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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?

Luboš Motl
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    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. – Peter Shor Feb 26 '11 at 14:23
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    This question really boils down to a previous question, namely the conservation of energy "after the measurement", see http://physics.stackexchange.com/questions/4047/energy-conservation-and-quantum-measurement – Luboš Motl Feb 26 '11 at 16:54
  • @Peter Shor, Thanks. It seems to be pretty straightforward. –  Feb 26 '11 at 18:59
  • @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. –  Mar 03 '11 at 20:53
  • For teleportation, where the two systems are separated, the Hamiltonian will be of that form. – Peter Shor Mar 03 '11 at 22:39
  • 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. – Steve Flammia Jun 04 '12 at 05:28

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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.

Carl Brannen
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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.

Terry Bollinger
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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.

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    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. – Emilio Pisanty Jul 20 '12 at 15:47