<p>Yes.  If the trace distance of two density matrices is equal to 1, then they have orthogonal supports, and therefore they are simultaneously diagonalizable.</p>

CC BY-SA 2.5
2010-09-16T08:00:31.157Z
I guess the answer is yes, but I don't know the proof.
Jeremy Yan
CC BY-SA 2.5
2010-09-16T08:13:07.517Z
The main idea of the proof which establishes two density matrices are totally distinguishable when the trace distance is one, is diagonalizing the *difference* of the two density matrices; but how to prove the very same basis diagonalizes the two density matrices themselves? Maybe these two density matrices are not diagonal with respect to this basis, but their difference is. Can anyone give some proof idea, or give some references to the proof? Thank you.
Jeremy Yan
CC BY-SA 2.5
2010-09-16T04:57:04.867Z
2010-09-16T04:57:04.867Z
Tsuyoshi Ito
<p>Here is one way to prove the fact you are interested in.</p>

<p>Suppose $\rho_0$ and $\rho_1$ are density matrices.  Like every other Hermitian matrix, it is possible to express the difference $\rho_0-\rho_1$ as
$$\rho_0-\rho_1 = P_0-P_1$$ for $P_0$ and $P_1$ being positive semidefinite and having orthogonal images.  (Sometimes this is called a Jordan-Hahn decomposition; it is unique and easily obtained from a spectral decomposition of $\rho_0-\rho_1$.) Note that the fact that $P_0$ and $P_1$ have orthogonal images implies that they are simultaneously diagonalizable, which I interpret is the property you are interested in.</p>

<p>The trace norm of the difference $\rho_0-\rho_1$ (as you define it, with the multiplicative factor 1/2), is given by $$\|\rho_0-\rho_1\|_{\text{tr}} = \frac{1}{2}\operatorname{Tr}(P_0) + \frac{1}{2}\operatorname{Tr}(P_1).$$  Under the assumption that this quantity is 1, we will conclude that $P_0=\rho_0$ and $P_1=\rho_1$, which proves what you want to prove.</p>

<p>To draw this conclusion, note first that $\operatorname{Tr}(P_0)-\operatorname{Tr}(P_1)=0$ and $\operatorname{Tr}(P_0)+\operatorname{Tr}(P_1)=2$, so $\operatorname{Tr}(P_0)=\operatorname{Tr}(P_1)=1$.  Next, take $\Pi_0$ and $\Pi_1$ to be the orthogonal projections onto the images of $P_0$ and $P_1$, respectively.  We have $$\Pi_0 (\rho_0 - \rho_1) = \Pi_0 (P_0 - P_1) = P_0$$ so $$\operatorname{Tr}(\Pi_0 \rho_0) - \operatorname{Tr}(\Pi_0 \rho_1) = 1.$$ Both $\operatorname{Tr}(\Pi_0 \rho_0)$ and $\operatorname{Tr}(\Pi_0 \rho_1)$ must be contained in the interval [0,1], from which we conclude that $\operatorname{Tr}(\Pi_0\rho_0)=1$ and $\operatorname{Tr}(\Pi_0\rho_1) = 0$.  From these equations it is not difficult to conclude $\Pi_0\rho_0=\rho_0$ and $\Pi_0\rho_1=0$, and therefore $P_0=\rho_0$ by the equation above.  A similar argument shows $P_1=\rho_1$.</p>

CC BY-SA 2.5
2010-09-18T02:56:51.900Z
Thank you, Prof. Watrous. Actually, I learn all these trace norm and density matrices stuff from your lecture notes.
Jeremy Yan
CC BY-SA 2.5
2010-09-18T02:59:10.227Z
I'd like to add that all stuffs discussed in this post can be found in Professor Watours' on-line lecture notes (lecture 3): http://www.cs.uwaterloo.ca/~watrous/quant-info/
Jeremy Yan
CC BY-SA 2.5
2010-09-16T16:01:14.363Z
2010-09-16T16:01:14.363Z
John Watrous
<p>I believe the answer to this question is well-known; but, unfortunately, I don't know.</p>

<p>In quantum computing, we know that mixed states are represented by density matrices. And the trace norm of the difference of two density matrices characterizes the distinguishability of the two corresponding mixed states. Here, the definition of <em>trace norm</em> is the sum of all eigenvalues of the density matrix, with an extra multiplicative factor <strong>1/2</strong> (in accordance with statistical difference of two distributions). It is well-known that, when the the difference of two density matrices is one, then the corresponding two mixed states are totally distinguishable, while when the difference is zero, the two mixed states are totally indistinguishable.</p>

<p>My question is, does the trace norm of the difference of two density matrices being one imply these two density matrices can be simultaneously diagonalizable? If this is the case, then taking the optimal measurement to distinguish these two mixed states will behave like to distinguish two distributions over the same domain with <em>disjoint</em> support.</p>

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2010-09-16T04:05:41.210Z
Could you define what a density matrix is ? is it just a positive definite matrix ?
Suresh Venkat
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2010-09-16T04:13:34.977Z
@Suresh: A density matrix is a hermitian, positive semidefinite matrix whose trace is equal to 1.
Tsuyoshi Ito
CC BY-SA 2.5
2010-09-16T04:17:41.143Z
The answer to the question is yes, because the trace distance being 1 implies that the two density matrices have orthogonal supports.
Tsuyoshi Ito
CC BY-SA 2.5
2010-09-16T04:45:47.430Z
@Tsuyoshi: Maybe you should write that comment as an answer?
Robin Kothari
CC BY-SA 2.5
2010-09-16T04:58:15.710Z
@Robin: Sure, done.
Tsuyoshi Ito
<quantum-computing><quantum-information>
Does the trace norm of the difference of two density matrices being one imply these two density matrices can be simultaneously diagonalizable?
Jeremy Yan
# Does the trace norm of the difference of two density matrices being one imply these two density matrices can be simultaneously diagonalizable?

(Asked by: username_0 on 2010-09-16)

I believe the answer to this question is well-known; but, unfortunately, I don't know.

In quantum computing, we know that mixed states are represented by density matrices. And the trace norm of the difference of two density matrices characterizes the distinguishability of the two corresponding mixed states. Here, the definition of *trace norm* is the sum of all eigenvalues of the density matrix, with an extra multiplicative factor **1/2** (in accordance with statistical difference of two distributions). It is well-known that, when the the difference of two density matrices is one, then the corresponding two mixed states are totally distinguishable, while when the difference is zero, the two mixed states are totally indistinguishable.

My question is, does the trace norm of the difference of two density matrices being one imply these two density matrices can be simultaneously diagonalizable? If this is the case, then taking the optimal measurement to distinguish these two mixed states will behave like to distinguish two distributions over the same domain with *disjoint* support.

**Tags:** quantum-computing, quantum-information
> **Comments:**
> **username_1**: Could you define what a density matrix is ? is it just a positive definite matrix ?  
> **username_2**: @Suresh: A density matrix is a hermitian, positive semidefinite matrix whose trace is equal to 1.  
> **username_2**: The answer to the question is yes, because the trace distance being 1 implies that the two density matrices have orthogonal supports.  
> **username_3**: @Tsuyoshi: Maybe you should write that comment as an answer?  
> **username_2**: @Robin: Sure, done.  

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## 2 answers

### Answer by username_4 on 2010-09-16. Score: 23 | Accepted answer

Here is one way to prove the fact you are interested in.

Suppose $\rho\_0$ and $\rho\_1$ are density matrices. Like every other Hermitian matrix, it is possible to express the difference $\rho\_0-\rho\_1$ as
$$\rho\_0-\rho\_1 = P\_0-P\_1$$ for $P\_0$ and $P\_1$ being positive semidefinite and having orthogonal images. (Sometimes this is called a Jordan-Hahn decomposition; it is unique and easily obtained from a spectral decomposition of $\rho\_0-\rho\_1$.) Note that the fact that $P\_0$ and $P\_1$ have orthogonal images implies that they are simultaneously diagonalizable, which I interpret is the property you are interested in.

The trace norm of the difference $\rho\_0-\rho\_1$ (as you define it, with the multiplicative factor 1/2), is given by $$\|\rho\_0-\rho\_1\|\_{\text{tr}} = \frac{1}{2}\operatorname{Tr}(P\_0) + \frac{1}{2}\operatorname{Tr}(P\_1).$$ Under the assumption that this quantity is 1, we will conclude that $P\_0=\rho\_0$ and $P\_1=\rho\_1$, which proves what you want to prove.

To draw this conclusion, note first that $\operatorname{Tr}(P\_0)-\operatorname{Tr}(P\_1)=0$ and $\operatorname{Tr}(P\_0)+\operatorname{Tr}(P\_1)=2$, so $\operatorname{Tr}(P\_0)=\operatorname{Tr}(P\_1)=1$. Next, take $\Pi\_0$ and $\Pi\_1$ to be the orthogonal projections onto the images of $P\_0$ and $P\_1$, respectively. We have $$\Pi\_0 (\rho\_0 - \rho\_1) = \Pi\_0 (P\_0 - P\_1) = P\_0$$ so $$\operatorname{Tr}(\Pi\_0 \rho\_0) - \operatorname{Tr}(\Pi\_0 \rho\_1) = 1.$$ Both $\operatorname{Tr}(\Pi\_0 \rho\_0)$ and $\operatorname{Tr}(\Pi\_0 \rho\_1)$ must be contained in the interval [0,1], from which we conclude that $\operatorname{Tr}(\Pi\_0\rho\_0)=1$ and $\operatorname{Tr}(\Pi\_0\rho\_1) = 0$. From these equations it is not difficult to conclude $\Pi\_0\rho\_0=\rho\_0$ and $\Pi\_0\rho\_1=0$, and therefore $P\_0=\rho\_0$ by the equation above. A similar argument shows $P\_1=\rho\_1$.

> **Comments:**
> **username_0**: Thank you, Prof. Watrous. Actually, I learn all these trace norm and density matrices stuff from your lecture notes.  
> **username_0**: I'd like to add that all stuffs discussed in this post can be found in Professor Watours' on-line lecture notes (lecture 3): http://www.cs.uwaterloo.ca/~watrous/quant-info/  

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### Answer by username_2 on 2010-09-16. Score: 10

Yes. If the trace distance of two density matrices is equal to 1, then they have orthogonal supports, and therefore they are simultaneously diagonalizable.

> **Comments:**
> **username_0**: I guess the answer is yes, but I don't know the proof.  
> **username_0**: The main idea of the proof which establishes two density matrices are totally distinguishable when the trace distance is one, is diagonalizing the *difference* of the two density matrices; but how to prove the very same basis diagonalizes the two density matrices themselves? Maybe these two density matrices are not diagonal with respect to this basis, but their difference is. Can anyone give some proof idea, or give some references to the proof? Thank you.  

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