Large deviation principle in wave-motion
3.2.1 Discussion
The wave equation in one dimension driven by a white-Gaussian noise field has the form
with boundary conditions
u(t, 0) = u(t, L) = 0, t > 0
Here w is a zero mean Gaussian random field with covariance
E(w(t, x)w(t', a/)) = <5(t — t')6(x — x')
We can develop the solution in a half sine-wave Fourier series:
u(t,x) = «n(i)ÿn(æ) n>l
where
<£n(a;) = (2/L)1^2 sinÇmrx/L)
We also note that the random noise field w(t, x) can be expanded as w(t,x) =
n>l
Then, we require that
E(wn(t)wm(i')) = - t')
so that using
6(x - x) = ^n(^)d>n(a:')
n>l
we get the correct expression for the covariance of w. Thus, {wn(.)} are independent standard white Gaussian noise processes and can be represented as
w„(t) = B’n(t)
where {/?„(.)} are independent standard Brownian motion processes. Substituting this into the above wave equation gives us
u"(t) + w2un(t) = yëwn(t), n = 1,2,... where
a>n = mu-/ L
Equivalently, in terms of Ito stochastic differential equations,
dun(t) = vn(t)dt,dvn(t) = -WnUn(t)dt + /edBn(t),n > 1
Using Schilder’s theorem, we obtain the rate function of the processes un(t), 0 <
t
Z(un(t),n > 1,0 < t < T) =
T (1/2) £ [ (<(t)+^Wn(i))2dt n>lJo
Rate function for one dimensional stochastic filtering problems. The state process x(t) 6 R satisfies the sde
dx(t) = ft(x(t))dt + ^Jgt(x(t))dB(t)
and the measurement process is
dz(t) = x(t)dt + y/ecrdv(t)
where B.v are independent Brownian motion processes. If e = 0. the process x can be recovered exactly from the measurement z since there is no measurement noise. If 5 = 0, then the process x is non-random and can be uniquely determined from its initial value a;(0) by solving the differential equation
x'(t) = ft(x(t),t > 0
The EKF for this system is
dx(t) = ft(x(t))dt +
P ) = 2/f(£(t))F(t) +
We see that if e —> 0, then the second Ricatti equation gives F(t) —> 0 which means that the state estimate is perfect. This agrees with what we said earlier, namely, that zero measurement noise implies perfect state recovery. In general, we define
e(t) = x(t) — x(t)
and obtain approximately,
de(t) = —
= (/'(¿(t)) —
Since F(t) scales as e, we define
Q(i) = F(t)/e
Then the above equation along with the Ricatti equation can be expressed as
de(t) = (/'(i(t)) - cr-2Q(t))e(t)dt +
— ea~1 Q(t)dv(t),
Q'(t) = 2/t(i(i))Q(i) + -
This equation implies that conditioned on the state observer £’(t),0 < t < T, the rate function for e(t),0 < t < T is given by
Some more problems in Schrodinger-wave mechanics and Heisenberg-matrix mechanics with relevance to quantum information theory
3.3.1 Discussion
[a] Initially, the system is in a pure state f >. A random unitary operator U(0) where 0 has the probability distribution P(d0) acts on this state, then the output state is a mixed state:
P = I U(0)f X fU(0)*P(d0)
The entropy of the input state is zero while the entropy of the outptut state is non-zero. Such a situation occurs for example when the em field within a waveguide depends on random parameters like amplitudes and phases and this field is incident on an atom or a quantum harmonic oscillator that is initially in a pure state. The computations can be carried out in the interaction picture causing the state after interacting with the random em guide field to be mixed. Thus, the waveguide field described as a classical random em field pumps in entropy into the atomic system.
[b] A more accurate description of the above phenomena is as follows: The system is initially in the pure state f > while the photon bath is in another pure state |<7 > like for example, a coherent state. Then the initial state of the system and bath is the pure state f <8>g >= |/ > |.9 >. The atomic/system observables are defined in the Hilbert space { while the quantum em field observables are defined in the Hilbert spaced H, like for example, the Boson Fock space. The Hamiltonian of the system and bath is of the form
H(t) = Hs ® Ib + Is ® HB + Vsb(î)
where Ib is the identity operator in the bath space H. Is is the identity operator in the system Hilbert space f), Hs is the system Hamiltonian acting in f), Hb is the bath/em field Hamiltonian acting in H while Vsb(î) is the interaction Hamiltonian acting in the tensor product space In the interaction picture, the state of the system after time T is
PS(T) = TrB(U(T)f ®gxf® gU(T)*)
where
T
U(T) =T{exp(—i Î VSB(t)dt)},
Jo
with
VsB(t) = (exp(itHs) ® exp[itHB))VsB{t)exp{-itHs) ® exp(-itHB))
Writing
U(T) = uk ® wk k
where Uk is a linear operator acting in h while Wk is a linear operator acting in
H, the unitarity condtion on U(T) becomes
£ u^uk ® = i = £ umu*k ® wmwk
k,m k,m
and we find that
PS(T) = £ < > (Ukf >< fm
k, m
Once again we find that the initial system state is pure and hence has zero entropy while the final system state is mixed and therefore has non-zero entropy, implying therefore that the bath has pumped entropy into the system. It is possible to describe this latter quantum process using the former classical probabilistic method, ie, by deriving Ps(T) as the action of a classically random unitary operator acting on the initial system state alone. This is achieved as follows: The matrix ((< >)) is positive definite and hence we can
write its spectral decomposition as
< pIWWIs >= £p(Z)e(Z,/e)e(Z,m) l
where p(Z) > 0 and
^e(l,k)e(l',k) = 6(1,1'),
k
e(l, k)e(l, m) = 6(k, m)
I
Thus,
PS(T) = £ p(l)e(l,k)e(l,m)(Ukf X /|^)
l, k,m
= £p(Z)(W>i
where
k
It is easy to verify that by absorbing a positive factor from p(l) into V/, we can make Vi unitary operators and simultaneously guarantee 22;p(Z) = 1. We leave this as an exercise.
Remark: In Cq coding theory, we have a source emitting an alphabet x with probability p(x) and we encode this alphabet into a quantum state p(x) in a Hilbert space and transmit this state over a noiseless channel. The output stateis then p(x)p(x) and its entropy is H('£2xp(x')p(x)). The entropy of the output state given the input signal is JL,. p(æ)/f (p(-'r))- Thus, the information transmitted over the channel is
In case the transmitted states p(æ) are all pure, we have H(p(x) = 0 and the information transmitted becomes
Z(p,p) = H(£p(x)p[x))
X
The generalization to continuous random sources is elementary:
H( / p(0)p(0)d0) - / p(0)tf(p(0))d0
where p(f)) is the probability density of the emitted source signal 0.
3.3.2 Quantum hypothesis testing
Let p. a be two mixed states in a Hilbert space H.
We wish to design a POVM test T for deciding whether p or <7 is the true state. It is known from basic operator theory (See M.Hayashi, ”An introduction to quantum information”) that the optimal test that minimizes the probability of miss given an upper bound on the false alarm probability (the quantum Neyman-Pearson test) is of the form
T = {p — c.a > 0}
where {A > 0} for a Hermitian operator A, denotes the orthogonal projection onto the space spanned by the eigenvectors of A having positive eigenvalues. Then fixing the false alarm probability as e means that c must be selected so that the false alarm probability is
Pp = Tr(a{p — ca > 0}) = e
and the probability of miss is then
PI = Tr(p{p -ca < 0})
We have
Pm = Tr((p — c.a){p — c.a < 0}) + c.Tr(a.{p — c.a < 0}) < c.(l - Pp} = c.(l - e) ^/p{p-c. = ^{ps < csas}y/p ^{1 < < Csp<1-S)/2<7SP(1-S)/2,S > o This gives on taking trace, Pm < csTr{pi~s where Ds(pa) = —s~1.log(Tr(pl~sas')) Note that application of L’Hospital rule gives lims^0Ds(pa) = Tr(p(log(p) - log(a))) = £>(p|a) namely, the quantum relative entropy between a and p. So, we get the result that if log{c) = Ds(p| Pm < exp(-sô) On the other hand, we have for s > 0, e = Pp = 7V( < c-sTr(^^~s/2ps^~s/2^) = c~sTr^al~s ps) = c~sexp(—sDs(rr|p)) = exp(-s(Ds(ap) + /og(c))) Replacing s by 1 — s in this inequality with s < 1, we get Pf < ea>p(-(l - s)((s/(l - s))Z>s(p|cr) + Zop(c))) = exp(-s.£>s(p| It follows that assuming s € (0,1), by selecting c so that Zop(c) = Ds(pa) - 6 we get Pm < exp(-sô'), Pf < ea;p((l — s)<5 — Ds(rr|p)) Now consider the problem of testing the hypothesis between the tensor product state pP>n and where n = 1,2,.... Following the same logic, we get by considering the test {p®n — cn.a®n > 0}, the following results Pm = PM(n) < cns{Tr^-sas) = exp^log^ - Ds(pa))))n = exp(ns(log(c) - £>s(p| and PF = PF(n) < exp(n(—sDs(p(r) + (s - l)log(c))) It follows that by choosing /op(c) = Ds(p|<7) - 6 we get Pyi{n) < exp(-nsS), PF(n) < ea?p(n((l — s)8 — Ds{pa)) Letting 0 < <5 < L>s(p| we get the result that P.l(n). PF(n) —> 0,n —> oo These inequalities are valid for all s € (0,1). Moreover, from the above discussion, we have for all s > 0 and all c > 0 the following: i’w(n) < exp(ns(log(c) - Ds(pa))), PfIp) <= exp(-s(Ds(ap) + logic))) Thus, ltrnsupn^xn~' .loglPMln)) < s(log{c) - Ds( for all s > 0. For s —> 0+, this gives limsupn .,X/n~l .loglPyiin)) < —slog(c) — sDl limsupn^oon-1 .loglPF(n)) < — sDlpa) — slog(c) + o(s), 3.3.3 Some useful matrix inequalities in quantum information theory [1] S(p|<7) = Trlp.lloglp) - log(a))) is convex as a function of a pair of density matrices (p, a), ie, if 0 < a < 1, then + (1 - a)p2|a oS(pi|<7i) + (1 - a)S’(p2|o-2) This is deduced from Lieb’s inequality. Now, if P is an orthogonal projection, then for an operator X, its pinching by P is defined to be the operator PXP + ll- P)X{1- P) More generally, if P,Pn are orthogonal projections such that PiPj = 0, i / j and Pj = I, then the pinching of X by {Pj : 1 < ji < n] is defined to be the operator n cP(X) = £pjxp> j=i We claim that if p, a are density operators and {Pj} a pinching as above, then S(Cp(p)|CP(<7)) < 5(p||sipma) This inequality states that performing a quantum operation/measurement on two states cannot increase the relative entropy between them. This result is also known as ’’monotonicity of the quantum relative entropy”. Let P be an orthogonal projection. By an appropriate unitary transformation of our basis, we can assume that Accordingly define the unitary operator U=( 1 M 0 / Also if X is an operator of the same size, we can partition it in the same way as /Xu X12 A21 X22 ) We then find that pxp + (i - p)X(i - P) = ( xon ) = ±(UXU* + X) Hence, if f is an operator convex function, then we have the result that for any orthogonal projection P and operator X. we have f(PXP + (1 - P)X(1 - P)) = f{X + U2XU*) < (l/2)(/(PXP*) + /(X)) = (l/2)(Pf(X)P* + /(X)) = P/(X)P + (1 - P)/(X)(1 - P) Taking trace gives Tr(/(CP(X)) < Tr(/(X)) where More generally, let H be a finite dimensional Hilbert space and f : —> R be an operator convex function, ie, for Xi,Xk, Yi,..., Yk € J(aXm+(l-a)ym,m = 1,2,..., A) < af(Xm, m = 1,2,..., k)+(l-a)f(Ym, m = 1,2,..., A) then by the same argument as above, we have /(PXmP+(l-P)Xm(l-P),rn = 1,2,..., A) =f((UXmU*+Xm)/2,rn= 1,2,..., A) < (/(PXmPm=l,2,...,A)+/(Xm,m = 1,2,..., A))/2 = (/(Xm,m= l,2,...,A) + P/(Xln,m = 1,2,..., A)P‘)/2 which gives on taking trace, Tr(f(PXmP + (1 - P)Xm(l - P), m = 1,2,..., A)) < Tr(/(X„, m = 1,2,..., A)) We further have the following obvious result: Let f : B(T4)k —> R be unitarily invariant and convex. Then for any pinching { }, we have f(Cp(X1),...Cp(Xk)) In all these discussions, we assume Xj,Yj,j = 1,2, ...,A to be positive definite matrices. Here, cP(X) = £ pkxp, k This follows from the fact that any pinching is a product of 2 x 2 pinchings and if P is an orthogonal projection operator and U the associated unitary matrix, then /(PXP+(1-P)X(1-P)) = /((X+PXP‘)/2) < (/(X)+/(t/Xt/‘))/2 = /(X) since f is convex and unitarily invariant. Consider now the function S(x,y) = x.(log(x) - log(y)) defined on R+ x R+. This function is convex since its Hessian is > 0: Hs(x,y)=( S/x ) °,xy D,yy / _ ( 1/x -1/y -1/y x/y2 J We have Tr(Hs(x,y)) = 1/x + x/y2 > 0, detHs(x, y) = 0 proving that Hs(x,y) > 0. Lieb proved that for any matrix X and t € [0,1] the function /t(T, S) = Tr(XTtX*S1~t) is concave on the space of pairs (T, S) of positive definite matrices. It follows that F(T,S) = limt^0+ft(T,S)/t = Tr(Xlog(T)X* S) - Tr(XX* Slog(S)) is also concave. Taking X = I gives the result that S(S,T) = Tr(S.(ZOff(S)-MT))) is convex on the space (5. T) of pairs of positive definite matrices. We also note that S(S,T") is unitarily invariant, ie, if U is a unitary matrix, then S(USU*,UTU*) = S(S,T) Hence, the above pinching results are applicable and establish the fact that S(Cp(S),Sp(T)) which in the context of density matrices and quantum information theory implies that mutual information between two states cannot increase after performing a generalized measurement or a quantum operation. 3.3.4 Points to remember [1] Entropy can be pumped into a quantum system by coupling it to a noisy bath. For example, suppose that the em field within a waveguide consists of TE and TM waves with random amplitudes and phases. Then suppose we place an atom or a quantum harmonic oscillator within the guide so that it interacts with the random classical em field. The interaction Hamiltonian can is then a random operator built out of the position and momentum observables of the atom, the randomness arising due to the random amplitudes and phases of the guide’s em field. The Schrodinger unitary evolution operator in the interaction picture thus becomes a random operator and the action of this unitary operator on an initially pure atomic state will retain purity but this purity will be destroyed once we average the resulting state w.r.t. the probability measure of the random amplitudes and phases of the em fields. An approximate computation of the resulting mixed state of the atom can be made and its Von-Neumann entropy can be computed using approximate formulas for the logarithm of a perturbed matrix and also approximate perturbation theoretic formulas for the eigenvalues of a perturbed matrix. [2] A more accurate way to model the interaction of the guide’s em field with the atom is to regard the em field as a quantum em field built out of creation and annihilation operators with the em field bath being in a coherent state. The interaction Hamiltonian of this quantum em field with the atom can be expressed as a bilinear function of these field operators and the atomic observables. We can then compute the unitary evolution of the tensor product of the initial pure atomic state with the bath coherent state after time T. This state will again be pure but when we calculate its partial trace over the bath state, the result is a mixed state for the atomic system and its entropy can be approximately calculated using matrix perturbation theory. It should be noted that in the previous model, we assume that the em field is a classical random field while in the latter case, it is a quantum field. The role played by averaging over classical randomness in the previous case is played by partial tracing over the bath state in this case. This case gives a more accurate description of how nature works. [3] In the problem of discriminating between two quantum states p. a, we have to construct a decision operator 0 < T < I so that Tr(pT) is the probability of making a correct decision given that p is the true state and Tr( [4] Some other useful quantum information inequalities are the pinching inequality for operator convex functions and the monotonicity of the quantum relative entropy.
