Holevo's Theorem (Holevo, 1973) - Quantum Information Capacity

A classical bit has 2 states: 0 and 1.

{0,1}\{0,1\}

A Qubit has infinite distinct states.

Example

We can let there be three states in spin-1/2.

∣z+⟩british coming by land\left\lvert z+ \right\rangle\quad\text{british coming by land} ∣z−⟩british coming by sea\left\lvert z- \right\rangle\quad\text{british coming by sea} ∣x+⟩british not coming\left\lvert x+ \right\rangle\quad\text{british not coming}

So i can measure SzS_z to be either +ℏ/2+\hbar/2 or −ℏ/2-\hbar/2. If it is +ℏ/2+\hbar/2, then the british are not coming by sea. If it is −ℏ/2-\hbar/2, then the british are not coming by land.

I can also measure SxS_x to be either +ℏ/2+\hbar/2 or −ℏ/2-\hbar/2. If it is +ℏ/2+\hbar/2, we don’t get any extra information as it could equally be either three. If it is −ℏ/2-\hbar/2, then the british are coming either by land or sea.

How reliably can we distinguish between a Set of quantum states that serve as signals?

Consider Alice and Bob. Alice sends quantum state to Bob and this state is living in some Hilbert space HH

∣ψ⟩∈Hdim(H)=d\left\lvert \psi \right\rangle\in H\quad dim(H)=d

Alice encodes NN equally likely messages

{∣α⟩}α=1N\{\left\lvert \alpha \right\rangle\}_{\alpha=1}^N

Bob makes measurements in some basis

{∣k⟩}k=1d′\{\left\lvert k \right\rangle\}_{k=1}^{d'}

Bob measures a larger Hilbert space H′H' because he doesn’t know the Hilbert space Alice is using.

H⊂H′dim(H′)=d′>dH\subset H'\quad dim(H')=d'>d

Bob still needs to make a determination. But he still needs to make a decision on which of the NN messages sent.

Let

k=1,2,3,4,⋯ ,d′d′>Nk=1,2,3,4,\cdots,d'\quad d'>N

for some of these outcomes you can get maybe you assign some like 1,21,2 to be message 1 C1C_1 and 33 to message 2 C2C_2, etc. then you need to cover all possibilities of the message Alice sends. You want to map the outcome to messages. Some of the bins may be empty.

The success probability for Bob to correctly distinguish the different messages.

Ps=∑α=1NP(α)∑k∈CαP(outcome k∣message α was sent)P_s=\sum_{\alpha=1}^NP(\alpha)\sum_{k\in C_\alpha}P(outcome~k|\text{message $\alpha$ was sent})\\

Recall Born Rule that

=∑α=1N1N∑k∈Cα∣⟨k∣α⟩∣2=\sum_{\alpha =1}^N\frac{1}{N}\sum_{k\in C_\alpha}|\left\langle k|\alpha \right\rangle|^2

We want to try to upperbound this. Let {∣ϕn⟩}r=1d\{\left\lvert \phi_n \right\rangle\}_{r=1}^d be some orthonormal basis of HH

We know that

Π∣α⟩=∣α⟩\Pi\left\lvert \alpha \right\rangle=\left\lvert \alpha \right\rangle

And also

(Π∣α⟩)†=∣α⟩†(\Pi\left\lvert \alpha \right\rangle)^\dagger=\left\lvert \alpha \right\rangle^\dagger ⟨α∣Π†=⟨α∣\left\langle \alpha \right\rvert\Pi^\dagger=\left\langle \alpha \right\rvert ⟨α∣Π=⟨α∣\left\langle \alpha \right\rvert\Pi=\left\langle \alpha \right\rvert

So

Ps=1N∑α∑k∈Cα⟨α∣k⟩⟨k∣α⟩P_s=\frac{1}{N}\sum_\alpha\sum_{k\in C_\alpha}\left\langle \alpha|k \right\rangle\left\langle k|\alpha \right\rangle\\ =1N∑α∑k∈Cα⟨α∣Π∣k⟩⟨k∣Π∣α⟩=\frac{1}{N}\sum_\alpha\sum_{k\in C_\alpha}\left\langle \alpha|\Pi|k \right\rangle\left\langle k|\Pi|\alpha \right\rangle\\ =1N∑α∑k∈Cα⟨α∣(Π∣k⟩⟨k∣Π)∣α⟩=\frac{1}{N}\sum_\alpha\sum_{k\in C_\alpha}\left\langle \alpha \right\rvert(\Pi\left\lvert k \right\rangle\left\langle k \right\rvert\Pi)\left\lvert \alpha \right\rangle =1N∑α∑k∈Cαtr(⟨α∣Π∣k⟩⟨k∣Π∣α⟩)= \frac{1}{N}\sum_\alpha \sum_{k\in C_\alpha}tr(\left\langle \alpha \right\rvert\Pi\left\lvert k \right\rangle\left\langle k \right\rvert\Pi\left\lvert \alpha \right\rangle)\\

Using trace cyclic property

=1N∑α∑k∈Cαtr(∣α⟩⟨α∣⋅Π∣k⟩⟨k∣Π)= \frac{1}{N}\sum_\alpha \sum_{k\in C_\alpha}tr(\left\lvert \alpha \right\rangle\left\langle \alpha \right\rvert\cdot \Pi\left\lvert k \right\rangle\left\langle k \right\rvert\Pi)\\ ≤1N∑α∑k∈Cαtr(Π∣k⟩⟨k∣Π)\leq \frac{1}{N}\sum_\alpha \sum_{k\in C_\alpha}tr(\Pi\left\lvert k \right\rangle\left\langle k \right\rvert\Pi)\\ ≤1N∑α∑k∈Cαtr(Π2⋅∣k⟩⟨k∣)\leq \frac{1}{N}\sum_\alpha \sum_{k\in C_\alpha}tr(\Pi^2\cdot\left\lvert k \right\rangle\left\langle k \right\rvert)\\ ≤1Ntr(Π2∑α∑k∈Cα∣k⟩⟨k∣)\leq \frac{1}{N}tr(\Pi^2\sum_\alpha\sum_{k\in C_\alpha}\left\lvert k \right\rangle\left\langle k \right\rvert)\\ ≤1Ntr(Π2I)\leq \frac{1}{N}tr(\Pi^2I)\\ ≤1Ntr(Π)\leq\frac{1}{N}tr(\Pi)\\

Using trace cyclic property and the fact that tr(⟨ϕr∣ϕr⟩)=⟨ϕr∣ϕr⟩tr(\left\langle \phi_r|\phi_r \right\rangle)=\left\langle \phi_r|\phi_r \right\rangle as it is a 1×11\times 1 matrix.

≤1Ntr(∑r=1d∣ϕr⟩⟨ϕr∣)\leq\frac{1}{N}tr(\sum_{r=1}^d\left\lvert \phi_r \right\rangle\left\langle \phi_r \right\rvert) ≤1N∑r=1dtr(⟨ϕr∣ϕr⟩)\leq\frac{1}{N}\sum_{r=1}^dtr(\left\langle \phi_r|\phi_r \right\rangle) ≤1N∑r=1d⟨ϕr∣ϕr⟩\leq\frac{1}{N}\sum_{r=1}^d\left\langle \phi_r|\phi_r \right\rangle ≤1N∑r=1d1\leq\frac{1}{N}\sum_{r=1}^d1 ≤dN\leq\frac{d}{N}

This means the probability of a successful decoding is at most the number of dimensions of the Hilbert space divided by the number of possible messages Alice can send