Complex coefficents in density operator expansion?

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SUMMARY

The discussion centers on the formulation of a density operator for a quantum state represented as |\psi\rangle = \frac{1}{\sqrt{2}} ( |+\rangle + i |-\rangle ). The correct density operator is derived as \rho(t) = \frac{1}{2} ( |+\rangle \langle + | + |- \rangle \langle - | + i(|-\rangle \langle + | - |+\rangle \langle -|), which includes complex coefficients. The proposed solution from the exam, lacking the complex terms, is incorrect as it fails to maintain the hermitian property of the density matrix.

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Hey, I recently had an exam where the quantum state were on the form

[tex]|\psi\rangle = \frac{1}{\sqrt{2}} ( |+\rangle + i |-\rangle )[/tex]

Here I formed the density operator for the pure state

[tex]\rho(t) = |\psi\rangle \langle \psi| = \frac{1}{2} ( |+\rangle + i |-\rangle )( \langle +| - i \langle -| ) = \frac{1}{2} ( |+\rangle \langle + | + |- \rangle \langle - | + i(|-\rangle \langle + | - |+\rangle \langle -|)).[/tex]

However in the solutions for the exam the complex i's were not there, i.e the solutions states that

[tex]\rho = \frac{1}{2} ( |+\rangle \langle + | + |- \rangle \langle - | + |-\rangle \langle +| - |+\rangle \langle - |).[/tex]

Have I missed something here or is the suggested solution erroneous? Is there a reason why a density operator expansion should not have complex coefficients?
 
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Your solution seems correct - I've got i's on anti-diagonal as well. The second expression cannot be right, because the density matrix has to be hermitian.
 

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