Density Matrix (pure state) Property

Toten
Messages
3
Reaction score
0
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!
 
Physics news on Phys.org
Even if the density matrix is defined for a ensemble of spinors, the property holds true. How can you prove it?
 
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.
 
arkajad said:
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!
 
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...

Similar threads

Replies
21
Views
3K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
17
Views
4K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
6
Views
3K
Back
Top