Decomposing a density matrix of a mixed ensemble

In summary: This is just a consequence of the Gram-Schmidt orthogonalization process. Any set of vectors can be transformed into an orthogonal set, and any linear combination of orthogonal vectors is a sum of projections onto those vectors. So by looking at the eigenvalue problem, we're just looking for a set of orthogonal eigenvectors with corresponding coefficients that add up to 1. This is a standard problem in linear algebra, and the solution is just a matter of finding the eigenvalues and eigenvectors of the matrix \rho.
  • #1
Gabriel Maia
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I'm trying to solve a problem where I am given a few matrices and asked to determine if they could be density matrices or not and if they are if they represent pure or mixed ensembles. In the case of mixed ensembles, I should find a decomposition in terms of a sum of pure ensembles. The matrix I'm having trouble with is this one

[tex] \rho = \left[\begin{array}{ccc}\frac{1}{2} & 0 & \frac{1}{4} \\ 0 & \frac{1}{4} & 0 \\ \frac{1}{4} & 0 & \frac{1}{4}\end{array}\right] [/tex]

I know it's a mixed ensemble density matrix because $Tr(\rho^2)<1$, but how can I decompose if I don't even know how big is this sum? I mean, any number of pure states may compose a mixed ensemble since they do not need to be orthogonal. How can I approach this?Thank you very much.
 
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  • #2
Gabriel Maia said:
I'm trying to solve a problem where I am given a few matrices and asked to determine if they could be density matrices or not and if they are if they represent pure or mixed ensembles. In the case of mixed ensembles, I should find a decomposition in terms of a sum of pure ensembles. The matrix I'm having trouble with is this one

[tex] \rho = \left[\begin{array}{ccc}\frac{1}{2} & 0 & \frac{1}{4} \\ 0 & \frac{1}{4} & 0 \\ \frac{1}{4} & 0 & \frac{1}{4}\end{array}\right] [/tex]

I know it's a mixed ensemble density matrix because $Tr(\rho^2)<1$, but how can I decompose if I don't even know how big is this sum? I mean, any number of pure states may compose a mixed ensemble since they do not need to be orthogonal. How can I approach this?Thank you very much.

Is this homework? If so, it should be in the homework forum.

What you're looking for is an orthonormal basis [itex]|\psi_j\rangle[/itex] such that [itex]\rho = \sum_j p_j |\psi_j\rangle \langle \psi_j|[/itex], where the [itex]p_j[/itex] are all real, and add up to 1. Now consider the product:

[itex]\rho |\psi_k\rangle = \sum_j p_j |\psi_j\rangle \langle \psi_j | \psi_k \rangle = p_k |\psi_k\rangle[/itex]

where for the last equality, we used that the kets [itex]|\psi_j\rangle[/itex] are orthonormal, meaning that [itex]\langle \psi_j | \psi_k \rangle = \delta_{jk}[/itex]

So what that means is that the basis vector [itex]|\psi_k\rangle[/itex] and the coefficient [itex]p_k[/itex] are solvable as an eigenvalue problem:

[itex]\rho |\psi_k\rangle = p_k |\psi_k\rangle[/itex]

Do you know how to find the eigenvalues and corresponding eigenvectors of a square matrix?
 
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  • #3
Gabriel Maia said:
I know it's a mixed ensemble density matrix because $Tr(\rho^2)<1$, but how can I decompose if I don't even know how big is this sum? I mean, any number of pure states may compose a mixed ensemble since they do not need to be orthogonal. How can I approach this?

The problem only asks you to find a sum, so any sum will do. Their purifications to a larger Hilbert space with fixed dimension will all be related by unitary matrices via the HJW theorem: https://arxiv.org/abs/quant-ph/0305068
 
  • #4
stevendaryl said:
Is this homework? If so, it should be in the homework forum.

What you're looking for is an orthonormal basis [itex]|\psi_j\rangle[/itex] such that [itex]\rho = \sum_j p_j |\psi_j\rangle \langle \psi_j|[/itex], where the [itex]p_j[/itex] are all real, and add up to 1. Now consider the product:

[itex]\rho |\psi_k\rangle = \sum_j p_j |\psi_j\rangle \langle \psi_j | \psi_k \rangle = p_k |\psi_k\rangle[/itex]

where for the last equality, we used that the kets [itex]|\psi_j\rangle[/itex] are orthonormal, meaning that [itex]\langle \psi_j | \psi_k \rangle = \delta_{jk}[/itex]

So what that means is that the basis vector [itex]|\psi_k\rangle[/itex] and the coefficient [itex]p_k[/itex] are solvable as an eigenvalue problem:

[itex]\rho |\psi_k\rangle = p_k |\psi_k\rangle[/itex]

Do you know how to find the eigenvalues and corresponding eigenvectors of a square matrix?

By considering that [itex] \langle \psi_j|\psi_{k} \rangle = \delta_{j\,k}[/itex] aren't we assuming that all the states that form the density matrix are orthogonal?
 
  • #5
Gabriel Maia said:
By considering that [itex] \langle \psi_j|\psi_{k} \rangle = \delta_{j\,k}[/itex] aren't we assuming that all the states that form the density matrix are orthogonal?

Well, yes. But if there is any solution, there is a solution of that form.
 

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