QuantumPhysics.dvi

The combinations of projectors obeys

tr

(

PiPj′

)

≥ 0 (16.68)

As a result, every term in the double sum is either positive orzero. The equality in the lemma is
achieved if and only if every term separately vanishes for alli,j,
(
p′j−pi−piln

(

p′j

pi

))

tr

(

PiPj′

)

= 0 (16.69)

If tr(PiPj′) 6 = 0, then we must havep′j =pi, even whenpi= 0. Now fix the indexi, and letj
run throughj = 0, 1 , 2 ,···,N′. If tr(PiPj′) = 0 for all valuesj, then dim(Pi) = 0, sincePj′span
the entire Hilbert space. Thus, we must havei= 0, and dim(P 0 ) = 0, in which case the result is
trivial. Thus, for alli 6 = 0, there must be a uniquejsuch that tr(PiPj′) 6 = 0, and for whichpi=p′j.
Reversing the argument, fixingj, and lettingirun through the valuesi= 0, 1 , 2 ,···,N, we see
that everyj 6 = 0, there must be a uniqueisuch that tr(PiPj′) 6 = 0 andp′j=pi. Given the ordering
(16.65) that we assumed, this implies a one-to-one and onto map between thepiand thep′i, so that
N′=N. As a result we have

ρ=

∑N

i=1

piPi

∑N

i=0

Pi=I

ρ′=

∑N

i=1

piPi′

∑N

i=0

Pi′=I (16.70)

where

tr(PiPj′)

{

= 0 j 6 =i

6 = 0 j=i

(16.71)

Finally, using these results, we compute

tr

(

PiPi′

)

= trPi−

∑

j 6 =i

tr(PiPj′) (16.72)

Each term under the sum on the rhs vanishes, so that tr (Pi(I−Pi′)) = 0 As a result, we have
Pi′=Pi, and thusρ′=ρ, which completes the proof of the lemma.

16.7.4 Completing the proof of subadditivity

To complete the proof of 6, we proceed as follows. First, define the density operators of the
subsystemsaandbby

ρa = TrHb(ρ)

ρb = TrHa(ρ) (16.73)