2. Particle is position measured with observable X (measurement-1)
  3. Particle is now in a Position Eigenstate
  4. Particle is now Energy measured (measurement-2)
  5. Particle is in state $\psi_{21/2}$

In this example energy has been given to the particle.

Firstly this is similar to an atom's electron becoming excited. If this has happened via a collision or photon absorption during either measurement, then the particle will only absorb an integral quantums worth of energy.

However since the Hamiltonian in this case is $H=1/2(p^2 + x^2)$, then $[X,H]=\hbar p\neq0$. So the quantum formalist can simply take the Copenhagen view that since the observables dont commute no guarantees can be given about their eigenvalues staying the same after such measurement history. And no further explanation as to "why" need be given beyond "quantum fluctuation" (or "quantum randomness" if you prefer). Thus there is no "microtheory" explaining where the energy (in this case - maybe extra momentum in another case) came from.

Further considerations: (1) if one entertains a "hidden variable" philosophy then this experiment needs to be explained by that means. (2) This experiment has similarities to the quantum vacuum explanations for spontaneous emission. If such a model could apply here then we could say that the particle has had a random payoff from the quantum vacuum, perhaps mediated by one of the measurements.

Also any entanglement between measurement instrument and particle is supposed to have ended shortly after measurement as a result of decoherence between a classical object and a quantum one.

  • $\begingroup$ I ask again, in the scenario the particle is measured to have increased its energy following the second measurement. Really! This is not some uncertainty principle energy fluctuation. It really has!! So (i) what system lost energy so that the particle would gain it; (ii) what was the mechanism? No answer so far seems very clear on this. $\endgroup$ – Nigel Seel Jan 28 '11 at 17:34
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    $\begingroup$ It is the measuring device that supplies energy while "localization" of particle. You can consider its influence as an additional time-dependent external "squeezing" potential. $\endgroup$ – Vladimir Kalitvianski Jan 28 '11 at 18:31
  • $\begingroup$ Yes the issue is that without a description of the classical measurement we have some possibilities, but photon absorption is the most likely (discussed in that paragraph). The photon would come from the measuring instrument. $\endgroup$ – Roy Simpson Jan 29 '11 at 10:50

Thinking about your question has led me to the following conclusion:

Suppose there is a a conserved quantity $Y$, and an isolated system $S$. Let's also suppose the isolation is temporarily lifted to measure an observable $\widehat{X}$ of $S$. This measurement is performed by a measuring apparatus $A$. For simplicity, let's lump together the rest of the universe outside the system including the apparatus and the environment into $A$. Let's further assume we have a tensor product structure between $S$ and $A$ for the Hilbert space. Let $\widehat{Y}_S$ and $\widehat{Y}_A$ be the restriction of $\widehat{Y}$ to $S$ and $A$ respectively. Then, the sum $\widehat{Y}_S + \widehat{Y}_A$ has to be conserved.

Now, let's suppose $\widehat{Y}_S$ and $\widehat{X}$ don't commute. Let's see what happens if we further make an assumption very common in measurement theory that for systems starting out in an eigenstate of $X$, a perfect measurement will leave it in an eigenstate of $X$ with the same eigenvalue. The Hilbert space of $A$ decomposes into eigenspaces of a pointer operator $\widehat{P}$ such that after a measurement of an eigenstate of $X$, $A$ ends up in an eigenstate of $\widehat{P}$ with an eigenvalue equal to the eigenvalue of $X$.

As $\widehat{Y}_S$ and $\widehat{X}$ don't commute, there exists an eigenvalue of $X$ such that all nonzero eigenstates of it aren't also $\widehat{Y}_S$ eigenstates. With all these assumptions in place, an initial state of $\sum_n c_n \left| x_n \right\rangle$ will evolve into $$\sum_n c_n \left| x_n \right\rangle \otimes \left| \chi_n \right\rangle$$ where $\left| \chi_m \right\rangle$ and $\left| \chi_n \right\rangle$ are orthogonal if $m \neq n$ because they have different pointer values. Let's further assume $A$ is initially in an eigenstate of $\widehat{Y}_A$. Assume that the system is initially in an eigenstate of $\widehat{Y}_S$ with eigenvalue $y_m$ which isn't also an eigenstate of $X$. Then, this state can be expressed as $\sum_n U_{mn} \left| x_n \right\rangle$ with $U_{mn}$ being nonzero for at least two different values of $n$. The final state is $$\sum_{np} U_{mn} U_{pn}^* \left| y_p \right\rangle \otimes \left| \chi_n \right\rangle$$ If we look at the contribution to the sum for one of the values of $n$ for which $U_{mn}$ is nonzero, we find the combined $S+A$ system can't be in an eigenstate of $\widehat{Y}$. Precisely because the $\chi$'s are orthogonal, the contributions from different $n$'s can't cancel, and this remains true for the wavefunction as a whole.

But of course, $\widehat{Y}$ generates a symmetry, and $\widehat{X}$ is not invariant under this symmetry. So, to measure $X$, the initial state of the apparatus can't possibly be invariant under this symmetry either. So, we shouldn't expect $A$ to start in an eigenstate of $\widehat{Y}_A$. The measurement process itself is expected to change $\widehat{Y}_S$, but we have a compensating change in $\widehat{Y}_A$.


If I understand correctly, the key question in the topmost post is in this sentence "measurement-1 which probed the position of the particle can’t have delivered to it a precise amount of energy".

Now, what I am thinking is that, if you really measure the position of a system (or a component of a system), then measure the energy of this system again, will you really get a discrete spectrum of energies (in the case of a simple harmonic oscillator)?

I guess not, because measuring the position can perturb the overall momentum of the system, thus changing the translational energy, which has a continuous spectrum.

So in the case of harmonic oscillator, after measuring its position, the wave function will not be a superposition of the discrete eigenfunctions, but rather an integral of a continua of eigenfunctions, which include the cases with nonzero translational energies.

But I believe the point here is not really about "position measurement". I mean, maybe position measurement is just a bad example. One could devise other types of measurements to collapse an energy-eigenstate into a superpositional one (of discrete energy levels). If the measurement is designed careful enough, I believe the process by which a discrete amount of energy is transferred to or from the system being measured will become very clear.


I read the original question and all six answers above. I do not think that these answers directly answer the question. There is energy transferring from the apparatus to the particle, amounting to an uncontrollable disturbance of the momentum, indicated also by the Heisenberg Uncertainty Principle in the context of the measuring position of an electron using a light beam focusing on it. The original question is, how the energy transformation is made during the measurement? Or, how the energy flows from the apparatus to the particle? Or, is there any dynamical mechanism? As far as I know, there is NO answer within the current formalism of quantum mechanics.


A QM wave function describes an ensemble of measurements, not the single measurement output. The corresponding potential does not describe the perturbation due to measurement. If you only measure the system energy, it will be $E_0$, of course, because you measure the eigenstate of the corresponding operator.

But the coordinate operator does not commute with the Hamiltonian, the ground state is not its eigenstate so you will get dispersion in coordinate measurements. While measuring coordinate you may introduce a perturbation that changes the initial state. Now you may have a superposition of eigenstates of Hamiltonian. No wonder you may then find energies different from the ground state.

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    $\begingroup$ Yes, we do indeed find energies different than the ground state after the fairly-precise position measurement, and we may even measure one. Looks like the system suddenly changed its energy, doesn't it. How could that work? Well, the particle in the SHO is NOT a closed system under measurement so the delta in energy must have occurred in the measurement process itself. OK, now re-read the question (last para). $\endgroup$ – Nigel Seel Jan 28 '11 at 16:18
  • $\begingroup$ No, there is no entanglement. You have just to think of a measurement of an observable as of "projection" of the given state to an eigenstate of the observable and leaving the system in the latter state. In case of coordinate measurement, the system left in the eigenstate of coordinate evolves somehow with time. Anyway, it is a superposition of different energetic states now. When you measure another observable in this new state, you "project" it to its own eigenstate and leave it in that. $\endgroup$ – Vladimir Kalitvianski Jan 28 '11 at 16:39
  • $\begingroup$ Dear downvoters, tell me where I am wrong, please. I would like to learn. $\endgroup$ – Vladimir Kalitvianski Jan 31 '11 at 16:22

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