The trace of the square of a density matrix(which is also called the Purity of the quantum state) is (lower)bounded by the inverse of the dimension of the Hilbert space and (upper) bounded by 1.

I think that after diagonalizing the density matrix of a maximally mixed state, the resulting matrix is a scalar matrix(all the diagonal elements are equal).This is because(say) the ith diagonal element is the probability of finding the system in the state i, and if all the probabilities are equal, then the state is maximally mixed.Is my reasoning correct?Is the basis in which the density matrix diagonal an orthogonal basis?

My understanding is that the Purity of the quantum state is equal to the lower bound when the quantum state in question is 'maximally mixed'.How do i find it's lower bound?I have tried proving the inequality by expanding the trace, but am not able to see an obvious way of finding it's lower bound.


1 Answer 1


Is the basis in which the density matrix diagonal an orthogonal basis?

Note that if the density operator is $1/N$ times the identity, then it is diagonal in every basis. On the other hand, every density operator on a finite dimensional Hilbert space is diagonal in an orthonormal basis: this follows from the spectral theorem, which asserts that this is exactly equivalent to require that the operator is hermitian. In general you can write the density operator in a unique way as a sum: $$\hat \rho = \sum _i \lambda _i \hat P _i,$$ where $\hat P _i$ is the projection on the eigenspace $\Lambda _i$ of the eigenvalue $\lambda _i$, the $\lambda _i$'s are distinct, $0\leq \lambda _i$ and $\sum _i \lambda _i \cdot \text {dim} \Lambda _i =1$. Then, you can arbitrarily fix an orthogonal basis on each $\Lambda _i$ and assign to each vector of such a basis the weight $p_i=\lambda _i$.

How do I find it's lower bound?

You want to prove that: $$\sum _{i=1}^N p_i^2 \geq 1/N$$ where $$\sum _{i=1}^Np_i=1.$$

The simplest method is to apply Jensen's inequality with the convex function $x\mapsto x^2$, to the sum $\sum _i p_i/N$ ($1/N$ being the weight of $i$):

$$\dfrac{1}{N^2}=(\sum _i \dfrac{p_i}{N})^2\leq \dfrac{1}{N}\sum _i p_i ^2.$$

As an alternative method (which is far more general), you can apply Lagrange's multipliers theorem: it states that a necessary condition for a local minimum of a smooth function $f\colon \mathbb R ^N \to \mathbb R$, restricted to a smooth submanifold $M$ of $\mathbb R ^n$, is that $\nabla f(x_0)\perp T_{x_0}M$ at the point of minimum $x_0$.

Define $$\overline M_N =\{(p_1,\dots,p_N)\in \mathbb R ^N \,\, \sum _i p_i =1, \,\, 0\leq p_i \leq 1\}.$$ Since $\overline M _N$ is compact, $f(\{p_i\})= \sum _i p_i ^2$ has a minimum in $\overline M _N$. First, assume that the minimum lies in the interior $M_N$ of $\overline M _N$. Then, Lagrange's theorem states that, at the minimum: $$\nabla f=(p_1,p_2,\dots,p_N)\propto (1,1,\dots,1),$$ which, together with $\sum _i p_i =1$ gives $p_i=1/N$ and $\sum _i p_i^2=N(1/N)^2=1/N$.

Now, if $N=1$ the lower bound is trivial. By induction, we see that on the boundary of $\overline M _N$, $f$ is lower bounded by $1/(N-1)$, so the minimum is actually in the interior $M _N$. This completes the proof.


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