Can Bell's Theorem Disprove This Density Matrix Representation?

In summary, the conversation discusses the representation of joint probabilities in quantum mechanics and the statistical mix of pure states. Bell's theorem is not directly related to this representation, and any set of normalized kets can be used, not just a basis. The density matrix is a pure state, and Bayes's theorem can be used to determine the probability of measurement results.
  • #1
jk22
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With ##\rho=\sum_i p_i|\Psi_i\rangle\langle\Psi_i|##If the ##p_i=|\langle\Psi|\lambda_i\rangle|^2## are taken as joint probabilities given by quantum mechanics for the singlet state in EPRB then this cannot represent a statistical mix (classical) of those states because of Bell's theorem ?
 
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  • #2
I don't see what Bell's theorem has to do with it. You know the density matrix is a pure state since it represents the singlet state. Choosing a basis which obscures this fact doesn't mean that it ceases to be a pure state.
 
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  • #3
I also don't see what this has to do with Bell's theorem. You can use any set of normalized kets ##|\psi_i \rangle## (it doesn't need to be a basis nor need the vectors be orthogonal to each other). Now
$$\hat{\rho}=\sum_i p_i |\psi_i \rangle \langle \psi_i|$$
is a statistical operator, if it's self-adjoing (##\Rightarrow p_i \in \mathbb{R}##), positive semidefinite (##\Rightarrow p_i \geq 0##) and normalized, ##\mathrm{Tr} \hat{\rho}=1##. To see what this means for the ##p_i## just take an arbitrary orthonormal basis ##|u_n \rangle## to evaluate the trace:
$$\mathrm{Tr} \hat{\rho} = \sum_n \langle u_n |\hat{\rho} u_n \rangle = \sum_i p_i \sum_n \langle \psi_i|n \rangle \langle n|\psi_i \rangle = \sum_i p_i \langle \psi_i|\psi_i \rangle=\sum_i p_i \stackrel{!}{=}1.$$
The heuristic meaning of this is most simply seen by the following gedanken experiment. Assume that Alice can prepare a particle in the pure states defined by the ##|\psi_i \rangle##. Alice now sends Bob particles prepared in these states. For each particle Alice chooses randomly with probability ##p_i## to send Bob a particle prepared in state ##|\psi_i \rangle## at each time, and that's all Bob knows about this ensemble of particles sent to him. So he'll describe the statistics of this ensemble with the above given statistical operator. This you can also formally show by using the Shannon-Jaynes principle of maximum entropy.

More intuitively you can argue with Bayes's theorem. Say Bob makes a complete measurement on each of the particles Alice sends him. E.g., he measures a complete minimal set of compatible observables ##A_k## (##k \in \{1,\ldots,N \}##. Then the probability to get the measurement result ##(a_1,\ldots,a_N)## should be given by
$$P(a_1,\ldots,a_N)=\langle u_{a_1,\ldots,a_n}|\hat{\rho} u_{a_1,\ldots,a_n} \rangle, \qquad (*)$$
where ##|u_{a_1,\ldots,a_n} \rangle## are the uniquely (up to an irrelevant phase factor) defined simultaneous eigen states of the operators ##\hat{A}_k## with eigenvalues ##(a_1,\ldots,a_k)##.

To see this just consider what Alice does. She prepares particles in the states ##|u_i \rangle## with probability ##p_i##. Supposed a specific particle sent to bob is prepared in this state. Then the probability for measuring ##(a_1,\ldots,a_N)## for Bob's measurement is
$$P(a_1,\ldots,a_n|u_i)=|\langle u_{a_1,\ldots,a_N}|u_i \rangle|^2.$$
Now the probability that Alice sends a particle prepared in state ##u_i \rangle## is ##p_i##. So due to Bayes the probability to measure ##(a_1,\ldots,a_N)## for the ensemble of particles sent by Alice is
$$P(a_1,\ldots,a_n)=\sum_i p_i P(a_1,\ldots,a_n|u_i),$$
but this is indeed given by (*).
 
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  • #4
vanhees71 said:
You can use any set of normalized kets (it doesn't need to be a basis nor need the vectors be orthogonal to each other)
Yes, I should have been a little more careful with language and not used "basis."
 

Related to Can Bell's Theorem Disprove This Density Matrix Representation?

1. What is a density matrix?

A density matrix is a mathematical representation of the density operator, which describes the quantum state of a system. It is used to calculate the probability of finding a particle in a particular state, as well as to study the evolution of a system over time.

2. How is a density matrix different from a wave function?

A wave function is a mathematical function that describes the quantum state of a system at a single point in time. A density matrix, on the other hand, takes into account the entire history of a system and can provide information about the system's past and future states.

3. What is the physical significance of the diagonal elements in a density matrix?

The diagonal elements of a density matrix represent the probabilities of finding a particle in a particular state. This means that the square of the magnitude of the diagonal element is equal to the probability of finding the particle in that state.

4. How does the density matrix relate to the concept of entanglement?

The density matrix is essential for understanding and quantifying entanglement between particles. It allows us to calculate the amount of entanglement between two or more particles and study how it changes over time.

5. What are the applications of density matrix in quantum information theory?

The density matrix is a fundamental tool in quantum information theory, used in various applications such as quantum computing, quantum cryptography, and quantum communication. It is also used in studying quantum systems and understanding their behavior and properties.

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