# Eigenvalues of a quantum state after partial tracing

I am interested in the smallest nonzero eigenvalue of a quantum state.

1. Does this eigenvalue always increasing after a partial trace i.e. the smallest nonzero eigenvalue of $$\rho_A$$ is always larger than that of $$\rho_{AB}$$? Intuitively, this seems like it must be true since the reduced state has fewer eigenvalues but its trace is still 1. On average, the eigenvalues must be larger. How can one prove this statement (or perhaps a more refined version of it)?
2. Is there any bound on how much larger this eigenvalue can get? A trivial bound is that it cannot increase by more than 1 but is there anything better that can be said?
• The smallest non-zero eigenvalue of the density operator corresponding to the state $\tfrac{1}{\sqrt2}(|00\rangle + |11 \rangle)$ is $1$ but the eigenvalues of its reduced density operators are $1/2$. So in this case the smallest non-zero eigenvalue has decreased. In fact by the Schmidt decomposition this is true for any pure entangled bipartite state. Sep 22, 2020 at 22:03
• Thank you - that makes sense! Sep 23, 2020 at 10:31
• This statement will hold if you consider the smallest eigenvalue (possibly zero). You can see this by noting that the partial transpose is a CP map and applying it to the operator $\rho_{AB} - \lambda_{\min} I$ which is positive semidefinite. Sep 23, 2020 at 11:10
• *Partial trace... not partial transpose Sep 23, 2020 at 12:10

There are no effective bounds on how small or how large the smallest non-zero eigenvalue can become upon taking the partial trace, other than the obvious constraint that it must be in $$(0, 1]$$.

Let $$\lambda_{min}(\rho)$$ denote the smallest non-zero eigenvalue of operator $$\rho$$.

Fix a real $$p \in (0, \frac{1}{2}]$$ and consider the pure two-qubit state

$$|\psi\rangle = \sqrt{p}|00\rangle + \sqrt{1 - p}|11\rangle.$$

Note that $$\rho_{AB} = |\psi\rangle\langle\psi|$$ has eigenvalues $$0, 0, 0, 1$$ so $$\lambda_{min}(\rho_{AB}) = 1$$.

On the other hand,

$$\rho_A = \mathrm{tr}_B(\rho_{AB}) = p|0\rangle\langle 0| + (1 - p)|1\rangle\langle 1|$$

has eigenvalues $$p, 1 - p$$ and so $$\lambda_{min}(\rho_A) = p$$.

Consequently, not only can partial trace decrease the value of $$\lambda_{min}$$ (as already noted in the comments), but there is no positive lower bound on $$\lambda_{min}(\mathrm{tr}_B\rho)$$. More precisely, for any $$\epsilon > 0$$ we can find a state $$\rho_{AB}^\epsilon$$ such that $$\lambda_{min}(\rho_{AB}^\epsilon) = 1$$ yet $$\lambda_{min}(\mathrm{tr}_B \rho_{AB}^\epsilon) = \epsilon$$.

In order to find a bound on how much larger the smallest non-zero eigenvalue can become after taking the partial trace, fix a real $$p \in (0, \frac{1}{2}]$$ and consider the state

$$\rho_{AB} = p|00\rangle\langle 00| + (1 - p)|01\rangle\langle 01| = |0\rangle\langle 0| \otimes (p|0\rangle\langle 0| + (1 - p)|1\rangle\langle 1|).$$

Note that its eigenvalues are $$0, 0, p, 1-p$$ and so $$\lambda_{min}(\rho_{AB}) = p$$.

On the other hand,

$$\rho_A = \mathrm{tr}_B(\rho_{AB}) = |0\rangle\langle 0|$$

has eigenvalues $$0, 1$$ and so $$\lambda_{min}(\rho_A) = 1$$.

Consequently, upon taking the partial trace $$\lambda_{min}$$ can increase all the way to the maximum $$1$$ regardless of $$\lambda_{min}$$ of the original state.