Information content of a qubit

[Sunny Singh]

TL;DR

I have read in Neilson and chuang that the information represented by a qubit is infinite because the state vector takes values in a continuous space. I have a doubt regarding this statement.

I have just started reading Neilson and chuang's book on quantum computing and two times already have they said that when a qubit is not observed, it can contain infinite information.
"How much information is represented by a qubit? Paradoxically, there are an infinite number of points on the unit sphere, so that in principle one could store an entire text of Shakespeare in the binary expansion of theta"
And again when explaining quantum teleportation, "even if Alice did know the state, describing it precisely takes an infinite amount of classical information since ψ takes values in a continuous space"
Now, doesn't the complete description of the qubit depend only on the two coefficients of its two computational basis states? If Alice did know the state, doesn't it mean that she just knew what these two complex coefficients are? Why will she ever need infinite classical information to explain the state? I understand that we need an ensemble of identically created qubits to determine the coefficients but even if we don't know the exact state of the qubit, we know that there's only two unknown complex numbers. I don't understand what do the authors mean by a qubit having enough space to store an entire book only because those two complex coefficients takes values in a continuous space?

 

[vanhees71]

It depends a bit on how you define "information". The standard way is the Shannon-Jaynes-von Neumann entropy, which measures the "surprise" when measuring the spin (if the qbit is realized as a spin state of a spin-1/2 particle or the polarization state of a single photon). It's defined by (in natural units with $k_{\text{B}}=1$)
$$S[\rho]=-\mathrm{Tr}[\hat{\rho} \ln \hat{\rho}],$$
where $\hat{\rho}$ is the statistical operator describing the state the spin is prepared in.

If you don't know anything about the spin, you have to use the maximum-entropy principle to make a choice for $\hat{\rho}$, which corresponds to the least prejudice, i.e., you must make $\hat{\rho}$ such that the entropy gets maximal under the only constraint you have, i.e., that $\mathrm{Tr} \hat{\rho}=1$. Since any spin-measurement has 2 outcomes (say, you meausure $\sigma_1$, it can only take two values $\pm 1/2$). This implies that the maximum-entropy statistical operator is
$$\hat{\rho}=\frac{1}{2} (|1/2 \rangle \langle 1/2| + |-1/2 \rangle \langle -1/2|)=\frac{1}{2} \hat{1}.$$
The maximum entropy thus is
$$S[\rho]=\ln 2.$$
That's the maximum information you can get out of measuring a q-bit, i.e., measuring one spin component when you don't know nothing about the preparation of the system before measuring.

The point is that you cannot by a single spin measurement determine the full state $\hat{\rho}$. For that you need a preparation of many systems in the same state $\hat{\rho}$ and measure a certain set of incompatible spins to figure out the state. For a good discussion on "state determination", see Ballentine, Quantum Mechanics.

 

[PeroK]

There is an infinite amount of information in a single real number. Or, to put it another way: suppose I want to send you a book with up to a million pages (let's say up to a billion characters), then I can do that with 8 billion bits.

Every book, therefore, can be represented by a unique number between zero and one. Or, alternatively, an angle between $0$ and $2\pi$.

I could in principle send you a single picture of an arrow at some angle and that single angle would encode 8 billion bits, say. If you can measure the angle accurately enough - to 12 decimal places, or whatever.

A qubit has, theoretically, an infinite number of possible states. And, theoretically, each state could represent a different piece of data. And, just like the above, any finite amount of information could be encoded in a single qubit state.