Mixed versus Pure Quantum States for the Singlet

[knowwhatyoudontknow]

I have some basic questions about mixed states and entanglement.

1. Do mixed states always imply that the states are entangled and vice versa?

2. Can mixed states ever be separable?

3. Does interference have anything to do with entanglement?

In terms of Density Matrices, ρ = |ψ><ψ|:

4. The singlet can be written as the superposition of basis states |ψ> = 0|uu> + 1/√2|ud> - 1/√2|du> + 0|dd> to obtain ρ² = ρ and Tr(ρ²) = 1 indicating it is a pure state. On the other hand, it can be written as a statistical ensemble to produce a mixed state with ρ² ≠ ρ and Tr(ρ²) < 1. Is there any contradiction in this?

 

[Demystifier]

1. No.
2. Yes.
3. Yes.
4. Yes.

 

[vanhees71]

A quantum state is uniquely determined always by a statistical operator $\hat{\rho}$, i.e., a positive semidefinite self-adjoint operator with trace 1.

Ad 1.) I don't know, how you come to this conclusion. Maybe you have in mind is that if you have a quantum system consisting of two parts described by Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$, then the space of the total system is the direct product of these two Hilbert spaces $\mathcal{H}=\mathcal{H}_A \otimes \mathcal{H}_B$. Then an entangled state is any pure state for which $|\psi \rangle$ is a normalized vector in $\mathcal{H}$ that is not a product state. Then the reduced state of the systems are always mixed and not pure states, i.e., with $\hat{\rho}=|\psi \rangle \langle \psi|$
$$\hat{\rho}_A=\mathrm{Tr}_B \hat{\rho} = \sum_{\alpha,\alpha' \beta} |alpha \rangle \langle \alpha\beta|\hat{\rho}|\alpha',\beta \rangle \langle \alpha'|,$$
where $|\alpha \rangle$ and $|\beta \rangle$ denote arbitrary complete orthonormal states of $\mathcal{H}_A$ and $\mathcal{H}_B$, respectively and $|\alpha \beta \rangle \equiv |\alpha \rangle \otimes |\beta \rangle$. If $\hat{\rho}$ is an entangled state than the reduced states of the parts of the system, $\hat{\rho}_A=\mathrm{Tr}_B \hat{\rho}$ and $\hat{\rho}_B=\mathrm{Tr}_A \hat{\rho}$, are not pure.

Ad 2.) A (mixed or pure) state $\hat{\rho}$ is called separable if there are states $\hat{\rho}_{Aj}$, $\hat{\rho}_{B_j}$ and $P_j \geq 0$ such that
$$\hat{\rho}=\sum_j P_j \hat{\rho}_{Aj} \otimes \hat{\rho}_{Bj}.$$
Otherwise by definition $\hat{\rho}$ is called and entangled state.

Ad 3.) In the sense that "interference" has to do with superposition yes since for a pure state is entangled its state ket cannot be written as a product state, i.e., if it's a proper superposition of product states.

Ad 4.) doesn't make sense. A state $\hat{\rho}$ is a pure state if and only if it is a projection operator, fulfilling $\hat{\rho}^2=\hat{\rho}$. That's equivalent to $\mathrm{Tr} \hat{\rho}^2=1$. Any state with $\hat{\rho}^2 \neq \hat{\rho}$ is different from any pure state and called mixed. Then $\mathrm{Tr} \hat{\rho}^2<1$.

 

[knowwhatyoudontknow]

Let me focus on 4 in the hope it will clarify 1, 2 and 3. Math aside, it seems I can write the singlet as either a pure state as a result of superposition or a mixed state if I consider it as a statistical ensemble with P = 0.5. Also, if I assume that superposition gives me a pure state, the ρ matrix has off diagonal terms which are related to interference which seems to contradict the idea that the state is not entangled. Clearly, I am misunderstanding something. I came across the following statement in my reading "A pure entangled state of the whole system can be described by a wave function, but the states of its subsystems cannot be described by wave functions and are mixed states". Maybe my confusion arises from the belief that entangled states cannot be pure.

 

[vanhees71]

Again you have $\hat{\rho}=|\psi \rangle \langle \psi$. That's a pure state. Any mixed state by definition is different from this pure state. I still don't understand what you are after. Please write clearly which other than the pure state you want to compare with this pure state.

It's clear that the single-spin states are not pure. These are given by the corresponding density matrices as explained in my previous posting:
$$\hat{\rho}_A=\hat{\rho}_B=\frac{1}{2} |u \rangle \langle u| + \frac{1}{2} |d \rangle \langle d|=\frac{1}{2} \hat{1}.$$
The proof is very easy by applying the formula given in my previous posting. It's a good exercise to get familiar with product spaces and partial traces.

The fact that the pure state of the total system is entangled, by definition of entanglement is reflected by the fact that the parts of the system are both in non-pure mixed states. In this case they are even in mixed states of maximum entropy. That's why one calls the pure state of the total system a maximally entangled state or a Bell state (to honor John Bell for his groundbreaking work making EPRs philosophical gibberish a scientifically decidable question, leading to full confirmation of quantum theory in comparison to an alternative unrealistic kind of theories, which EPR ironically dubbed "local realistic theories").

 

[knowwhatyoudontknow]

OK. I think my problem is understanding that states can be pure-separable, pure-entangled, mixed-separable and mixed-entangled. If I treat the singlet as a pure state it cannot be factored and therefore is entangled. If I treat it as a 50%/50% ensemble it is a mixed state that is also entangled because ρ cannot be factored. ρ for the superposition has off-diagonal elements representing interference. ρ for the statistical ensemble has no off-diagonal elements. Am I getting closer?

 

[PeterDonis]

it seems I can write the singlet as either a pure state as a result of superposition or a mixed state if I consider it as a statistical ensemble with P = 0.5.

This doesn't make sense. The term "singlet" denotes a definite state: a pure state that is an eigenstate of total spin with eigenvalue zero, in which two identical quantum particles are entangled and neither one has a definite spin on its own. You can't "write" such a state to be a mixed state; you can't change what a state is by writing it down one way rather than another.

 

[knowwhatyoudontknow]

Yes. I agree. Bad choice of words on my behalf. Indeed, the mixed state is not a singlet state.

 

[vanhees71]

I think the misunderstanding is, what I tried to explain above. You have to distinguish between the state of the total system of 2 spins, living in the Hilbert space $\mathcal{H}=\mathcal{H}_A \otimes \mathcal{H}_B$ of the two spins (of particle $A$ and particle $B$, where for simplicity I assume distinguishable particles like an electron and a positron to avoid the somewhat more subtle case of Bose-Einstein/Fermi-Dirac (anti-)symmetrization) and the "reduced states" of the single spins.

The two spins are in a pure state, described by the statistical operator $\hat{\rho}_{AB}=|\Psi \rangle \langle \Psi|$ with
$$|\Psi \rangle=\frac{1}{\sqrt{2}}(|u d \rangle-|d u \rangle),$$
to take the $S=0$ (singlet) state (total spin 0). Here I used the usual short-hand notation
$$|\sigma_1 \sigma_2 \rangle=|\sigma_1 \rangle \otimes |\sigma_2 \rangle.$$
The single particle's spin states are described by the reduced statistical operators,
$$\hat{\rho}_A=\mathrm{Tr}_{B} \hat{\rho}_{AB}, \quad \hat{\rho}_B=\mathrm{Tr}_A \hat{\rho}_{AB}.$$
It's clear that $\hat{\rho}_A$ is "living" in $\mathcal{H}_A$ and $\hat{\rho}_B$ in $\mathcal{H}_B$. Using the formulae for the partial traces defined above, leads to
$$\hat{\rho}_A=\frac{1}{2} \hat{1}_{A}, \quad \hat{\rho}_B=\frac{1}{2} \hat{1}_B,$$
i.e., to maximum-entropy states of totally unpolarized particles. That's a characteristics of "maximally entangled pure states", aka "Bell states".

 

[PeroK]

Clearly, I am misunderstanding something. I came across the following statement in my reading "A pure entangled state of the whole system can be described by a wave function, but the states of its subsystems cannot be described by wave functions and are mixed states". Maybe my confusion arises from the belief that entangled states cannot be pure.

The singlet is an entangled state. You can describe this as a state where each particle does not have a definite (pure) state of its own. This is the simplest way to think about it.

However, if you are clever, you may describe the state of each particle as a mixed state that depends is written in terms of the mixed state of the other particle. This is what @vanhees71 has done here:

The single particle's spin states are described by the reduced statistical operators,
$$\hat{\rho}_A=\mathrm{Tr}_{B} \hat{\rho}_{AB}, \quad \hat{\rho}_B=\mathrm{Tr}_A \hat{\rho}_{AB}.$$

 

[gentzen]

However, if you are clever, you may describe the state of each particle as a mixed state that depends on the mixed state of the other particle.

The mixed state of each particle does not depend on the mixed state of the other particle. And using the reduced statistical operator is just as clever or dumb as using the marginal distribution in classical probability theory. In classical probability theory too, it would be misleading to claim that the marginal distribution of each particle would depend on the marginal distribution of the other particle.

 

[PeroK]

The mixed state of each particle does not depend on the mixed state of the other particle. And using the reduced statistical operator is just as clever or dumb as using the marginal distribution in classical probability theory. In classical probability theory too, it would be misleading to claim that the marginal distribution of each particle would depend on the marginal distribution of the other particle.

Depends was a poor choice of word, perhaps, given its statistical meaning. What you're saying is that all the fancy footwork in post #9 only describes a mixed state for each of $A$ and $B$ but doesn't retain the entanglement or correlation between the states? Hence, not clever at all?

 

[gentzen]

What you're saying is that all the fancy footwork in post #9 only ...? Hence, not clever at all?

I am saying that post #9 neither intents to be fancy nor clever, but just tries to clarify the misunderstanding by explicitly writing down everything.

 

[PeroK]

I am saying that post #9 neither intents to be fancy nor clever, but just tries to clarify the misunderstanding by explicitly writing down everything.

If we say that particle A has a given mixed state and particle B has a given mixed state, then where is the information about entanglement and correlation of measurements?

 

[gentzen]

If we say that particle A has a given mixed state and particle B has a given mixed state, then where is the information about entanglement and correlation of measurements?

The information about entanglement and correlation of measurements is still in the statistical operator $\hat{\rho}_{AB}=|\Psi \rangle \langle \Psi|$ with $|\Psi \rangle=\frac{1}{\sqrt{2}}(|u d \rangle-|d u \rangle),$ that describes particle A and particle B together. The mixed states of particle A and particle B alone no longer contain the full information.

 

[PeroK]

The information about entanglement and correlation of measurements is still in the statistical operator $\hat{\rho}_{AB}=|\Psi \rangle \langle \Psi|$ with $|\Psi \rangle=\frac{1}{\sqrt{2}}(|u d \rangle-|d u \rangle),$ that describes particle A and particle B together. The mixed states of particle A and particle B alone no longer contain the full information.

That clears that up, thanks.

 

[vanhees71]

The mixed state of each particle does not depend on the mixed state of the other particle. And using the reduced statistical operator is just as clever or dumb as using the marginal distribution in classical probability theory. In classical probability theory too, it would be misleading to claim that the marginal distribution of each particle would depend on the marginal distribution of the other particle.

Of course, that's why the reduced state of a subsystem is defined in the way it is!