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Figure 102 Measurement of a superposition has the effect of forcing the system to decide on a particular state, with probabilities determined by the amplitudes
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second excited state, and so on So we could consider a k-level system consisting of the ground state and the rst k 1 excited states, and we could denote these by 0 , 1 , 2 , , k 1 The superposition principle would then say that the general quantum state of the system is 0 0 + 1 1 + + k 1 k 1 , where k 1 | j |2 = 1 Measuring the state of the system j=0 would now reveal a number between 0 and k 1, and outcome j would occur with probability | j |2 As before, the measurement would disturb the system, and the new state would actually become j or the jth excited state How do we encode n bits of information We could choose k = 2 n levels of the hydrogen atom But a more promising option is to use n qubits Let us start by considering the case of two qubits, that is, the state of the electrons of two hydrogen atoms Since each electron can be in either the ground or excited state, in classical physics the two electrons have a total of four possible states 00, 01, 10, or 11 and are therefore suitable for storing 2 bits of information But in quantum physics, the superposition principle tells us that the quantum state of the two electrons is a linear combination of the four classical states, = 00 00 + 01 01 + 10 10 + 11 11 , normalized so that x {0,1}2 | x |2 = 12 Measuring the state of the system now reveals 2 bits of information, and the probability of outcome x {0, 1} 2 is | x |2 Moreover, as before, if the outcome of measurement is jk, then the new state of the system is jk : if jk = 10, for example, then the rst electron is in the excited state and the second electron is in the ground state An interesting question comes up here: what if we make a partial measurement For instance, if we measure just the rst qubit, what is the probability that the outcome is 0 This is simple It is exactly the same as it would have been had we measured both qubits, namely,
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Recall that {0, 1}2 denotes the set consisting of the four 2-bit binary strings and in general {0, 1}n denotes the set of all n-bit binary strings
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Suppose we have two qubits, the rst in the state 0 0 + 1 1 and the second in the state 0 0 + 1 1 What is the joint state of the two qubits The answer is, the (tensor) product of the two: 0 0 00 + 0 1 01 + 1 0 10 + 1 1 11 Given an arbitrary state of two qubits, can we specify the state of each individual qubit in this way No, in general the two qubits are entangled and cannot be decomposed into the 1 1 states of the individual qubits For example, consider the state = 2 00 + 2 11 , which is one of the famous Bell states It cannot be decomposed into states of the two individual qubits (see Exercise 101) Entanglement is one of the most mysterious aspects of quantum mechanics and is ultimately the source of the power of quantum computation Pr {1st bit = 0} = Pr {00} + Pr {01} = | 00 | 2 + | 01 | 2 Fine, but how much does this partial measurement disturb the state of the system The answer is elegant If the outcome of measuring the rst qubit is 0, then the new superposition is obtained by crossing out all terms of that are inconsistent with this outcome (that is, whose rst bit is 1) Of course the sum of the squares of the amplitudes is no longer 1, so we must renormalize In our example, this new state would be 00 01 new = 00 + 01 | 00 | 2 + | 01 | 2 | 00 | 2 + | 01 | 2
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Finally, let us consider the general case of n hydrogen atoms Think of n as a fairly small number of atoms, say n = 500 Classically the states of the 500 electrons could be used to store 500 bits of information in the obvious way But the quantum state of the 500 qubits is a linear superposition of all 2500 possible classical states: x x
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