# Trace of density matrix square greater than 1?

I learned that if you have a density matrix $$\rho$$ then

• $$\mathrm{Tr}(\rho^2)=1 \Rightarrow$$ pure state

• $$\mathrm{Tr}\rho^2<1 \Rightarrow$$ mixed state

Can one have $$\mathrm{Tr}(\rho^2)>1$$? For example in a two state system: $$\rho²=\begin{pmatrix}\rho_{11}&\rho_{12}\\\rho_{21}&\rho_{22}\end{pmatrix}^2=\begin{pmatrix}\rho_{11}^2+\rho_{12}\rho_{21}&\rho_{11}\rho_{12}+\rho_{12}\rho_{22}\\\rho_{21}\rho_{11}+\rho_{22}\rho_{21}&\rho_{21}\rho_{12}+\rho_{22}^2\end{pmatrix}$$ from that follows $$\mathrm{Tr}(\rho^2)=\rho_{11}^2+2\rho_{12}\rho_{21}+\rho_{22}^2$$ which properties of the density matrix elements limit this expression to a maximum of 1? I know that the diagonal elements ($$\rho_{11}$$, $$\rho_{22}$$) represent probabilities, so they are limited. But I dont know about a limitation for the off diagonal elements.

• No, you should have learned that tr$\{\rho\}=1$ for all states due to normalization. For pure states tr$\{\rho^2\}=1$ and for mixed states tr$\{\rho^2\}<1$. The situation tr$\{\rho^2\}>1$ is then not possible. Commented Sep 28, 2022 at 8:45
• It is a simple property of any set of positive values that add up to one that their squares would add up to a value smaller than one. Commented Sep 28, 2022 at 8:59
• Note that $\rho_{12}\,\rho_{21}=|\rho_{12}|^2 \leq \rho_{11} \,\rho_{22}$, cf. here, and thus, regarding your $2\times 2$ example (where I assume you mean an ONB here): $\mathrm{Tr}\rho^2 \leq \rho_{11}^2 +2\rho_{11}\,\rho_{22}+\rho_{22}^2=(\rho_{11}+\rho_{22})^2 = (\mathrm{Tr}\rho)^2=1$. As shown in the linked answer, the equality holds if and only if $\rho$ is pure. Commented Sep 28, 2022 at 22:01
• You have three answers, so please consider to accept one if your question has been answered (you can also accept your own answer). If you think none of them answered your question, you should clarify exactly what is missing. Commented Oct 4, 2022 at 5:13

To show it you can use the following inequality

$$|\mathrm{Tr} (A^\dagger B) | \le \| A\|_1 \,\, \| B\|_\infty$$

Where $$\| A\|_1$$ is the trace norm and so it's one for a density matrix and $$\| B\|_\infty$$ is the sup norm, which for hermitian $$B$$ is the largest eigenvalue in modulus of $$B$$ and so it's bounded by one for a density matrix.

• If I $A^\dagger=B=\rho$ then $|Tr(\rho^2)\leq ||\rho^\dagger||_1 ||\rho||_\infty$ and from $||\rho^\dagger||_1=||\rho||_1=1$ and $||\rho||_\infty\leq 1$ follows that $|Tr(\rho^2)|\leq 1$ ? Commented Sep 28, 2022 at 9:50
• Yes precisely .
– lcv
Commented Sep 28, 2022 at 15:42
• Moreover, since $\rho$ is a positive matrix (it's hermitian and all its eigenvalues are non-negative) you indeed have $| \mathrm{Tr}(\rho^2) | = \mathrm{Tr}(\rho^2)$
– lcv
Commented Sep 28, 2022 at 15:44
• @Icv okay but I don't know how I can apply this inequality to a real case. I edited an example to my post Commented Sep 28, 2022 at 21:28
• A much simpler argument: diagonalizing $\rho$, we're interested in the sum of squares of eigenvalues $\in[0,\,1]$ (on which $x^2$ is contractive) that sum to $1$.
– J.G.
Commented Sep 28, 2022 at 21:33

"okay but I don't know how I can apply this inequality to a real case. I edited an example to my post" (OP in a comment to a different answer)

$$\rho²=\begin{pmatrix}\rho_{11}&\rho_{12}\\\rho_{21}&\rho_{22}\end{pmatrix}^2=\begin{pmatrix}\rho_{11}^2+\rho_{12}\rho_{21}&\rho_{11}\rho_{12}+\rho_{12}\rho_{22}\\\rho_{21}\rho_{11}+\rho_{22}\rho_{21}&\rho_{21}\rho_{12}+\rho_{22}^2\end{pmatrix}$$

By the spectral theorem, every density operator can be written in some basis as: $$\hat\rho = \sum_{n} p_n|\phi_n\rangle\langle\phi_n|\;.$$ That is, a basis can be chosen such that $$\rho$$ is diagonal. And we have $$\sum_n p_n= 1$$

For your "real case" example, the matrix representation of the density operator in the basis where it is diagonal is: $$\rho =\begin{pmatrix}p_{1}&0\\0&p_{2}\end{pmatrix}$$ and $$Tr(\rho)$$ is clearly equal to 1.

$$Tr(\rho^2) = \sum_n p_n^2\;,$$ which is clearly less than or equal to 1.

Suppose that the density matrix is re-written in a different basis $$|\tilde \phi_n\rangle$$, where it does not appear manifestly diagonal: $$\hat \rho = \sum_{n}p_n|\phi_n\rangle\langle\phi_n| =\sum_{nij}p_n a_{ni}a_{nj}^*|\tilde \phi_i\rangle\langle\tilde \phi_j| \equiv \sum_{ij}\tilde\rho_{ij}|\tilde \phi_i\rangle\langle\tilde \phi_j|\;,$$ where $$\tilde\rho_{ij} = \sum_n p_n a_{ni}a_{nj}^*\;,$$ and where, we also know that the transformation matrix is unitary. I.e., $$\sum_i a_{ni}a_{mi}^* = \delta_{nm}\;,$$ which we know to be true since we require both bases to be orthonormal.

So then, let's compute the trace in the $$\tilde \phi$$ basis (we could use any basis really). The trace is: $$Tr(\rho) = \sum_{k}\sum_{ij}\tilde\rho_{ij}\langle\tilde\phi_k|\tilde \phi_i\rangle\langle\tilde \phi_j|\tilde \phi_k\rangle =\sum_i \tilde \rho_{ii}\;,$$ which is what we expect.

But now recall the above definition of $$\tilde \rho_{ij}$$ to see that: $$\sum_i \tilde\rho_{ii} = \sum_i \sum_n p_n a_{ni}a_{ni}^* =\sum_n p_n \delta_{nn} = \sum_n p_n = 1$$

• okay but now you only have shown that the trace is not greater than one in the basis where the density matrix is diagonal. but that does not show that this is the case for the density matrix represented in ANY possible basis. Commented Sep 29, 2022 at 11:45
• @peter mafai trace is basis independent. Commented Sep 29, 2022 at 13:20
• @petermafai See my update for an explicit demonstration of the invariance of the trace.
– hft
Commented Sep 29, 2022 at 16:02

Okay so I will try to answer my own question? Is this allowed?
If you diagonalize the given matrix you get the eigenvalues $$\lambda_{1,2}=\frac{\rho_{11}+\rho_{22}}{2}\pm\sqrt{\Big(\frac{\rho_{11}+\rho_{22}}{2}\Big)^2+\rho_{12}\rho_{21}}$$ and they are the probabilities of the system being in state $$|1\rangle$$ or $$|2\rangle$$ where these states are represented by the first and the second eigenvector. The probability of finding them in one of these two states must be equal to $$1$$. This comes from "classical" quantum mechanics. From that one gets $$\lambda_1+\lambda_2=1\Rightarrow\rho_{11}+\rho_{22}=1$$ but it is also that the sum of the squares must be less equal than $$1$$ because $$(\lambda_1+\lambda_2)^2=\lambda_1^2+\lambda_2^2+2\lambda_1\lambda_2=1\Rightarrow\lambda_1^2+\lambda_2^2\leq1$$ and from that follows $$\lambda_1^2+\lambda_2^2\leq1\Rightarrow\rho_{11}^2+\rho_{22}^2+2\rho_{12}\rho_{21}\leq1\Rightarrow Tr(\rho^2)\leq1$$ but I am not sure about the argumentation with "either state $$|1\rangle$$ or $$|2\rangle$$" because im not sure if one can just decompose a density matrix into several states with each state having a probability (of the system being in that state) attached to it. That would seem too great to be true.
I would be grateful for some feedback to my answer as this is the one I've ever written (did not thought the first answer would be to one of my own questions) . Is it correct? Is the diagonalization of a density matrix really as useful as I think?

• I don't know what problems you have in mind but states obtained by diagonalization of the density matrix may not be as useful in physical problems as you think. Lets assume that you expressed your problem in an energy eigenbasis of the Hamiltonian that governs the dynamics of your system. The basis that diagonalizes your density matrix will most likely be a linear combination of energy eigenstates, meaning that the coefficients will change with time for any non trivial dynamics and you would have to re-diagonalize at every point in time. Commented Sep 29, 2022 at 14:19
• It is allowed to answer your own question. In any case, what's wrong with the comment I left under the question? You don't have to diagonalize the density matrix to obtain the last inequality you've written here; as I've shown, that is a one line proof. That being said, diagonalization is useful. You want to prove a property of the density operator which is not dependent on the basis you might express the operator as a matrix; so in general it is a good idea to use a basis which simplifies the problem the most, which in this case is indeed the eigenbasis. Commented Sep 29, 2022 at 14:25