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1. Researchers set the value of qubit X to zero 2. They pressed rewind on all qubits 3. They scrambled the value of qubit X 4. They pressed “play” 5. They c

1. Researchers set the value of qubit X to zero 2. They pressed rewind on all qubits 3. They scrambled the value of qubit X 4. They pressed “play” 5. They checked the value of qubit X to see how close it was to zero. Turns out it was very very close to zero. This calls for an explanation of the recovery of the final value of qubit X despite the destruction of its initial state during step 3. The article proposes that the information necessary to recover the final value is encoded in the qubits with which X is entangled. I don’t know enough about QM to understand how the system is able to use this information to “set” the qubit to the value that it “should” be. Do the researchers’ actions influence the values of the qubits with which X is entangled in such a way that the forward evolution results in a system where X recovers its initial (final) value? My background is in deep learning. If I look at the activations in the final layer of a network, and want to achieve an output for a particular element, I can set that element to the desired value and run the network backward to reconstruct an input that would give that value. If I perturb that input slightly, the network will produce a slightly different activation pattern. If I take an adversarial approach, I can have what would appear to be an outsized influence on the final layer activations. I wonder if such an adversarial perturbation has a quantum analogue.

gus_massa 6y ago
Following your numeration, I'll try to ELI25: 0) Let's assume they have 4 qbits. Then the system has 2^4 = 16 possibilities, so the complete estate of the system is a complex vector of length 16 (v_0000, v_0001, v_0010, v_0011, ..., v_1110, v_1111) such that |v_0000|^2 + |v_0001|^2 + ... + |v_1110|^2 + |v_1111|^2 = 1 And |v_abcd|^2 is the probability that if you measure the value of all the qbits you get the value a in the first one, b in the second one, c in the third and d in the fourth. When you measure the value of all the bits, the system "collapses" and randomly select one of the possibilities using this probabilities. (Yes, it is weird.) 1) Let's assume that X is the first qbit. They made a system where the first qbit is zero. This is means that the last 8 coefficient of the vector, i.e. v_1000, ..., v_1111 are zero. So if you measure the system you always will get a 0 in the first qbit. They probably made an easier initial configuration, where one of the coefficients is one and all the other are zero, but the important part is that at least all the coefficients of the vector where the first qbit is 1 have the value zero. 2) Letting the system evolve in time (without an intermediate measurement) is equivalent of multiplying the vector of length 16 by an unitary matrix. The coefficients of the matrix are determined by how you connect the quantum gates or how your quantum system is made. The important part is that is unitary. Reversing the time means multiplying by the inverse. If you make no measurements, yo can go(simulate) back in time and return, and you get v_new = U U^-1 v so v_new = v Here we must assume that the matrix U and U^-1 are not very diagonal or has block, sparse, whatever. Let's assume that it is full or almost full and the values of the vector are mixed thoughtfully by the matrix multiplication. 3) After going back in time, the state is w = U^-1 v. Then they measure one qbit. Let´s assume that it is the last one. When you measure only one qbit, you can get a 0 or a 1. Here you have to distinguish the coefficients where the fourth qbit is 0 and the coefficients where the fourth qbit is 1. The probability of getting 0 is related to the values of the coefficients where the fourth qbit is 0 |w_0000|^2 + |w_0010|^2 + ... + |w_1100|^2 + |w_1110|^2 and the probability of getting 1 is |w_0001|^2 + |w_0011|^2 + ... + |w_1101|^2 + |w_1111|^2 They sum 1, so it's a probability. As a side effect of the measurement, if you get a 0, all the coefficients where the fourth qbit is 1 get erased and only survive the coefficients where the fourth qbit is 1. You have to multiply all of them so the sum is again 1. But the important part is that half of the information is destroyed. If the result of the measurement is 1, then the other half of the vector of length 16 is erased. The important part is that in either case, half of the vector is erased. You get a new vector w_half. :) 4) Now you return from your trip to the past. This is like multiplying the new vector by U. So the result is v_new = U w_half = U Msm w = U Msm U^-1 v where Msm is the measurement operation. Now v_new is not equal to the initial v, because the measurement in the past is messing the calculation. 5) Now they measure the value of the first qbit in the new state. It is like in 3. The probability of getting a 0 is |v_new_0000|^2 + |v_new_0001|^2 + ... + |v_new_0110|^2 + |v_new_0111|^2 and the probability of getting 1 is |v_new_1000|^2 + |v_new_1001|^2 + ... + |v_new_1110|^2 + |v_new_1111|^2 The result of the article is that in this last measurement there is a high probability of getting a 0. Note that if you apply this calculation to the initial states made in 1) you get a 100% probability of getting a 0. So the idea is that the measurement in the simulated past has not messed too much this value. The important part is you can't use any unitary matrix U, it must be not very sparse so U^-1 distribute the initial information in all the coefficients, and when half of them are erased, you don't loose too much information.

gus_massa 6y ago

Following your numeration, I'll try to ELI25: 0) Let's assume they have 4 qbits. Then the system has 2^4 = 16 possibilities, so the complete estate of the system is a complex vector of length 16 (v_0000, v_0001, v_0010, v_0011, ..., v_1110, v_1111) such that |v_0000|^2 + |v_0001|^2 + ... + |v_1110|^2 + |v_1111|^2 = 1 And |v_abcd|^2 is the probability that if you measure the value of all the qbits you get the value a in the first one, b in the second one, c in the third and d in the fourth. When you measure the value of all the bits, the system "collapses" and randomly select one of the possibilities using this probabilities. (Yes, it is weird.) 1) Let's assume that X is the first qbit. They made a system where the first qbit is zero. This is means that the last 8 coefficient of the vector, i.e. v_1000, ..., v_1111 are zero. So if you measure the system you always will get a 0 in the first qbit. They probably made an easier initial configuration, where one of the coefficients is one and all the other are zero, but the important part is that at least all the coefficients of the vector where the first qbit is 1 have the value zero. 2) Letting the system evolve in time (without an intermediate measurement) is equivalent of multiplying the vector of length 16 by an unitary matrix. The coefficients of the matrix are determined by how you connect the quantum gates or how your quantum system is made. The important part is that is unitary. Reversing the time means multiplying by the inverse. If you make no measurements, yo can go(simulate) back in time and return, and you get v_new = U U^-1 v so v_new = v Here we must assume that the matrix U and U^-1 are not very diagonal or has block, sparse, whatever. Let's assume that it is full or almost full and the values of the vector are mixed thoughtfully by the matrix multiplication. 3) After going back in time, the state is w = U^-1 v. Then they measure one qbit. Let´s assume that it is the last one. When you measure only one qbit, you can get a 0 or a 1. Here you have to distinguish the coefficients where the fourth qbit is 0 and the coefficients where the fourth qbit is 1. The probability of getting 0 is related to the values of the coefficients where the fourth qbit is 0 |w_0000|^2 + |w_0010|^2 + ... + |w_1100|^2 + |w_1110|^2 and the probability of getting 1 is |w_0001|^2 + |w_0011|^2 + ... + |w_1101|^2 + |w_1111|^2 They sum 1, so it's a probability. As a side effect of the measurement, if you get a 0, all the coefficients where the fourth qbit is 1 get erased and only survive the coefficients where the fourth qbit is 1. You have to multiply all of them so the sum is again 1. But the important part is that half of the information is destroyed. If the result of the measurement is 1, then the other half of the vector of length 16 is erased. The important part is that in either case, half of the vector is erased. You get a new vector w_half. :) 4) Now you return from your trip to the past. This is like multiplying the new vector by U. So the result is v_new = U w_half = U Msm w = U Msm U^-1 v where Msm is the measurement operation. Now v_new is not equal to the initial v, because the measurement in the past is messing the calculation. 5) Now they measure the value of the first qbit in the new state. It is like in 3. The probability of getting a 0 is |v_new_0000|^2 + |v_new_0001|^2 + ... + |v_new_0110|^2 + |v_new_0111|^2 and the probability of getting 1 is |v_new_1000|^2 + |v_new_1001|^2 + ... + |v_new_1110|^2 + |v_new_1111|^2 The result of the article is that in this last measurement there is a high probability of getting a 0. Note that if you apply this calculation to the initial states made in 1) you get a 100% probability of getting a 0. So the idea is that the measurement in the simulated past has not messed too much this value. The important part is you can't use any unitary matrix U, it must be not very sparse so U^-1 distribute the initial information in all the coefficients, and when half of them are erased, you don't loose too much information.