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Conservation of energy in quantum teleportation

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Consider the quantum state teleportation protocol of Bennett et. al. How does one prove that this protocol would never violate the conservation of energy? At the face of it, it doesn't seem to be something obvious, as the various measurements, communications, rotations etc. don't seem to be able to account for the differences in the Hamiltonians of Alice's and Bob's labs. The state $\psi$ might have an expected energy $E_1$ in Alice's lab and a completely different expected energy $E_2$ in Bob's. Where does the difference come from?

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user Siddharth Muthukrishnan
asked Feb 26, 2011 in Theoretical Physics by Siddharth Muthukrishnan (25 points) [ no revision ]
During the teleportation protocol, Alice and Bob perform local quantum operations (the various measurements, rotations, etc.) in both of their labs, all of which conserve energy, but may add/subtract energy from the various states they are operating on.

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user Peter Shor
This question really boils down to a previous question, namely the conservation of energy "after the measurement", see physics.stackexchange.com/questions/4047/…

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user Luboš Motl
@Peter Shor, Thanks. It seems to be pretty straightforward.

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user Siddharth Muthukrishnan
@Peter @Lubos Just wanted a clarification as this would completely answer my question: So, if we have a bipartite Hilbert space, then will the Hamiltonian of the system always be of the form $H_A \otimes H_B$? I suppose the reason for this is that in quantum mechanics the Hilbert space of a quantum system is generated from the eigenvectors of the Hamiltonian (via the Schrodinger equation) and so the Hamiltonians being in the tensor product form is equivalent to the postulate that a combined system has the tensor product structure.

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user Siddharth Muthukrishnan
For teleportation, where the two systems are separated, the Hamiltonian will be of that form.

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user Peter Shor
Correction: The Hamiltonian for two isolated systems will be of the form: $H = H_A \otimes 1\hspace{-3pt}\mathrm{l} + 1\hspace{-3pt}\mathrm{l} \otimes H_B$. That is, it will be a sum of terms, not a product of terms.

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user Steve Flammia

3 Answers

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In quantum mechanics energy is an operator. If an isolated system is an eigenstate of energy, then any measurement of the system will give the same result. Such an eigenstate does conserve energy; the energy does not change with time of measurement.

On the other hand, if an isolated system is not in an eigenstate of energy (for example, if it is a superposition of states with different energies), then measuring the energy can give more than one answer. In such a system, energy is still conserved in that the probabilities of the various energies stay the same. This just follows from writing the state as a superposition over energy eigenstates.

Whether you call that feature "energy conservation" or not depends on the definitions of words not physics. I don't have a preference either way.

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user Carl Brannen
answered Aug 20, 2012 by Carl Brannen (240 points) [ no revision ]
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It appears the Bob and Alice have to use identical systems in order to successfully transmit Bob's state $|\phi\rangle$ to Alice. After transmission (I prefer this term over "teleportation") Alice's state is also $|\phi\rangle$, so I don't see where the question of conservation of energy arises.

What does matter is that in order to "teleport" one qubit between two classical observers, one maximally entangled Bell pair is required. This pair is not entangled any more after the state is transmitted. In order to teleport $X$ qubits, one therefore needs $X$ Bell pairs. Work is done every time a entangled pairs are created from a reservoir of unentangled particles and thus an energy cost is associated with each Bell pair and consequently with each qubit that is teleported.

So there is a consideration of energy but in regards to the cost of the resource which allows the process to occur. But energy is conserved locally at every step of the protocol as @Peter Shor mentioned.

Also see this paper by Masahiro Hotta for some nice ideas on how one can teleport energy between two systems in this manner.

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user user346
answered Feb 26, 2011 by Deepak Vaid (1,975 points) [ no revision ]
Bob and Alice don't have to use identical systems but only, really, systems of the same dimension, since all equidimensional Hilbert spaces are isomorphic. (Keep in mind that what's being teleported is the information content of the state.) Thus different systems could correspond to wildly different energy differences.

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user Emilio Pisanty
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Since energy (really mass-energy) is an absolutely conserved quantity, it too will be entangled by the state teleportation system.

That means that whenever your final measurements are compared between the two remote locations, you will always find that mass-energy has been conserved. While in principle that does indeed provide a way to achieve at least a random transfer of energy from one point to another at superluminal speed, the catch is that the largest quantity you can transfer by such a method can never be larger than the amount of energy that from a classical perspective was part of the unresolvable uncertainty in the energy levels when you first set up the remote sites. That energy uncertainty will trace back to the moment when you first separated the entangled components, as with spin uncertainty. So: It's fascinating, but no free lunch.

This post imported from StackExchange Physics at 2014-04-01 17:37 (UCT), posted by SE-user Terry Bollinger
answered Mar 22, 2012 by Terry Bollinger (110 points) [ no revision ]

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