TITLE: Purifying a mixed state using a Hilbert space of smaller dimensions
QUESTION [1 upvotes]: If I have a mixed state $\rho \ \epsilon \ H^A$ and I want to purify this using the Hiblert space $H^B$. I know this can be done if $dim(H^B) \geq dim(H^A)$. But can I do this if $dim(H^B)<dim(H^A)$? I tried to redo the proof I found on Wikipedia for purification state. I looks like it doesn't work.
The problem is: If $dim(H^B) > dim(H^A)$ and I find $\psi \ \epsilon \ H^A \bigotimes H^B$, if I trace out system A, I will get some $\sigma \epsilon H^B$. Now I can pose the question of starting out with a mixed state $\sigma \epsilon H^B$ and trying to use $H^A$ to purify it, which should be possible. It's $\psi$! But $dim(H^A)<dim(H^B)$. Where am I going wrong?

REPLY [4 votes]: Given a mixed state $\rho$ with rank $r$ on any (finite-dimensional) Hilbert space $\mathcal H_A$, it can be purified using a purifying system $\mathcal H_B$ of dimension $\mathrm{dim}\,\mathcal H_B\ge r$.  
That a purification with $\mathrm{dim}\,\mathcal H_B=r$ exists can be seen using the spectral decomposition $\rho=\sum p_i|\psi_i\rangle\langle\psi_i|$, and constructing a purification $\sum \sqrt{p_i} |\psi_i\rangle|i\rangle$.  
That this dimension is minimal can be seen by noting that for any purification with purifying system of dimension $r'<r$, the Schmidt decomposition of the purification must have a Schmidt number $s\le r'<r$, and thus, the rank of the reduced density operator is $s<r$, in contradiction to the assumption.
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