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It is known that for a classical system the amount of information needed to store its state is the same as the amount of information that can be stored in that system. This amount is equal to Shannon's entropy or its classical generalizations.

On the other hand the amount of information needed to describe the state of a quantum system is greater than the amount of information that can be stored in such quantum system.

My impression is that to describe the state of a quantum system of $N$ qubits one needs $2^N$ bits of information, on the other hand one can only store $N$ bits of information inside such system, which is in my impression equal to von Neumann's entropy of the system.

Thus I have two possible explanations of such discrepancy.

a) The true entropy of a quantum system is exactly von Neumann's entropy. The difference is only due the fact that a classical storage can only independently operate with digits not less than one bit, thus to store, say 4 variables of 1/4 bit each a classical system needs 4 bits (at least one bit for each variable) while a quantum system can use parts of a digit independently and thus store all 4 variables in one binary digit.

b) The true entropy of a quantum system of $N$ qubits is exactly $2^N$ bits but the amount of extractable information is only $N$ bits.

Which explanation do you consider more correct or are there possible other explanations?

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1 Answer 1

The entropy of $N$ qubits is always $N\log(2)$. That's pretty much by definition. One must carefully refine the definition in quantum mechanics. When $N$ qubits are measured, what we get is the same amount of information as in the case when we measure $N$ classical bits.

On the other hand, when we're preparing a state of $N$ qubits, we may choose $2^N$ complex coefficients of the individual $2^N$ states. Well, only $2^N - 1$ of these complex numbers are physical because one (real) overall normalization has to be fixed to be one and one (real) overall angle (phase) has no physical impact.

However, there is no sense in which even a conceptually misleading calculation could produce $2^N$ bits from $N$ qubits. You may have incorrectly assumed that all the amplitudes in front of the $2^N$ basis vectors of the Hilbert space for $N$ qubits have to be either $0$ or $1$. But it's not the case: they're arbitrary complex numbers - only constrained by the overall normalization of the state.

So to precisely specify the state, you need to choose a point in ${\mathbb{CP}}^{2^N - 1}$. That's a continuous space so "classically" you need an infinite amount of information to specify a point in it. However, there's no physical sense in which the state actually carries an infinite amount of information.

Everything that can be measured on one copy of such a system only reveals the information of $N$ bits. Their being quantum means that the initial state may be prepared in continuously many ways but this "information" we put into the preparation of the initial state can't be gotten back. Being quantum also means that the measurement may in principle measure the qubits with respect to a generally rotated basis. And the operations of the quantum computer may be arbitrary unitary transformations, too.

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One minor point: $N$ bits (or $N\log{2}$ nats if you prefer) is an upper bound on the entropy of $N$ qubits, but it is only the case that the von Neumann entropy of an $N$-level system is $N$ when that system is in a completely mixed state. When describing the number of bits which may be encoded (and subsequently extracted) from a system, the measure you want to use is the log of the dimension of the system, not the entropy. A Bell state has 4 orthogonal vectors on which 2 bits may be encoded, but the entropy of a particular Bell state is zero. –  John Schanck Jun 10 '11 at 13:40
@John Schanck I think that $N$ bits are equivalent to $N / \log 2$ nats. –  mmc Jun 10 '11 at 22:47
So it seems that Luboš supports answer 1 and John supports answer 2? –  Anixx Jun 13 '11 at 16:47
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