QuantumPhysics.dvi

Henceforth, we shall assume thatpi>0, and omit those contributions for whichpI= 0 from the
sum. The numbersis then the common rank ofρaandρb, which is nothing but the Schmidt
number of the state|Ψ〉,

s= rank(ρa) = rank(ρb) (17.28)

• If s = 1, the state|Ψ〉 is non-entangled. In this case, the density operators ρa andρb
  correspond to pure states ofHaandHbrespectively.


• Ifs >1, the state|Ψ〉isentangled. In this case, the density operatorsρa andρbdo not
  correspond to pure states ofHaandHbrespectivelybut represent a mixture instead.

17.5 Entanglement entropy

The Schmidt number gives a discrete measure of the degree of entanglement of a bipartite division.
But there is also value in having a continuously varying measure of the entanglement of a state.
This is provided by theentanglement entropy. Recall that the state|Ψ〉in the full Hilbert space
is pure, so its statistical entropy vanishes. But when the state is considered from the point of view
of the bipartite division into systemsaandb, it makes sense to talk about entropiesS(ρa) and
S(ρb) associated with the density operatorsρaorρb. By the Schmidt purification theorem, we have
S(ρa) =S(ρb), so that we may consistently defined an entanglement entropy by

Sentang≡S(ρa) =S(ρb) =−

∑s

i=1

pilnpi (17.29)

As the state|Ψ〉approaches a non-entangled state|Ψ;j〉(in whichpi= 0 for alli 6 =j, andpj= 1),
the entanglement entropy of|Ψ〉tends to 0 in a continuous manner.

17.6 The two-state system once more

Consider the following pure state in the system of two spin 1/2 particles,

|Ψ〉= cosθ|z+,a;bz−,b〉+ sinθ|z−,a;z+,b〉 (17.30)

The density operators of the subsystemsaandbare readily evaluated,

ρa = cos^2 θ|z+,a〉〈z+,a|+ sin^2 θ|z−,a〉〈z−,a|

ρb = cos^2 θ|z−,b〉〈z−,b|+ sin^2 θ|z+,b〉〈z+,b| (17.31)

Note the reversal of probability assignments for spin + and spin−. The Schmidt number is

• s= 1 when sin(2θ) = 0: the state|Ψ〉is not entangled;


• s= 2 when sin(2θ) 6 = 0: the state|Ψ〉is entangled.

The entanglement entropy is given by

Sentang=−cos^2 θln(cos^2 θ)−sin^2 θln(sin^2 θ) (17.32)

Whenθ→0, for example, we approach the non-entangled state|z+,a;bz−,b〉, and the entangle-
ment entropySentang∼−θ^2 lnθ^2 tends to zero in a smooth way.