What Conditions Must a Matrix Meet to Be a Density Matrix?

[natugnaro]

What are the conditions for some matrix to be a density matrix ?
I know of these conditions: 1.) \rho=\rho^{2}
2.) Tr(\rho)=1 (for pure state)

Is this all ?

 

[Fredrik]

See section 2.3 in Ballentine.

 

[arkajad]

\rho\geq 0,\; tr(\rho )=1.

 

[natugnaro]

ok, thanks.

As an example let's say I have a matrix 1/2((1 ,a),(a,1))
of single spin 1/2 particle at rest.
Then for this matrix to be a density matrix conditions (from Ballentine 2.10,2.11,2.12) give me that a=0,
and for this matrix to be a pure state matrix condition \rho^{2}=\rho gives me again that a=0.
Is this a right conclusion ?

sorry for latex problems.

 

[arkajad]

First of all your matrix must be nonnegative. For 2x2 matrices hermitian that is equivalent to:

1) trace nonnegative (=sum of eigenvalues)
2) determinant nonnegative (=product of eigenvalues)

Assuming a is real, your matrix is hermitian.

Trace of your matrix is 1 - that's good.
Determinant is (1-a^2)/4. So, you know that a must by <= 1.

Now, you notice the determinant =0 means that one of the eigenvalues is 0. Since trace is 1, the other must be 1. That looks like in that case we have a projection operator on a 1-dimensional subspace - a pure state.

That suggests that you have made some mistake when calculating using the condition for purity. Probably you have forgotten the factor 1/2 in front of the matrix. You did not square it or something like that. Do it again carefully.

 

[natugnaro]

Eigenvalues of my matrix are l1=(1-a)/2 and l2=(1+a)/2,

1) trace nonnegative (=sum of eigenvalues)
this reduces to (1-a)/2 + (1+a)/2 which gives 1>=0, this condition is met.

2) determinant nonnegative (=product of eigenvalues)
(1-a)/2*(1+a)/2>=0 gives a is in the range -1 to 1 (-1 and 1 included).

so for my matrix to be a density matrix a can be in the range -1 to 1, this meets conditions
2.10-2.12 from Ballentine(Fredrik), and conditions 1) & 2) arkajad as I can see.

For pure state matrix (yes I have forgotten the factor 1/2 ) a can be either 1 or -1, since in both cases condition for pure state matrix are met.
Is this now correct on the first sight ?

 

[arkajad]

Good. But remember, trace a determinant conditions are sufficient only for 2x2 matrices. In general, assuming that \rho is hermitian, to check that \rho\geq 0 you must check either that (\psi, \rho \psi)\geq 0 for all \psi or that all eigenvalues of \rho are nonnegative. Sometimes \rho may be given as
\rho=\sum_i A_i^\dag A_i - then it is automatically nonnegative. A necessary (but in general not sufficient) condition for a density matrix to describe a pure state is that its determinant is zero - because at least one of its eigenvalues must be zero.

 

[natugnaro]

one more thing, what are some applications of density matrices in practice ?
(I mean in experimental physics)