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Stephen Tashi

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A quantum system goes from an uncertain to a certain state upon measurement.This indicates a decrease of entropy

From the perspective of the single variable being measured, the prior and posterior distributions can have a different entropy, but what variables are involved in your definition of the entropy of a "quantum system"? Are you only considering a single variable?

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The answer to both questions is - no. This is consistent with the 2nd law of thermodynamics, provided that you only compare one

Above you compared a high entropy state

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$$S=-\mathrm{Tr} \hat{\rho} \ln \hat{\rho},$$

where ##\hat{\rho}## is the statistical operator of the system. For a pure state, where ##\hat{\rho}=|\psi \rangle \langle \psi |## you have ##S=0##.

Now if you make a measurement, it depends on what happens to the system. If you make an ideal and complete von Neumann filter measurement and only keep states of a particular outcome you have prepared the system in a new pure state, and the entropy doesn't change. If you make an ideal incomplete ideal filter measurement the entropy increases, because then you get a non-pure state.

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Strilanc

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Actually, collapse causes an increase in entropy. Quantum mechanics without collapse is deterministic: if you know the input state you know the output state. With collapse the output is probabilistic: single input state is mapped into multiple possible output states. The entropy of the output is higher than the entropy of the input.

When you learn the result of a measurement, it doesn't fix this problem. The entropy of the universe has still gone up. It's just that the mutual information between you and the system also went up. It's important to keep those two concepts (mutual information vs entropy of a process) separate in your mind.

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The entropy of the system being measured has gone down--the information content of the qubit has reduced.By mutual information,do you mean the knowledge of the state of the system gained by the observer?Actually, collapse causes an increase in entropy. Quantum mechanics without collapse is deterministic: if you know the input state you know the output state. With collapse the output is probabilistic: single input state is mapped into multiple possible output states. The entropy of the output is higher than the entropy of the input.

When you learn the result of a measurement, it doesn't fix this problem. The entropy of the universe has still gone up. It's just that the mutual information between you and the system also went up. It's important to keep those two concepts (mutual information vs entropy of a process) separate in your mind.

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As long as I am talking about the entropy of the universe, why can't I compare the (total)entropy before measurement to that after measurement?The answer to both questions is - no. This is consistent with the 2nd law of thermodynamics, provided that you only compare onemeasuredstate with anothermeasuredstate.

Above you compared a high entropy statebeforemeasurement with the final low entropymeasuredstate. Those states belong to different categories (the second is measured while the first is not), so they should not be compared in a verification of the 2nd law.

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The precise answer depends on the choice of interpretation of quantum mechanics. An answer that does not depend much on interpretation would be that the state before measurement represents possibilities, while the measured state represents the actual reality.As long as I am talking about the entropy of the universe, why can't I compare the (total)entropy before measurement to that after measurement?

In fact, there is a simple classical analogue. Consider a coin flip experiment. Before you look at the coin, you don't know whether it is heads or tails. This uncertainty corresponds to entropy ##{\rm ln}2##. But when you look at it then you know whether it is heads or tail, which reduces entropy to ##{\rm ln}1=0##.

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Just a question--what is the von Neumann entropy of a qubit?

$$S=-\mathrm{Tr} \hat{\rho} \ln \hat{\rho},$$

where ##\hat{\rho}## is the statistical operator of the system. For a pure state, where ##\hat{\rho}=|\psi \rangle \langle \psi |## you have ##S=0##.

Now if you make a measurement, it depends on what happens to the system. If you make an ideal and complete von Neumann filter measurement and only keep states of a particular outcome you have prepared the system in a new pure state, and the entropy doesn't change. If you make an ideal incomplete ideal filter measurement the entropy increases, because then you get a non-pure state.

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From the moment a coin is tossed to the point it falls on the ground, it obeys well defined Newton's laws and is in a definite orientation at all times.So, there is no uncertainty in its state.Besides, when it falls, work is done by the ground to stop its motion and bring about its orientation to up or down!The precise answer depends on the choice of interpretation of quantum mechanics. An answer that does not depend much on interpretation would be that the state before measurement represents possibilities, while the measured state represents the actual reality.

In fact, there is a simple classical analogue. Consider a coin flip experiment. Before you look at the coin, you don't know whether it is heads or tails. This uncertainty corresponds to entropy ##{\rm ln}2##. But when you look at it then you know whether it is heads or tail, which reduces entropy to ##{\rm ln}1=0##.

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There is no uncertainty inSo, there is no uncertainty in its state.

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Two things--1. We can talk about the entropy of a system before measurement and after measurement.There is no uncertainty initsstate, but there is uncertainty inyour knowledgeof its state. Some interpretations of QM say that quantum uncertainty is also uncertainty of your knowledge. Other interpretations deny it, but as I said, a detailed answer in the quantum case depends on the choice of interpretation.

2. The entropy reduces even for a classical system upon measurement--now is there any work done?

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Sure, but we cannot1. We can talk about the entropy of a system before measurement and after measurement.

No work is done. A work is associated with2. The entropy reduces even for a classical system upon measurement--now is there any work done?

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When you said ln2 before coin toss, we did quantify entropy before measurement.Sure, but we cannotmeasureentropy of a system before measurement. You have probably heard that measurement in QM plays a much more important role than measurement in classical mechanics.

So, non-thermal entropy can be reduced upon measurement.Thermal entropy(associated with phase space only) can not be reduced without work being done--right?No work is done. A work is associated withthermalentropy, while here we are talking about a reduction of anon-thermalentropy.

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Quantification is not the same as measurement.When you said ln2 before coin toss, we did quantify entropy before measurement.

It's true for some kinds of non-thermal entropies, not for any non-thermal entropy.So, non-thermal entropy can be reduced upon measurement.

Not quite. Are you familiar with the following thermodynamic relations?Thermal entropy(associated with phase space only) can not be reduced without work being done--right?

$$dU=dQ-PdV$$

$$dS=\frac{dQ}{T}, \;\; dW=PdV$$

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Please explain further.Not quite. Are you familiar with the following thermodynamic relations?

$$dU=dQ-PdV$$

$$dS=\frac{dQ}{T}, \;\; dW=PdV$$

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##dW## is the work, ##dS## is the change of entropy. Taking ##dW=0## gives ##dU=TdS##, meaning that entropy can change without work.Please explain further.

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This is just the first law of thermodynamics--it doesn't prohibit unrestrained conversion of internal energy to dQ(which equals TdS only for an isothermal process).##dW## is the work, ##dS## is the change of entropy. Taking ##dW=0## gives ##dU=TdS##, meaning that entropy can change without work.

Coming back to my original question, because of $$\delta p$$ and $$\delta q$$, a quantum system's phase space should be bigger than that of a classical system--so does it have a higher entropy?

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The set of all quantum states is the Hilbert space, not the phase space. So one cannot conclude that it has a higher entropy.Coming back to my original question, because of ##\delta p## and ##\delta q##, a quantum system's phase space should be bigger than that of a classical system--so does it have a higher entropy?

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We agreed that the thermal entropy is related to the phase space only.A quantum system is not without a phase space--so I am asking if a quantum system occupies a bigger phase space and has a higher thermal entropy?The set of all quantum states is the Hilbert space, not the phase space. So one cannot conclude that it has a higher entropy.

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I didn't say that I agree with that. But I didn't explicitly complain about your statement because it didn't seem essential at that time.We agreed that the thermal entropy is related to the phase space only.

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Logically correct, but it doesn't answer my question.I didn't say that I agree with that. But I didn't explicitly complain about your statement because it didn't seem essential at that time.

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As far as I know, there is no simple general answer. Quantum entropy is sometimes bigger and sometimes smaller than its classical counterpart.Logically correct, but it doesn't answer my question.

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The von Neumann entropy refers to state of a quantum system. So the correct question to ask is what's the Neumann entropy for a given state of a qubit. A qubit is a system with a two-dimensional Hilbert space of states. The most general state is given by a statistical operator, and we can work in a given basis (e.g., if we consider the spin of a spin-1/2 particle the standard eigenbasis ##|+1/2 \rangle, \quad -1/2,\rangle## of the operator ##\hat{\sigma}_3##). The statistical operator is given as a positive semidefinite hermitean matrix ##\rho_{ij}## with trace 1, and the entropy is defined asJust a question--what is the von Neumann entropy of a qubit?

$$S=-k_{\text{B}} \mathrm{Tr} \hat{\rho} \ln \hat{\rho}.$$

Now you can always diagonalize the matrix representing the statistical operator, then it's juse given by the corresponding matrix elements ##\rho_{ij}'=\mathrm{diag}(p,1-p)##, and the (basis independent!) entropy is given by

$$S=-k_{\text{B}} [p ln p + (1-p) \ln(1-p)].$$

Of course ##p \in [0,1]## and by definition for ##p=0## or ##p=1## you have to interpret ##x \ln x## for ##x \rightarrow 0## to be ##0##.

You can prove that you have maximal information about the system if and only ##p=0## or ##p=1##, i.e., if the qubit is prepared in a pure state. The entropy becomes maximal for ##p=1/2##, as is expected for thermal equilibrium.

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