Off diagonal element of density matrix

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SUMMARY

The discussion focuses on the off-diagonal elements of the density matrix for a two-level quantum system, specifically defining the elements as \(\rho_{12} = |2\rangle\langle 1|\) and \(\rho_{21} = |1\rangle\langle 2|\). It elaborates on deriving the density operator from a pure state \(f = a|0\rangle + b|1\rangle\), resulting in an off-diagonal term \(ab^*|0\rangle\langle 1|\). Additionally, for a mixture of pure states \(f_1\) and \(f_2\) with probabilities \(p_1\) and \(p_2\), the off-diagonal contribution is expressed as \(p_1\langle f_1 |0\rangle\langle 1|f_1\rangle + p_2\langle f_2 |0\rangle\langle 1|f_2\rangle\). The calculations presented are accurate and confirm the properties of density matrices in quantum mechanics.

PREREQUISITES
  • Understanding of quantum mechanics, specifically two-level systems
  • Familiarity with density matrices and their properties
  • Knowledge of bra-ket notation and inner products
  • Basic concepts of quantum state mixtures and probabilities
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  • Study the derivation of density matrices in quantum mechanics
  • Learn about the implications of off-diagonal elements in quantum coherence
  • Explore the role of density matrices in mixed states and entanglement
  • Investigate applications of density matrices in quantum information theory
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Quantum physicists, researchers in quantum mechanics, and students studying quantum information theory will benefit from this discussion on density matrices and their off-diagonal elements.

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For two level system , let denotes the ground state as 1 and exctied state as 2, for writing the office off-diagonal matrix element for the density operator, shall it be

\rho_{12} = |2\rangle\langle 1|

and

\rho_{21} = |1\rangle\langle 2|

?
 
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From a pure state f = a|0> + b|1> one gets the density opérator
(a|0> + b|1>)(a*<0| + b*<1|)
with a off diagonal part ab*|0><1|
the off diagonal matrix element is ab* = <f |0><1|f> (with |0> and |1> orthogonal)
With a mixture of pure state f1,f2 with probabilities p1, p2
we get p1<f1 |0><1|f1> + p2<f2 |0><1|f2>
I hope there is no error in my answer!
 

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