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I am going through this video of Quantum Computing for Computer Scientists. I am not able to understand the entanglement state of qubits.

Is entanglement just an operation, or is it a state which can be stored. If it's a state which is stored then how can we send two qubits far apart? But if it is just an operation then isn't the operation somehow affected the probability from 50% to 100% to superimposed states of qubits.

Also, what if the operation has 50% chance of assigning the probability. So when measuring multiple times we tend to get 50% probability of qubits collapsing to one of the states.

user1271772 No more free time
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rusty
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3 Answers3

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It’s a property of the state that is stored. But it’s not a physical change that relates to a force: it is not that you entangle qubits, and that means that they are physically joined. The two physical qubits remain as independent physical entities that can move around separately, just with a joint state. (At least, not usually. In things like covalent bonds, that’s exactly what’s happening!)

Incidentally, try not to talk too much about “probability” as it can be very misleading in a quantum scenario, especially when you start talking about entanglement. Talk about probability amplitudes where possible, and keep probabilities for being associated with the outcomes of measurements. Otherwise the differences between entanglement and classical correlation will be impossible to distinguish.

DaftWullie
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  • Okay, So it seems it's a information exchange between two qubits. Am I correct in thinking this way?

    According to you, it looks like a frequentists approach to probability, Maybe cause we only have experiments to rely on. Is there a probability theory for quantum which is well accepted?

     – rusty Aug 01 '18 at 16:46
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    @Rusty Well, there is some sort of information exchange going on, but it would be more accurate to talk about correlation. If you talk about communication, you can easily mislead yourself into thinking it can help you communicate faster than the speed of light (it can't). But to demonstrate that there is some communication, think about the process of dense coding: initially, an entangled state is shared. Later, that lets Alice send one qubit to Bob, and that achieves an equivalent of 2 bits of communication. – DaftWullie Aug 02 '18 at 07:31
  • @Rusty I don't think I'm exactly extolling a frequentist approach. You have to be aware that the nature of probability in quantum is quite different to classical. Classically, there is an absolute answer: you've rolled a die and it's definitely in some state, you just don't know what it is. Before you roll, if you knew exactly the state of the whole Universe, you could predict with certainty what the outcome would be. Probabilities reflect that lack of knowledge. In quantum, if you know the probability amplitudes, you know everything that can possibly be known about the state.... – DaftWullie Aug 02 '18 at 07:34
  • ... but you still can't predict all the outcomes. To reiterate, probability amplitudes are the fundamental thing, and probabilities are derived from them (with a corresponding loss of information). There is definitely a frequentist element appearing in the probabilities. But, probabilities are also used to represent subjective states of knowledge. For example, when talking about the part of a state that Bob holds in a 2-qubit entangled state, he has one description (as a density matrix), but Alice can have another description depending on what she's done to her part (measured, got some result) – DaftWullie Aug 02 '18 at 07:38
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entanglement is neither an operation nor a state, it is a concept describing a particular family of states. Here is a short explanation of this concept.

Let's limit ourselves to 2 qubits stored on 2 registers A and B :

  • case 1 : you prepare independently the qubit A in some state (a1|0> + a2|1>), then you do the same with the qubit B (b1|0> + b2|1>). In that case the state of your total system is described as the concatenation of the state of the system A and the state of the system B. This case is intuitive, there is no entanglement between the registers A and B -> we call it a separable state. We can write the state of the system as a1*b1|00> + a1*b2|01> + a2*b1|10> + a2*b2|11> (tensor product of states on A and B)

In quantum mechanics, there exists other states for the total system AB that cannot be described as the concatenation of the states of the subsystems A and B. This phenomenon has no equivalent in classical mechanics and is not very intuitive. This leads us to :

  • case 2 : you start with some state k1|0> + k2|1> on the register A and |0> on register B, then you apply a quantum operator known as CNOT that will "expand" the contents of the two superposition terms of register A to register B. The resulting state k1|00> + k2|11> on system AB cannot be written as a product (tensor product) of two separate states on A and B. We call it an entangled state.

One particularity of the entangled states is how it behave wrt measurement : in an entangled state, the outcome of a measurement in one of the subsystems can influence the outcome of a measurement on the other. For example with the state in case 2, you have :

  1. 50% chance of measuring 1 on register A
  2. 50% chance of measuring 1 on register B
  3. but if you have already measured the register A and you have gotten 1, then you have a 100% chance of measuring 1 on register B

This last item "3" would not be true if the state wasn't entangled, in which case the outcome of the measurement on register A couldn't have influenced the probabilities of the outcome on register B.

SilentMan
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To add on to what @DaftWullie said, in my mind it might make more sense to think of entanglement as a property of a multi-qubit state, instead of as a state itself. There is no special "Entangled" state, it's just that some states are entangled and some are not.

Dripto Debroy
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