# Quantum indistinguishability using density operators

There is something that bugs me concerning the use of density matrices. For instance, to argue that quantum teleportation does not spread an information faster than light, Nielsen and Chuang state the following:

Measuring in Alice's computational basis, the state of the system after the measurement is:

$$|00\rangle[\alpha|0\rangle+\beta|1\rangle]$$ with probability $$\frac14$$

[...]

$$|11\rangle[\alpha|1\rangle-\beta|0\rangle]$$ with probability $$\frac14$$

They then proceed to compute $$\rho^{AB}$$ and trace out Alice's system to find $$\rho^B=\frac12I$$, indicating that Bob has no information whatsoever on the state until Alice gives him the results she got with her measurement.

I struggle to understand how to compute the density matrix here, or more exactly why "the system is in state x with probability y" allows to compute it. Once Alice measures the state, she knows the state Bob's system is in. I reckon that Bob doesn't, but what about the following scenario?

Let us call the previous Alice and Bob Alice1 and Bob1 and the following ones Alice2 and Bob2.

Alice2 prepares a $$n$$-qubit basis state $$|x\rangle$$. She sends this register to Bob2, who measures it (to learn $$|x\rangle$$ with probability 1 since $$|x\rangle$$ is a basis state) creates a $$n$$-qubit basis state $$|h\rangle$$ chosen uniformly at random, CNOTs it with Alice2's register and sends back to Alice2 her register, without telling her what $$|h\rangle$$ is. Hence, Bob2 knows that the total system is in state $$|x\oplus h\rangle|h\rangle$$. In particular, these registers being not entangled, Bob2 knows exactly the state Alice2's system is in, just like Alice1 knew Bob1's state. However, from the point of view of Alice2, the system can be in any basis state with uniform probability. Hence (this is where, I think, the error is), her density matrix is identical to the state of a uniform superposition of basis states, that is: $$\rho^{A_2}=\sum_i|i\rangle\langle i|$$. There are now two contradictory things that come to my mind:

1. Since Alice2's density matrix is the same as if she created the state $$\mathbf{H}|0\rangle$$, applying $$\mathbf{H}$$ and measure should return $$|0\rangle$$ with probability 1. However, Bob2 knows that it will give $$|0\rangle$$ with probability $$2^{-n}$$. Hence, it is high likely that Alice won't obtain what she expects, leading to an inconsistency.
2. When carrying out the computation, here's what I got. We have (up to a normalisation constant): $$\mathbf{H}=\sum_{a,b}(-1)^{a\cdot b}|a\rangle\langle b|\,.$$ Hence, after having applied $$\mathbf{H}$$, the density matrix becomes: $$\mathbf{H}\rho^{A_2}\mathbf{H}^\dagger=\sum_{a,b,x,y,i}(-1)^{a\cdot b}(-1)^{x\cdot y}|a\rangle\langle b|i\rangle\langle i|x\rangle\langle y|$$ which can be reduced to: $$\mathbf{H}\rho^{A_2}\mathbf{H}^\dagger=\sum_{a,y,i}(-1)^{(a\oplus y)\cdot i}|a\rangle\langle y|$$ which can be reduced to: $$\mathbf{H}\rho^{A_2}\mathbf{H}^\dagger=\sum_{a}|a\rangle\langle a|\,.$$ hence Alice, will measure $$|0\rangle$$ with probability $$2^{-n}$$ which is consistent with Bob's point of view. However, this raises the inverse problem: why, if the density matrix is equal to the one of $$\mathbf{H}|0\rangle$$, don't we get $$\mathbf{H}\rho^{A_2}\,\mathbf{H}^\dagger=|0\rangle\langle0|$$? (This question also applies generally, since, I think, I started with the density matrix corresponding to $$\mathbf{H}|0\rangle$$, I expected the result to be $$|0\rangle\langle0|$$)

I think these inconsistencies comes from a misunderstanding of mine at some point of the density matrix properties, but I can't see at which point is my reasoning flawed (though I have a guess, see in the text).

• Your point: "Since Alice2's density matrix is the same as if she created the state 𝐇|0⟩, applying 𝐇 and measure should return |0⟩ with probability 1." (or the working out just above) is where you make the mistake. Take everything to be qubits and $|x\rangle = |0\rangle$ for simplicity. Then Bob prepares $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ and sends the first qubit to Alice. Tracing out Bob's system we find $\rho_{A_2} = I/2$ (which is a uniform mixture not a uniform superposition of basis states). So $\rho_{A_2} = I/2 \neq H|0\rangle$. Mar 25 at 12:25
• For some extra clarity (again using qubits for simplicity). A uniform superposition (in the computational basis and up to phases) is a state $|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$, in density matrix form this is $|\psi\rangle \langle \psi| = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}$. On the other hand a uniform mixture (operationally equivalent to uniformly choosing one of the basis states to prepare -- then forgetting the preparation) is given by the maximally mixed state $\rho = \begin{pmatrix} 1/2 &0 \\0&1/2\end{pmatrix}$. Mar 25 at 12:34
• Oh indeed, I didn't take care of the entanglement! Thanks! And your second comment also answers my second question. Would you mind putting it as an answer so that I can accept it? Mar 25 at 12:39

The mistake occurs when you compute the reduced state on Alice2's system. For simplicity we'll assume all systems are qubits and in addition $$|x\rangle = |0\rangle$$. In this setting, Bob2 prepares the state $$|\phi\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11 \rangle)$$ and sends the first qubit to Alice. Tracing out the system of Bob2 we find $$\rho_{A_2} = \mathrm{Tr}_{B_2}[|\phi\rangle \langle \phi|] = I/2$$ where $$I$$ is the identity matrix. Thus $$\rho_{A_2} = I/2 \neq H|0\rangle$$, Alice2's reduced state is a uniform mixture of the basis states $$\{|0\rangle \langle 0 |, |1\rangle \langle 1 |\}$$ and not a uniform superposition of $$\{|0\rangle , |1 \rangle \}$$.
To give an example of the difference between uniform superpositions and uniform mixtures, a uniform superposition is by nature a pure state, e.g. for a $$d$$-dimensional system we have (up to phases) $$|\psi\rangle = \sum_i \frac{1}{\sqrt{d}} |i \rangle$$. On the other hand a uniform mixture, which must be done in the density matrix formalism, is given by $$\rho = \sum_i \frac{1}{d} |i \rangle \langle i|$$. Operationally, you can think about a uniform mixture as resulting from choosing a basis state uniformly at random, preparing that state and then forgetting the outcome. Moreover, we see that $$|\psi \rangle \langle \psi | = \sum_{i,j} \frac{1}{d} |i \rangle \langle j| \neq \sum_i \frac{1}{d} |i \rangle \langle i | = \rho,$$ (well unless $$d=1$$, but that's a bit boring).