Controlled
Quantum
Dynamics
Quantum lost property:
An operational meaning
for the Hilbert-Schmidt product?
Matthew F. Pusey, Terry Rudolph
Introduction
Suppose you are given a quantum system prepared, with equal
probability, in state ρ or σ. After a measurement, your probability
of guessing the correct state depends on the trace distance
1
δ = tr |ρ − σ| .
2
Suppose ρ is prepared by sampling from a distribution {pi} and
P
P
preparing ρi, such that ρ =
i piρi and similarly σ =
j q j σj .
If in each run you are given the classical information (i, j), your
average probability of success depends on
X
∆=
piqj δ(ρi, σj ).
i,j
We prove that
∆≤
p
1 − tr(ρσ)
and conjecture that for all ρ and σ, there exist decompositions
that saturate this bound.
tr(ρσ) is known as the Hilbert-Schmidt inner product. To our
knowledge, this is the first time this inner product between two
states has appeared in a simple operational setting.
arXiv:1208.2550
An equivalent conjecture: existence of
unbiased decompositions
A decomposition
into
pure
states
ρ
=
|ψ
i
hψ
|
i
i
i
and σj = φj φj saturates the bound if and
only if
ψi φj is independent of i, j,
so we conjecture that such a decomposition
exists for all ρ, σ. Such decompositions may
be useful in other settings.
The evidence
Numerics indicate that such decompositions
can always be found.
We have analytic proofs for when the rank of
ρ is at most 2, and for when ρ is the maximally
mixed state (in the latter case the unbiased decompositions are simply mutually unbiased
bases).
Ideas for a proof or a counter-example are very
welcome!
Special case: qubit states
Rotate so that
ρ~ = (0, 0, r)
and
~σ = (sx, 0, sz ).
Then can take
p
ρ~1 = 0, 1 − r2, r ,
p
ρ~2 = 0, − 1 − r2, r ,
q
~σ1 =
1 − s2z , 0, sz
and
q
~σ2 = − 1 − s2z , 0, sz .
It’s easy to check that these are valid
decompositions into pure states, and
ρ~i · ~σj = rsz which is independent of i
and j as required.