quantum mechanics - 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)