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I'm trying to understand the difference between the cloning of a quantum state (which is not allowed in QM) and a measurement. I have a specific example in mind, and am trying to make sense of it in terms of entropy and information transfer.

Example (Based on Schrodinger's Cat):

Consider two spin-entangled electrons in which one is know to have spin up and the other has spin down. A detector is set up which will measure the spin of the particle which is then connected to a sealed, evacuated box with a vial of argon atoms. The vial has a volume V1 and the box has a volume of V2, and there are N argon atoms in the vial. If a spin up particle is detected, the detector loosens the rubber cork on the vial, allowing the argon atoms to escape into the box. If a spin down particle is detected, nothing happens.

Analysis based on entropy change:

The von Neumann entropy of the electron is initially S_e_init is ln(2). The initial entropy of the detector, vial and gas is S_init. After the electron strikes the detector, and is measured, it has an entropy of zero (it's spin has been measured - and we are only counting the spin entropy). The entropy of the detector, vial and gas is S_final, which will vary depending on the measured spin of the electron.

So, the entropy change of the electron is ln(2), and the entropy change of the detector is unknown, but I imagine it could be made quite small, as well as that of the rubber cork in the vial (perhaps a different mechanism could be used), but we can at least define the entropy change of the argon gas, which will be (on average):

0.5*N*k*ln(V2/V1)

Now, if a similar measurement is performed on the other entangled electron, the same result will be obtained, except that if one electron is spin up, the other must be spin down, so one vial will have retained the argon, and in the other it will have leaked out via free-expansion into the box. In any case, the total entropy change of the argon in both places would be:

N*k*ln(V2/V1)

Suppose that V2/V1=2. Then the average entropy change in the argon gas would be N*k*ln(2), and the entropy change of the electron would be k*ln(2), so to perform a single measurement, there is a thermodynamic cost of (N-1)*k*ln(2). This seems pretty costly for measuring a single particle, but is it accurate? The states of both pieces of measuring equipment are theoretically entangled with the electron after measurement, and they are entangled with each other, because if the argon gas has expanded in one, it has not done so in the other. Is it correct to compute the classical entropy as done above in this case, or is there more to it? If it is correct, how small can the entropy change of the measuring device be made?

If both of the measuring systems are entangled after the measurement, aren't all of the argon atoms entangled then also? If so, then wouldn't this be the same as cloning the state of the electron's spin into each argon atom? I must be making some mistakes in my logic here, because it just doesn't seem to work out right. Can someone tell me where I went wrong?

Example (Based on Schrodinger's Cat):

Consider two spin-entangled electrons in which one is know to have spin up and the other has spin down. A detector is set up which will measure the spin of the particle which is then connected to a sealed, evacuated box with a vial of argon atoms. The vial has a volume V1 and the box has a volume of V2, and there are N argon atoms in the vial. If a spin up particle is detected, the detector loosens the rubber cork on the vial, allowing the argon atoms to escape into the box. If a spin down particle is detected, nothing happens.

Analysis based on entropy change:

The von Neumann entropy of the electron is initially S_e_init is ln(2). The initial entropy of the detector, vial and gas is S_init. After the electron strikes the detector, and is measured, it has an entropy of zero (it's spin has been measured - and we are only counting the spin entropy). The entropy of the detector, vial and gas is S_final, which will vary depending on the measured spin of the electron.

So, the entropy change of the electron is ln(2), and the entropy change of the detector is unknown, but I imagine it could be made quite small, as well as that of the rubber cork in the vial (perhaps a different mechanism could be used), but we can at least define the entropy change of the argon gas, which will be (on average):

0.5*N*k*ln(V2/V1)

Now, if a similar measurement is performed on the other entangled electron, the same result will be obtained, except that if one electron is spin up, the other must be spin down, so one vial will have retained the argon, and in the other it will have leaked out via free-expansion into the box. In any case, the total entropy change of the argon in both places would be:

N*k*ln(V2/V1)

Suppose that V2/V1=2. Then the average entropy change in the argon gas would be N*k*ln(2), and the entropy change of the electron would be k*ln(2), so to perform a single measurement, there is a thermodynamic cost of (N-1)*k*ln(2). This seems pretty costly for measuring a single particle, but is it accurate? The states of both pieces of measuring equipment are theoretically entangled with the electron after measurement, and they are entangled with each other, because if the argon gas has expanded in one, it has not done so in the other. Is it correct to compute the classical entropy as done above in this case, or is there more to it? If it is correct, how small can the entropy change of the measuring device be made?

If both of the measuring systems are entangled after the measurement, aren't all of the argon atoms entangled then also? If so, then wouldn't this be the same as cloning the state of the electron's spin into each argon atom? I must be making some mistakes in my logic here, because it just doesn't seem to work out right. Can someone tell me where I went wrong?

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