In this paper, pages 6-9, the author discusses mixed states for two entangled qubits. After solving for Alice's density matrix, the author keeps referring to Alice's qubit state as a mixed state. But, I believe Alice's, or even Bob's, individual qubit states are pure in their own respective basis, and the only mixed state is when we consider the quantum state of the two qubits together. I present the following argument.

Per my current understanding, a mixed state and a pure state satisfy the following:

  1. the state $|\Psi\rangle = |\psi\rangle_1\otimes|\psi\rangle_2\otimes...\otimes|\psi\rangle_n$ is a mixed state and lives in a Hilbert space $\mathcal{H}=\mathcal{H_1}\otimes\mathcal{H_2}\otimes...\otimes\mathcal{H_n}$. In other words, mixed states only exist when talking about multiple Hilbert spaces, and the notion of mixed states for one given Hilbert space $\mathcal{H}_i$ does not make sense.
  2. the density matrix of a pure state, denoted as $\hat{\rho}$, satisfies $\hat{\rho}^2=\hat{\rho}$.

Question 1: Are these criteria correct? In particular, I'm interested to gauge the truth of condition (1).

Assuming my stated criteria are correct, I continue:

On page 7, the author calculates the density matrix of Alice's qubit as $$\hat{\rho}_A=\sum_{i_1 i_2j}a_{i1,j}a_{i2,j}^{\dagger}|i_1\rangle \langle i_2|. \tag{1}$$

Let us calculate $\hat{\rho}_A^2$ (I'm going to use different variables so it's easier to follow the math, and also let $a_{ij}=a_ib_j$):

$$\begin{align} \hat{\rho}_A^2&=\left(\sum_{ikj}a_ia_k^{\dagger} b_j b_j^{\dagger}|i\rangle \langle k|\right)\left(\sum_{mnp}a_ma_n^{\dagger} b_p b_p^{\dagger}|m\rangle \langle n|\right) \\ &= \sum_{ikjmnp} a_ia_k^{\dagger} b_j b_j^{\dagger}a_ma_n^{\dagger} b_p b_p^{\dagger}\ |i\rangle \langle k|m\rangle \langle n|\\ &=\sum_{ikjnp} a_ia_k^{\dagger} b_j b_j^{\dagger}a_ka_n^{\dagger} b_p b_p^{\dagger}\ |i\rangle \langle n|\\ &=\sum_{inp} a_ia_n^{\dagger} b_p b_p^{\dagger}\ |i\rangle \langle n| \sum_{kj}a_ka_k^{\dagger} b_j b_j^{\dagger} \tag{2}\\ &=\sum_{inp} a_ia_n^{\dagger} b_p b_p^{\dagger}\ |i\rangle \langle n|\\ &=\sum_{i_1i_2p} a_{i1,p}a_{i2,p}^{\dagger} |i_1\rangle \langle i_2| \tag{3}\\ &=\hat{\rho}_A \end{align}$$

In Eq (2), $\sum_{kj}a_ka_k^{\dagger} b_j b_j^{\dagger}=1$ because our mixed state of 2-qubit system is written as $|\Psi\rangle = \sum_{kj}a_kb_j\ |k\rangle \otimes |j\rangle$, and sum of probabilities is one.

In Eq (3), I have renamed indices to match Eq (1).

Alright, so now I have shown (hopefully with no math/physical errors) that the quantum state of Alice's qubit must be a pure state. Additionally, this would make sense because her qubit's state belongs to one Hilbert space so there's no mixing. Moreover, her qubit's state can be written as $$|\psi\rangle_A = \frac{1}{\sqrt{2}}\left[|\uparrow\rangle_A + |\downarrow\rangle_A\right], \tag{4}$$ which is a pure state representation.

Question 2: Have I made a mistake somewhere? If not, why does the author refer to Alice's quantum state as a mixed state?

I think there is a big hole in my understanding of mixed/pure states, because everywhere I search, it seems that entangled qubits are mixed even for Alice and Bob. If this is so, I'd appreciate any overall clarifications that could help me better understand this topic.


No, the concept of a mixed state is different from what you have. It is not a state composed of states from different Hilbert spaces. A state that one can write as a tensor product of pure states (subsystems) is still a pure state. One can show that by using the $\rho^2=\rho$ test. (Each subsystem will only contract on subsystems from the same Hilbert space.)

The derivation that you give is not valid, because of the assumption you make that one can write $a_{ij}=a_i b_j$. By making that assumption, you basically assume that the state is pure. In general one cannot make that assumption. So, a mixed state is given by $$ \rho=\sum_i |\psi_i\rangle P_i \langle \psi_i| , $$ where $P_i$ represents probabilities, so that $$ \sum_i P_i = 1 . $$

Another way to look at it is to write down the state in its most general form $$ \rho=\sum_{ij} |\phi_i\rangle a_{ij} \langle \phi_j| . $$ Here, $a_{ij}$ represents a matrix (the density matrix in case $|\phi_i\rangle$ represents some orthogonal basis). One can diagonalize this matrix (using singular value decomposition). It then reproduces the previous form of the density operator. The diagonal elements then represent the probabilities. If the state is pure, there would be only one nonzero diagonal element, which would be equal to 1. In that case $a_{ij}=u_i u_j^{\dagger}$ so that $$ |\psi_1\rangle = \sum_i |\phi_i\rangle u_i . $$

Mixed states often arise when a pure state interacts with an environment. Say we start with a pure state $$ |\psi\rangle = \frac{1}{\sqrt{2}}( |\phi_1\rangle + |\phi_2\rangle) . $$ Due to some interaction with an environment it becomes entangled with the environment $$ |\psi'\rangle = \frac{1}{\sqrt{2}}( |\phi_1\rangle|e_1\rangle + |\phi_2\rangle|e_2\rangle) . $$ However, since we do not observe the effect in the environment, we need to trace out these environmental degrees of freedom. So the result is a mixed state $$ \rho=\text{tr}_E\{|\psi'\rangle\langle\psi'|\} = \frac{1}{2}( |\phi_1\rangle\langle\phi_1| + |\phi_2\rangle\langle\phi_2|) . $$ This is only a rather simple example, but hopefully it explains the mechanism.

As for a reference, a very good book is: Quantum Computation and Quantum Information, by Isaac Chuang and Michael Nielsen. Although it seems to be about quantum computing, it covers much the basics.

  • $\begingroup$ Thank you for catching my mistake. I see now how assuming $a_{ij}=a_ib_j$ would lead to a pure state, as we've seen above already. Given my notion of mixed states was wrong, I guess now I should ask: what is a mixed state, and in what systems do they arise? Also, if you have good references, I'd appreciate you linking them. I googled this and I mostly found discussions from a quantum computing perspective from computer science departments, which usually had sloppy notations and were not as in depth in the theory as one would expect in a physics text. $\endgroup$ – Ptheguy Sep 24 at 17:08
  • $\begingroup$ Thanks. I added a link to the wikipedia page, gave an explanation, and added a reference. $\endgroup$ – flippiefanus Sep 25 at 3:53

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