# Energy transfer using quantum entanglement

Can we transfer energy from one place to another separated by arbitrarily large distances without any time lag?

For instance, if Alice and Bob are two observers making measurements having a singlet state, they synchronize their clocks and then go to sufficiently large distances both of them apply a magnetic field of equal magnitude (say B) in the positive Z direction. Now, if Alice makes a measurement in the X direction, he will measure a probability of a half for it to be up in that direction but he also changes the direction of the spin of the electron towards the right which implies that Bob's electron will face the left direction that is turning from positive Z direction ( a 90 degrees) which involves emission or absorption of energy. In this way, we are actually transferring energy very efficiently and very quickly (instantaneously).

Just a proof of Lumo's answer. The proof is rather trivial you should see this at least once. It assumes you know about density matrices and partial traces though. Suppose we have two people, Alice and Bob, with two quantum systems $A$ and $B$ respectively. The joined system $AB$ is initially prepared in a state described by a density matrix $\rho$ which could be entangled. Then Alice and Bob fly away in rocket ships to opposite sides of the galaxy. The expected outcome of any observable $O_B$ that Bob can measure on his system is then given by:

$$\left< O_B \right> = \mathrm{Tr}\left\{\rho O_B\right\}= \mathrm{Tr_B}\left\{\left(\mathrm{Tr_A}\rho\right) O_B\right\}= \mathrm{Tr_B}\left\{\rho_B O_B\right\}$$

where $\mathrm{Tr}_{A,B}$ means the partial trace of the degrees of freedom in $A,B$ respectively, and $\rho_B$ is the reduced density matrix of $B$. These are really just definitions, but the second equality holds because $O_B$ only acts on Bob's system. All of the measurements Bob can do are encoded in his reduced density matrix $\rho_B$. In particular, if Bob wants the energy of his system he can just use the Hamiltonian $O_B=H_B$. And as discussed here is makes no difference to Bob whether Alice does a measurement on her state or not, as long as she doesn't communicate the result classically.

Now, instead of measuring the state immediately we let Alice and Bob evolve their systems with unitary operators $U_{A,B}$ before measuring. Any unitaries whatsoever: magnetic fields, throwing the system in a microscopic black hole and catching the Hawking radiation, Schrodinger's cat set ups, anything. The key thing is that $U_A$ acts only on the $A$ degrees of freedom and $U_B$ acts only on the $B$ degrees of freedom. This is a consequence of the locality of interactions, which is a true fact about any accepted model of particle physics (because these are all effective quantum field theories). This changes the density matrix to

$$(U_A \otimes U_B) \rho (U_A \otimes U_B)^\dagger,$$

and Bob's reduced density matrix to

$$\mathrm{Tr}_A\left\{(U_A \otimes U_B) \rho (U_A \otimes U_B)^\dagger\right\}=U_B \mathrm{Tr}_A\left\{U_A^\dagger U_A \rho\right\}U_B^\dagger=U_B \rho_B U_B^\dagger.$$

In other words it doesn't matter to Bob what unitary Alice applies to her system. All that he can measure depends only on what he has done ($U_B$). You can even generalise this argument to the case where Alice does a measurement (projecting her system onto an outcome $a$ with a projection operator $\Pi_a$), and then evolves with a different unitary depending on the outcome of the measurement ($U_{Aa}$). So long as she doesn't communicate her measurement result ($a$) to Bob there is nothing he can do to extract information about it from the $B$ system. This is a very general result.

No, one can't transfer energy – or any other quantity that would carry information – instantaneously. This follows from special relativity. Quantum entanglement doesn't change anything about that.

The point is that if you're only interested in the measurements that Bob can make (e.g. whether he can see the energy of his object systematically increase), all of the quantum predictions are given by the density matrix for B that doesn't depend on anything in the A region at all. So the measured spin or measured energy is undetermined and partly random in general but all the probabilities are calculable from the initial state before any measurement and won't be affected by any decisions made by Alice at all.

## protected by ACuriousMind♦Jun 28 '17 at 11:07

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