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Background

Let's say I have a Hamiltonian $\hat H$ (assume the Schrodinger equation) and it be in an arbitrary eigen-energy state:

$$ \hat H_{\text{system}} | m \rangle = E_m |m \rangle$$

And I want to measure the momentum of the system without perturbing the Hamiltonian. We know this will force the wave-function into a momentum eigenket:

$$ \hat p |p_j \rangle = p_j | p_j \rangle $$

I thought what would be the average energy cost of this measurement:

$$ \langle p_j | \hat H | p_j \rangle = \langle p_j |\hat H ( \sum_{n} |n \rangle \langle n | ) |p_j \rangle = \sum_{n} E_n \langle n |p_j \rangle \langle p_j |n \rangle = \sum_{n} E_n |\langle n |p_j \rangle |^2 $$

Hence, the difference in energies for the $2$ states are:

$$\Delta E = \sum_{n} E_n |\langle n |p_j \rangle |^2 - E_m $$

I think this energy comes from another system (that of the experimenters?)

Question

Is the above derivation correct? (If I am wrong what is the proper energy cost for a measurement on average?) Can I use this to say "the measurement only makes sense if there is more than one system?" (otherwise I'm not sure where the energy for the measurement would come from)

More Anonymous
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  • The is no state such that $\langle \psi | H | \psi \rangle < \langle 0 | H | 0 \rangle$, so the problem in the second part of your question simply never occures. To see this not that $\langle \psi | H | \psi \rangle = \sum_n E_n |\langle n|\psi\rangle|^2$. Since $E_n \ge E_0$, $|\langle n|\psi\rangle|^2 \ge 0$ and $\sum_n |\langle n|\psi\rangle|^2 = 1$, so the minimum value occures for $|\langle n|\psi\rangle| = \delta_{n,0}$ that is $|\psi\rangle = |0\rangle$ – By Symmetry Jun 04 '19 at 15:47
  • face palm ... Forgot about that ... Perhaps I can edit to ask the minimum energy cost of measurement? – More Anonymous Jun 04 '19 at 15:48
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    how do you define energy cost here? You seem to be just calculating the average energy of momentum eigenstates – glS Jun 04 '19 at 16:28
  • There is no summation on $j$ .. .So it is for a particular momentum state ... So one can define the average cost for a particular momentum eigenstate as the difference of energies in states: $\langle p_j | \hat H | p_j \rangle - \langle m | \hat H | m \rangle $ – More Anonymous Jun 04 '19 at 16:44
  • Actually i think if you do same calculation for all other momentum eigenstates ($|p_i>, |p_m>, etc$) by computing their $\Delta E^2$ s you can add them together to get the variance. So what you are doing here is basically measuring the distance (error) of each data relative to average, not what you think. – Paradoxy Jun 20 '19 at 21:58
  • @Paradoxy doesn't the measurement force the wavefunction into one eigenstate? I'm not sure what relevance the calculation of the variance has to do with my assertion. And to be clear about what I think would be: "On average the system energy changes (this does not happen during unitary evolution). To make it relevant to the conservation of energy I can always take ensemble data and focus on the average energy of a particular eigenstate and see the system independently does not obey the conservation of energy" – More Anonymous Jun 20 '19 at 22:08
  • Maybe I'm being thick headed and not getting your point but why would you want to do it for all eigenstates? – More Anonymous Jun 20 '19 at 22:09
  • Imagine a group of electrons with different momentums i consider this as my system, if you measure their momentums, every time you get a different value but on the other hand on average energy of the system will be constant. It's not because you add energy to the system every time, this variation in energy comes from the variance of the system itself. Just imagine a classic example of this where your interaction doesn't matter, that might help you to understanding my comment – Paradoxy Jun 20 '19 at 22:27
  • Yes .. But I can always group the observations when they have the same momentum (a particular eigenstate) and I would see an average energy cost for the measurement which is what I believe I have calculated. Also not sure what you mean by "variance of the system itself"? – More Anonymous Jun 20 '19 at 22:32
  • There is no energy cost here like you have defined above. See, in classical example, when your interacation doesn't matter, you still get different values for every measurements as such you can do what you have done here for classical case and still argue that this formula is "energy cost" but it isn't! We clearly know that in classical case there is nothing such as adding energy to the system and etc. If you write something similar for classical case, it would only mean error of each data relative to average, i am suggesting that same logic applies for quantum case, for that formula of course – Paradoxy Jun 20 '19 at 22:45
  • Sorry the word you had used previously was "classic example" not "classical case". I'm not sure you can force the electrons into a particular eigenstate in the classical case. Though I think it would be best if you answered my question (preferable with some equations). – More Anonymous Jun 20 '19 at 22:50

1 Answers1

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Edit: this is an answer now, not a comment.

Assume Schrödinger equation, we use Hamiltoian operator on wave function $$\hat{H}|\psi>=E_n|\psi>$$ What actually we get from it would be a set of eigenstates and eigenvalues such that $$|\psi>=\sum_{n}E_n|\psi_n>$$ using only Hamiltonian itself on the system does not leave an arbitrary state, rather it will give a set of eigenstates, so there is in fact a measurement before your very first paragraph.

The next thing that comes to mind would be the fact that Hamiltonian and momentum operator do not commute so Let's say that your arbitrary state after measurement of the energy is $|\psi_1>$. Then the set of your eigenstates for momentum will become $$|p'_i>=|p_i><p_i|\psi_1>$$

However you have chosen an eigenstate $|p_j>$ from the set of $|p_i>$ eigenstates with assuming

I can always group the observations when they have the same momentum (a particular eigenstate) and I would see an average energy cost for the measurement

By doing so, you can neglect $<p_i|\psi_1>$ from my third equation, because after all, you are grouping only one particular state. That'd mean we don't care about its likelihood at all

Hence

$$ \langle p_j | \hat H | p_j \rangle = \langle p_j |\hat H ( \sum_{n} |\psi_n \rangle \langle \psi_n | ) |p_j \rangle = \sum_{n} E_n \langle \psi_n |p_j \rangle \langle p_j |\psi_n \rangle = \sum_{n} E_n |\langle \psi_n |p_j \rangle |^2 $$

And also $$\Delta E = \sum_{n} E_n |\langle \psi_n |p_j \rangle |^2 - E_m $$ are both correct. There is a catch though!

As you can see there should be 3 consecutive measurements, 1.energy 2.momentum 3.energy measurements. Do note that in the very first measurement where you force the system to goes into one particular eigenstate $|\psi_1>$, what actually you are doing is interacting with the system. i.e there is an energy cost for your very first measurement itself that you can't "measure" it because you have no idea about previous state of the system. Thus your final equation does in fact tell us about the energy cost between two particular states, rather than energy cost of measurements as a whole. You can't be sure about the exact influence of your measurements on the system, because if you were, you could remove it from your calculation and claim that "the original state of the system was something" which goes against the uncertainty principle. The catch is your first measurement.

Paradoxy
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  • Alright, I think I get your point. However, I think my argument still holds. We have during the measurement $| m \rangle \to | p \rangle$ (where the $m$ is the m'th energy eigenstate and the probability of this $ \langle m | p \rangle|^2$). Now, for the $2$'nd measurement we have $|p \rangle \to |n \rangle$ with probability of $|\langle p| n \rangle|^2$ for $n \neq m$ the conservation of energy is in trouble.Note: $p$ is a momentum eigenstate. – More Anonymous Jun 21 '19 at 04:45
  • @MoreAnonymous There should be 3 measurements though, one for energy (to put the system in AN arbitrary state) one for momentum and again one for energy. There is no problem with conservation of energy at all, because when you do your momentum measurements, you will destroy whatever information you ever had from energy of the system(again momentum and energy don't commute), it's not possible to compare m and n states. Just change momentum and energy measurement with position and momentum measurement if you don't understand what i am saying – Paradoxy Jun 21 '19 at 09:25
  • I'm not saying I'm doing these measurements all at the same time ... I am gaining information of the system through successive measurements ... I'm not sure what "you will destroy whatever information you ever had from energy of the system" means. I can easily record the information about the energy of measurement $1$ and compare it to the measurement of $3$. Note: Conservation of energy would be meaningless if I cannot compare the energy of states at different times. (I'm not sure how to formulate it without doing that) – More Anonymous Jun 21 '19 at 09:49
  • @MoreAnonymous, I know these measurement are consecutive, that's what i have written in my comment and post after all, conservation of energy in quantum mechanics as much as i know is being held by Heisenberg uncertainty, that is $\Delta E \Delta t≥const$, roughly speaking if you compare your energy information in different times, they should obey this inequality, if not you can say that conservation of energy is violated. That's what i meant above, in quantum mechanics a constant energy is not necessary, rather we have an inequality because of information lost – Paradoxy Jun 21 '19 at 09:55
  • Since, we have $3$ measurements I can take measurements at times: $t_1 \to - \infty$, $t_2 \to 0$ and $t_3 \to + \infty$. This should considerably reduce the amount of variance in $\Delta t$ and allow you to make the statement I am without worrying about the standard deviation. Also I would like to know which energy time uncertainty u refer to specifically ... See https://physics.stackexchange.com/questions/53802/what-is-delta-t-in-the-time-energy-uncertainty-principle – More Anonymous Jun 21 '19 at 10:10
  • @MoreAnonymous If you choose your times such that, you should also consider evolution of the system itself. i.e if you leave a system from $t_1$ to $t_2$ system has enough time to change its eigenstate from m to the set of eigenstates that it had first, see my second formula, it will become like that. Also the accepted answer in your link is the best one by far. Imo This tell us if you measure energy in different times, you should always expect a deviation from what you have recorded before, and it's not because you put some energy to the system. – Paradoxy Jun 21 '19 at 10:39
  • Let us continue this discussion in chat. – Paradoxy Jun 21 '19 at 10:47
  • "f you leave a system from $t_1$ to $t_2$ system has enough time to change its eigenstate from m to the set of eigenstates that it had first" .. about this ... If we assume Unitary evolution there is no way a energy eigenstate will turn into a momentum eigenstate ... Ofcourse, there is a measurement which is usually considered instantaneous ... "if you measure energy in different times, you should always expect a deviation from what you have recorded before" again I do not where this is coming from ... Feel free to edit equations in your answer and refer them? – More Anonymous Jun 21 '19 at 10:47
  • @MoreAnonymous check my answer now. I have edited it. – Paradoxy Jun 28 '19 at 08:19