Density Matrix (pure state) Property

[Toten]

Hello.

I need some help to prove the first property of the density matrix for a pure state.
According to this property, the density matrix is definite positive (or semi-definite positive). I've been trying to prove it mathematically, but I can't.

I need to prove that |a|^2 x |c|^2 + |b|^2 x |d|^2 + ab*c*d + a*bcd* >= 0,
where a, b, c, and d are complex numbers, and * means the complex conjugate.

My Density Matrix M (2x2) has the following elements:
m11 = |a|^2, m12 = ab*, m21 = a*b, m22 = |b|^2

And (c; d) is just a ket in the C2 space.

I hope you can help me. Thanks you!

 

[Toten]

Even if the density matrix is defined for a ensemble of spinors, the property holds true. How can you prove it?

 

[arkajad]

Probably you have that $|a|^2+|b|^2=1$, because trace should be 1.

You matrix is Hermitian. Trace is the sum of eigenvalues, determinant is the product of eigenvalues. The determinant of your matrix is zero. Therefore one of the eigenvalues is zero, the other must be 1. Therefore your matrix is semi-positive definite.

 

[Toten]

Probably you have that $|a|^2+|b|^2=1$, because trace should be 1.

You matrix is Hermitian. Trace is the sum of eigenvalues, determinant is the product of eigenvalues. The determinant of your matrix is zero. Therefore one of the eigenvalues is zero, the other must be 1. Therefore your matrix is semi-positive definite.

Thank you! I hadn't realize that the determinant of the density matrix is zero. This gave me an idea on how to reorganize the terms in the following expression:

|a|^2 x |c|^2 + |b|^2 x |d|^2 + ab*c*d + a*bcd*
= aa*cc* + bb*dd* + ab*c*d + a*bcd*
= ac*a*c + bd*b*d + ac*b*d + bd*a*c
= ac*(a*c + b*d) + bd*(b*d + a*c)
= ac*(a*c + b*d) + bd*(a*c + b*d)
= (ac* + bd*)(a*c + b*d)
= (ac* + bd*)(ac* + bd*)*
= |ac* + bd*|^2, which is always greater than or equal to zero.

I don't know if this is the easiest way to prove it, but it is definitely one way.

Thanks for your help!