Undergrad On uniqueness of density matrix description as mixed state

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The discussion centers on the uniqueness of the basis that diagonalizes a density matrix \rho. In a two-dimensional case, if the eigenvalues p and q are equal, \rho can be diagonalized in any basis; if they are unequal, a unique basis exists, subject to permutations and phase factors. For arbitrary dimensions, the condition for uniqueness is that the density matrix has no multiple eigenvalues. The conversation also clarifies that while entries of a density matrix are real-valued, they do not contain phase information, and changing the length of basis vectors does not affect linear independence. Ultimately, the trace of the density matrix remains a basis-independent concept.
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If you have a density matrix \rho, there is a basis |\psi_j\rangle such that
\rho is diagonal in that basis. What are the conditions on \rho such that the basis that diagonalizes it is unique?

It's easy enough to work out the answer in the simplest case, of a two-dimensional basis: Then \rho can be represented as a 2x2 matrix. It will have two eigenvalues, p and q. If p = q, then \rho is diagonal in every basis. If p \neq q, then there is a unique basis (up to permutations of elements and overall phase factor) that diagonalizes \rho.

Is there some result that is similar for bases of arbitrary dimension? What's the condition on \rho such that there is a unique way to diagonalize it?
 
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The basis is never unique as one can always change the length or the phase of the basis vectors. The necessary and sufficient condition for this being the only freedom is that the matrix has no multiple eigenvalues.
 
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A. Neumaier said:
The basis is never unique as one can always change the length or the phase of the basis vectors.
Aren't entries of density matrix real valued? Then there is no phase.
And what do you mean by changing length of basis vectors? If intensity in one output of PBS is twice the intensity of other output then it is so. You can not change that by mathematical manipulations.
 
zonde said:
Aren't entries of density matrix real valued?
No. Any positive semidefinite Hermitian matrix of trace 1 qualifies as a density matrix.

zonde said:
And what do you mean by changing length of basis vectors?
Changing the length of a basis vector preserves linear independence, hence we get another basis. This is a mathematical fact independent of physics. (Maybe stevendaryl intended to have an orthonormal basis - where lengths are fixed at 1 - but he didn't say so.)
 
A. Neumaier said:
No. Any positive semidefinite Hermitian matrix of trace 1 qualifies as a density matrix.
Okay. So let me restate my question:
Aren't entries of diagonal density matrix real valued?
A. Neumaier said:
This is a mathematical fact independent of physics.
If a mathematical fact disagrees with empirical fact then this particular mathematical fact does not describe particular physical situation and is irrelevant.
A. Neumaier said:
Changing the length of a basis vector preserves linear independence, hence we get another basis.
Does it preserve condition that the trace is fixed at 1?
 
zonde said:
Aren't entries of diagonal density matrix real valued?
Yes, but they do not contain the phase information.
zonde said:
Does it preserve condition that the trace is fixed at 1?
The trace is a basis-independent concept.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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