Linear algebra for quantum information theory

3.6.1 Discussion

Let A be an n x n Hermitian matrix. Let ei, ...,en be a complete orthonormal set of eigenvectors of A with eigenvalues Ai > A2 > ... > An:

>= AJej >,j = < efc|ej >= 6(k, j)

Let Ad be any k dimensional subspace of Cn. Then >Vf has a non-zero intersection with Vfc = span{ek, Cfc+i, ...,en} since the latter is n — k + 1 dimensional and

dim(Ad nV*.) = dimM + dimVk — dim^M + Vfc) >k + n — k+1—n = 1

It follows that

. || a; || = 1 < IT, A/T A/,.

On the other hand, choosing Ad = span{ei, we get

. || a; || = 1 < IT, A/T > A/,.

We have thus proved the fundamental inequality,

Alternately, suppose Ad is any n — k+ 1 dimensional subspace of Cn. Then, Ad has a non-zero intersection with span{ei, ...,efc} and hence

||x||=i < x.Ax >> Afc

On the other hand choosing Ad = span{ek, et+i, ...,en}, we get

maa;3.gJvi,||a:||=i < x.Ax > A/.

Thus, we get

A?+imn3.’3.^^j|a.||=i x,Ax > A^

These variational formulas can be used to approximately determine the energy levels of a quantum system given its Hamiltonian after appropriate truncation.

Now let A be as above and consider the 2n-dimensional vector space AV where V = Cn. This means that

AV = Ce^A/lV


Consider the operators

Ar(A) = (j) ^nfc-1 A A A I*n~k,r = 0,1, ...,n k=0

Ar(A) acts in ArV and an orthonormal basis for ArV is given by

Br = {c(n, r)eil A ... A ejr : 1 < ¿i < ¿2 < ... < ir < n}


|c(n,r)|2^ = 1

Here, we define

A ... A ur = 52 sgn(a')ual ® ... uar


It is easy to see that B, is a complete onb of eigenvectors of A,. (A) with eigenvalues

Aq + ... + Xir

for the eigenvector

c(n, r)eij A ... A

It follows that the maximum eigenvalue of Ar(A) equals Ai + ... + Ar and hence by the above variational principle applied to Ar(A), we get

mmEa.£Arv,||:r||=i < A.r(^A^x >— Ai H-... 4_ Ar

Note that we also obviously have the above equality if the maximum is taken over all x of the form x = fi A /2 A ... A fr where fj € V. < fj fi >= 6(j, I). This results in the following variational principle:

r r

< fj’Afj >= 52

j=i 3=1

This is called Ky-Fan’s maximum principle.

Some other inequalities related to quantum information theory:

[a] Matrix Holder inequality : If A, B are positive definite matrices and p,q> 1,1/p + 1/q = 1, then

Tr(AB) < (Tr(Ap))1/p.(Tr(B))1/q

To prove this, we use Hadamard’s three line theorem: If f(z) is non-negative, analytic and bounded in the strip 0 < Rez < 1, then for z = x + iy with a; € [0,1], we have

fV) <


Mo = supy&J(iy),Mx = supyt&f(l + iy)

Now put p = 1/z

x_ Tr{AzBl~z)

Then, f(z) satisfies the above requirements. We have

/(i?/) = l,|/(l + ii/)|

So, we get by applying the theorem,

f(x) < l,a? € (0,1)

which results in the desired inequality.


Transmission lines and waveguides-Questions

[1] Derive formulas for the distributed parameters R. L, G, C of a coaxial transmission line with infinitesimal parameters e, p, a when the radii of the inner and outer core are respectively a and b.

[2] Assume that a cylindrical cavity resonator is made of a perfect conductor of length d and radius R. The region 0 < z < dx is filled with water while the region di < z < d is filled with air. Assume water has permittivity and permeability (fi,pi) while air has the corresponding parameters as (f2-. M2). Denote the cylindrical components of the electric field and magnetic field components in the two regions respectively by (Epk E^k Ezk^), , Hzk^),

k=l,2 respectively. Assume that in the two regions, the dependences on z of the Hz components of the TE modes are of the form sin(az) and sm(/3(d — z). This would guarantee that Hz vanishes at z = 0 and z = d. Express Ep(p, , z), Ep(p, , z), H(p, d>, z) in the two regions in terms of Hz in the two regions. Hence, set up the appropriate Helmholtz equations for Hz in both the regions and obtain the general solutions for the TE mode by applying the boundary conditions that Ep. E^ vanish at z = 0, d and are continuous at z = ¿i, Hp.H,-, are continuous at z = dx (no surface current can exist at a dielectric interface), and finally, pHz is continuous at z = dx. Repeat for the TM modes.

[3] Calculate the far field radiation pattern produced by a rectangular cavity resonator with all walls being perfect electric conductors when the oscillation frequency is

w[mnp] = (m2/a2 + n2/fe2 + p2/d2)1/2^^

Before presenting your final expression for the far field Poynting vector, present the expressions for (a) the electric and magnetic field components within the resonator as a function of (x,y, z,t), and (b) the induced surface current density on the surface walls of the resonator.

[4] A lossy transmission line has distributed parameters R. L. G, C. The load is /¿(w). Write down the expressions for (1) the propagation constant q(w), (2) The VSWR after defining it, assume a lossless line for this case, (3) The distance between two successive voltage maxima and between two voltage minima, assume a lossless line for this case (4) the distance of the first voltage maximum and first voltage minimum from the load, assume a lossless line for this case. Explain how you would calculate the wavelength of the propagating wave in the lossless case as well as the reflection coefficient (both magnitude and phase) in terms of the VSWR and the quantities defined in (3) and (4). Finally, if the frequency of operation changes from uj to w', compute the new VSWR in terms of the original VSWR at a distance of d from the load assuming a lossless line and purely resistive load.

[5] (a) Write down a series expansion for the inverse Fourier transform of

where 7(0?) = y/(R + jivL)(G + jcvC) by writing it as

7(W) = > RV(1 - jR/a>(l - jG/cvC)

and assuming that R/L.G/C << |w|. Use generalized functions for expressing your answer, (b) Repeat (a) without making any assumption. For this, partition the frequency line into three parts assuming R/L > GfC and doing a Taylor-Laurent expansion for q(co') in each of the three regions. Use this formula to obtain a series expansion for the voltage as a function of z and time t for a lossy line when the input source voltage is vs(t) an source resistance is Rs.

hint: Use the fact that (Jw)rF(w) has inverse Fourier transform drf(t)/dtr while (jw)~rF(w) has inverse Fourier transform equal to the r-fold integral of /(t) from zero to t. You will also have to use the formula for the inverse Fourier transform of w(w — wo) where u is the unit step function. This is obtained by applying duality to the formula

Fu(f) = l/jcj + Trd(tv’)

[6] A transmission line of length d and characteristic impedance Zo(w) is terminated by a load of Zl(uj). Two stubs of length L.L2 respectively terminated by impedances of ZlJjjj) and Z[_2(uj) are located at distances of d and ¿2 respectively from the load. Find (a) The VSWR at all points along the line, (b) the impedance seen at the input z = d of the line, (c) Two relations between Li,L2, di,d2 for matching to a line of characteristic impedance Zg(w). Assume the frequency of operation cv to be fixed.

[7] If a rectangular cavity resonator supports the modes (ni,mi,pi), (n2,m2,P2) and (ns, m3,ps) with respective oscillation frequencies of wi,W2,W3, then determine the dimensions a, b, d of the resonator.

[8] Solve the transmission line equations approximately upto 0(6) for inhomogeneous distributed parameters, ie, the line equations taking into account line loading are

Vz(u,z) + (Ro + jwL0)I(w,z) + 6(Rx(z) + jwCi(z))I(w, z) = J(cc,z),

I.z(cj,z) + (Gq + juCo)V(

Assume source conditions

V(w,0) = Vs(cc), V(u,d) = ZL(cc)I(u,d)

3.7.1 Points to remember

[1] The Tx line equations are first order coupled pde’s in the voltage and current along the line as a function of (t,z). These are derived by applying the KCL and KVL to an infinitesimal section of the line with each infinitesimal section consisting of a series resistance and inductance and a parallel conductance and capacitance. Elimination of the current variable leads to the voltage variable satisfying a damped wave equation and vice versa. If the resistance and conductance per unit length are neglected ie the line is lossless, then these damped wave equations become the usual one dimensional wave equation.

[2] By taking Fourier transforms w.r.t. the time variable in the Tx line equations, we get a second order ordinary linear ordinary differential equation in the spatial variable for the voltage and current which can be solved immediately to obtain the line voltage and current in the form of a forward wave and a backward wave. These forward wave will be damped with spatial variation from the source while the backward wave will be amplified. For a lossless line no damping or amplification will be there, ie, the propagation constant will be purely imaginary.

[2] The reflection coefficient at any point on the line is the ratio of the amplitude of the backward wave to that of the forward wave. At the load end, the reflection coefficient admits a simple expression in terms of the load impedance and the characteristic impedance of the line. As we move away from the load, this reflection coefficient varies as exp(2yz) where 7 is the propagation constant. At any point on the line, the relation between the input line impedance and the reflection coefficient is the same as it is at the load end. This enables us to compute the input impedance at any point on the line.

[3] The VSWR is the ratio of the maximum voltage amplitude to the minimum voltage amplitude along the line. For a lossless line, it as a simple expression in terms of the magnitude of the reflection coefficient at the load end. The phase of the reflection coefficient at the load end for a lossless line and the propagation constant determines the location of the voltage maxima and minima w.r.t. the load. If these positions are known, we can then calculate the propagation constant and the phase of the reflection coefficient. Therefore, for a lossless line, knowing the VSWR and the location of the first voltage maxi-mum/minimum as well as the distance between successive voltage maxima or minima, we can determine the reflection coefficient at the load end and the propagation constant and hence the reflection coefficient anywhere along the line and hence the input impedance at any point on the line.

[4] If a line is terminated by a load equal to its characteristic impedance, then there is no reflected wave, ie, all the power on the line is transmitted to the load. Thus, it is of interest to attach open and short circuited stubs at different points on a loaded line to make its input impedance match that of a another given line. This problem can be formulated analytically and solved graphically using the Smith chart.

[5] If we express the relationship between the normalized input impedance and the reflection coefficient in terms of its real part and imaginary part, then the constant resistance curves and constant reactance curves in the reflection coefficient plane are circles. This is the fundamental property on which the construction of the Smith chart is based. As we move away from the load, the curve traced by a point on the Smith chart is a circular arc moving clockwise provided that the line is lossless, while if it is lossy, the curve is an inward going spiral.

[6] Nonuniform lines are analyzed using perturbation theory for differential equations or better still using Fourier series expansions of the distributed parameters and the line voltage and current in terms of the spatial variable and then followed by perturbation theory.

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