Energy cost of the measurement without perturbing the system? 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)
 A: 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.
