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suma

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Is it not possible to deduce quantum states from a density matrix?

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- Thread starter suma
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In summary: A state vector can only represent a pure state, while a density matrix can represent both pure and mixed states.

- #1

suma

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Is it not possible to deduce quantum states from a density matrix?

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- #2

ShayanJ

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- #3

suma

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but if (a1 a2)'*(a1 a2) = (1 0; 0 0) and in general case there is no solution to find a1 and a2, is this correct?

thanks

- #4

naima

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suma said:

but if (a1 a2)'*(a1 a2) = (1 0; 0 0) and in general case there is no solution to find a1 and a2, is this correct?

thanks

Is (1 0; 0 0) a notation for the density matrix in a v1 v2 basis?

if yes this density matrix is |v1> < v1|

- #5

suma

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naima said:Is (1 0; 0 0) a notation for the density matrix in a v1 v2 basis?

if yes this density matrix is |v1> < v1|

hi,

this was just an example, the main question is whether states can be derived from density matrix

thanks

- #6

naima

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a pure state |v> in a 2 dimensional system has |v><v| for density matrix (1 0;0 0) but all density matrix are not like that.

take (1/3 0; 0 2/3) it is the density matrix 1/3 |v1><v1| + 2/3 |v2><v2| = 1/3 (1 0;0 0) + 2/3 (0 0; 0 1).

It is to be used when somebody gives you apure state v1 (or v2) with 1/3 (2/3) probability.

- #7

bhobba

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suma said:Is it not possible to deduce quantum states from a density matrix?

I am not sure what you mean.

But a quantum state, by definition, is a positive operator of unit trace which can represented as a matrix called the density matrix.

You can however find more detail such as what mixed and pure states are in Ballentine - Quantum Mechanics - A Modern Development - Chapter 2.

Its use is the Born Rule which says given an observable O these exists a positive operator of unit trace such that the expected outcome of O, E(O) = Trace (PO). By definition P is called the state of the system. Ballentine develops QM from just two axioms - the Born Rule is the second - the first is associated with any observation is a Hermitian operator whose eigenvalues are the possible outcomes.

To some extent Born's Rule is implied by the first axiom via Gleason's Theorem:

http://kof.physto.se/cond_mat_page/theses/helena-master.pdf

Thanks

Bill

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- #8

Zarqon

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suma said:hi,

this was just an example, the main question is whether states can be derived from density matrix

thanks

In general, a state vector cannot be deduced from a density matrix. The reason was already mentioned, it is that the density matrix is more general than the state vector. One example is mixed states, which can be described in the density matrix notation, but not as a state vector.

A density matrix is a mathematical representation of a quantum system that takes into account both the pure and mixed states of the system. It is a matrix of numbers that describes the probability of the system being in a particular state.

The density matrix is used to calculate the properties of a quantum system, such as the expectation value of observables and the evolution of the system over time. It is also used to study the entanglement and coherence of quantum states.

Yes, it is possible to deduce quantum states from a density matrix. This process is known as quantum state tomography and involves reconstructing the quantum state based on measurements of the system.

One of the main challenges is that there are an infinite number of possible quantum states that could produce the same density matrix. This makes the reconstruction process complex and requires a large number of measurements to accurately deduce the state.

Deducing quantum states from a density matrix has applications in quantum information processing, quantum communication, and quantum computing. It can also be used for characterizing and diagnosing quantum systems, as well as studying the dynamics of complex quantum systems.

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