Which operator is used for exponential?

6.9.4 An Alternative Expression for the Memory Functions

The time dependence of the fluctuating force f is modified by the projection operator Qˆ(k),which multiplies the Hermitian conjugate Smoluchowski operator in the operator exponential in the definition (6.164,173) of f. This complicates the evaluation of the memory functions (6.166,172), and it is desirable to have an alternative expression in which the Hermitian Smoluchowski operator is not modified by this projection operator.

Such an alternative expression can be derived for the Laplace transform of the memory function, which is defined as,

(6.180)M(k,z)=∫0∞dt M(k,t) exp{−izt}·

The variable z is the Laplace variable, conjugate to t. When the Laplace transform is known, the transformation may in principle be inverted to obtain the memory function M(k, t). Substitution of the definition (6.164) of the fluctuating force f into eq.(6.166) for the memory function and Laplace transformation yields,

(6.181)M(k,z)=∫0∞dt<exp{−izt+Qˆ(k)ℒˆS†t}f(k|X|0)∥f(k|X|0)><ρ(k|X)∥ρ(k|X)>=1<ρ(k|X)∥ρ(k|X)><1iz−Qˆ(k)ℒˆS†f(k|X|0)∥f(k|X|0)>,

where 1/(iz−Qˆ(k)ℒˆS†)is the inverse operator of iz−Qˆ(k)ℒˆS†(see exercise 6.9 for mathematical details). The operator 1/(iz−pˆ(k)ℒˆS†)is usually referred to as the modified resolvent operator, while 1/(iz−ℒˆS†)is referred to as the resolvent operator. We wish to express the modified resolvent operator entirely in terms of the resolvent operator, and use that result to obtain an expression for the memory function in terms of the resolvent operator only. First of all, it is easily verified that,

1iz−Qˆ(k)ℒˆS†−1iz−ℒˆS†=−1iz−ℒˆS†pˆ(k)ℒˆS†1iz−Qˆ(k)ℒˆS†·

For brevity we shall use the following short-hand notation for the resolvent and modified resolvent operator, respectively,

Aˆ=1iz−ℒˆS†,  Bˆ=1iz−Qˆ(k)ℒˆS†·

The above identity thus reads, Bˆ=Aˆ−Aˆpˆ(k)ℒˆS†Bˆ·Repeated application of this identity yields,

(6.182)Bˆ=Aˆ−Aˆpˆ(k)ℒˆS†[Aˆ−Aˆpˆ(k)ℒˆS†Bˆ]  =···  =Aˆ−Aˆ[Iˆ−(pˆ(k)ℒˆS†Aˆ)+(pˆ(k)ℒˆS†Aˆ)(pˆ(k)ℒˆS†Aˆ)−···]pˆ(k)ℒˆS†Aˆ  =Aˆ−Aˆ[Σn=0∞(−1)n(pˆ(k)ℒˆS†Aˆ)n]pˆ(k)ℒˆS†Aˆ·

Since by definition,

pˆ(k)ℒˆS†Aˆf(k|X|0)=ρ(k|X)<ℒˆS†Aˆf(k|X|0)∥ρ(k|X)><ρ(k|X)∥ρ(k|X)>,

it follows that,

[Σn=0∞(−1)n(pˆ(k)ℒˆS†Aˆ)n]pˆ(k)ℒˆS†Aˆf(k|X|0)=<ℒˆS†Aˆf(k|X|0)∥ρ(k|X)><ρ(k|X)∥ρ(k|X)>Σn=0∞(−1)n(pˆ(k)ℒˆS†Aˆ)nρ(k|X)=<ℒˆS†Aˆf(k|X|0)∥ρ(k|X)><ρ(k|X)∥ρ(k|X)>Σn=0∞(−1)n(<ℒˆS†Aˆρ(k|X)∥ρ(k|X)><ρ(k|X)∥ρ(k|X)>)nρ(k|X)=<ℒˆS†Aˆf(k|X|0)∥ρ(k|X)><ρ(k|X)∥ρ(k|X)>+<ℒˆS†Aˆρ(k|X)∥ρ(k|X)>ρ(k|X)·

In the last line the geometrical series is resummed. Use of this expression in eq.(6.182), and subsequent substitution into eq.(6.181) for the Laplace transform of the memory function finally gives the alternative expression we were after,

(6.183)M(k,z)=1<ρ(k|X)∥ρ(k|X)>[<1iz−ℒˆS†f(k|X|0)∥f(k|X|0)>−<ℒˆS†1iz−ℒˆS†f(k|X|0)∥ρ(k|X)><1iz−ℒˆS†ρ(k|X)∥f(k|X|0)><ρ(k|X)∥ρ(k|X)>+<ℒˆS†1iz−ℒˆS†ρ(k|X)∥ρ(k|X)>]·

This expression contains the Smoluchowski operator without being multiplied by the projection operator Qˆ(k),in contrast with the original expression (6.181).

The alternative expression for the self memory function Ms(k, z) is obtained from the above expression by simply replacing f by fs and rH by rH1. Note that <ρ1∥ρ1>=1,while <ρ∥ρ>=NS(k),with S(k) the static structure factor.

1.5.3 Beyond the quadratic polynomials

As noted in the previous paragraph, the homogeneous quadratic polynomials provide the basic mathematical models for first-order optical systems. Notably they represent the only Hamiltonian functions, for which the corresponding transfer operators can be given an exact analytic representation, typically in the form of a finite product of exponential operators. This is true for both z-independent and z-dependent Hamiltonians. Precisely, it is proved that the ray-transfer operator, we may denote by Mˆ2(z, zi), associated with quadratic Hamiltonians is exactly equal, in a neighbourhood of the identity, to the product of the Lie transformations, generated by the three basic monomials q2, p2 and qp. Explicitly we may write

(1.5.30)Mˆ2(z,zi)=eα(z,zi).Lˆq2/2eγ(z,zi).Lˆqpeβ(z,zi).Lˆp2/2,

where the z-depending coefficients α(z, zi), β(z, zi) and γ(z, zi) obey a set of nonlinear first-order differential equations, determined by the specific functional form of the Hamiltonian as a quadratic function of q and p.

Deviations from linearity, signified by powers of q or p higher than two in the optical Hamiltonian, can be accommodated by an approximate representation, which, preserving the factored product structure of the transfer operator, adds factors to the basic quadratic-Hamiltonian generated operator Mˆ2as [12.1]

(1.5.31)Mˆ(z,zi)=ċċċMˆm(z,zi)ċċċMˆ3(z,zi)Mˆ2(z,zi),m>2,

thus yielding a modular, easily manageable, expression for the transfer operator, in which each accuracy-improving factor links to the previously and independently established operator modulus. Each added factor, Mˆm(z, zi), m = 3, 4, …, in the conceptually infinite product (1.5.31) is in the form of a Lie transformation generated by a polynomial function of m-th degree. Clearly the number and the specific expressions of the added factors are determined by the required degree of accuracy and the functional form of the higher-order terms in the Hamiltonian. The analytical and numerical procedures to elaborate the representation (1.5.31) for Mˆ(z, zi) at the desired aberration order m are well documented in the literature [12, 14].

2.9.4 Additional Generators of Galilean Transformations

This section can be skipped on a first reading.

In Sec. 1.3.6, we discussed the generators for Galilean transformations of translations, rotations, and time translations. Here, we return to consider boost and acceleration transformations.

The unitary transformation of a state that corresponds to a boost of the velocity V≡drdt=r˙of a system by velocity v is Uv=eimr⋅v∕ℏ. I.e., the generator for velocity boosts is the quantum operator Q=mr(see Sec. 1.3.6). The operator Qgenerates a displacement of the velocity in the sense [see Eq. (1.48)],

(2.177)eimr⋅v∕ℏVe−imr⋅v∕ℏ=V−v,

or in momentum space,

(2.178)eimr⋅v∕ℏpe−imr⋅v∕ℏ=p−mv,

in a fashion similar to p generating a displacement in coordinate space,7

(2.179)e−ip⋅d∕ℏreip⋅d∕ℏ=r−d.

Note that the boost generator Q commutes with the operator r^, hence, so does the boost Uv,

(2.180)eimr⋅v∕ℏre−imr⋅v∕ℏ=r.

Note further that the boost operator changes the kinetic energy as follows,

(2.181)eimr⋅v∕ℏp22me−imr⋅v∕ℏ=(p−mv)22m,

In order to better understand how a boost transformation works, let us apply Uv≡eimr⋅v∕ℏto the plane wave state ψp(r,t)=Ceip⋅r∕ℏ−iEt∕ℏ:

(2.182)Uvψp(r,t)=eimr⋅v∕ℏCeip⋅r∕ℏ−iEt∕ℏ=Cei(p+mv)⋅r∕ℏ−iEt∕ℏ.

We conclude that the boost transformation operator Uv≡eimr⋅v∕ℏis a unitary operator that changes the momentum of the state of the system from p to p + mv (not p − mv). Since the kinetic energy of a plane wave ei(p+mv)⋅r∕ℏis

E′=(p+mv)22m=E+p⋅v+mv2∕2, we can write (2.182) as

(2.183)eimr⋅v∕ℏψp(r,t)=ei(p⋅v+mv2∕2)t∕ℏψp+mv(r,t).

We have used the fact that ψp+mv(r,t)=Cei(p+mv)⋅r∕ℏ−iE′t∕ℏto obtain the final equality in (2.183). Similarly, U−v=e−imr⋅vapplied to ψp(r,t)gives ei(−p⋅v+mv2∕2)t∕ℏψp−mv(r,t). These boosts affect the momentum (and velocity) rather than the position.

The unitary transformation that boosts the velocity V≡drdt=r˙of a state by velocity v and translates it by vt is

(2.184)UG=ei(mr−pt)⋅v∕ℏ.

The generator of this type of velocity boost is the operator G≡mr−pt. The operator G generates a displacement in velocity space in the sense,

(2.185)eimG⋅v∕ℏVe−iG⋅v∕ℏ=V−v,

or in momentum space,

(2.186)eiG⋅v∕ℏpe−iG⋅v∕ℏ=p−mv,

and generates a displacement in coordinate space,

(2.187)eiG⋅v∕ℏre−iG⋅v∕ℏ=r−vt.

Note that UG≠eimr⋅v∕ℏe−ipt⋅v∕ℏsince r and p do not commute. To transform the wave function ψ(r,t)into a frame moving with velocity v and displaced by the vector vt, one applies UGto the wave function.

If, in a coordinate system undergoing rotational acceleration, such as the coordinate axes rotating with the Earth, the position, velocity, and acceleration of a particle is given by r, v=r˙, and a=v˙=r¨, the velocity and acceleration of the particle in the space-fixed (inertial) coordinate system is vsf=v+Ω×rand asf=a+Ω×vsf=a+2Ω×v+Ω×(Ω×r). Here, Ωis the angular velocity of the rotating coordinate system. There are, of course, consequences if we want to describe the particle in the rotating coordinate system, e.g., the force in the accelerating coordinate system is given by F=Fsf−2mΩ×v−mΩ×(Ω×r), where Fsfis the force in the space-fixed coordinate system, the last term in this equation is the centrifugal force, and the next to last term is the Coriolis force. In quantum mechanics, the Hamiltonian in a rotating frame of reference also includes extra terms, Coriolis and centrifugal terms. The transformation operator to the rotating frame is given by UGin (2.184) with a velocity veffdetermined as follows. The velocity in the inertial frame is r˙sf=r˙+Ω×r, where Ωis the angular velocity, and Ω×ris the effective velocity of the rotating coordinate system, veff≡Ω×r. Hence, using (2.184), we find UG(veff)=eiG⋅veff∕ℏ=ei(mr−pt)⋅(Ω×r)∕ℏ=e−iΩ⋅(r×p)t∕ℏ; and the rotating frame,

(2.188)ψ′(r,t)=UGψ(r,t)=e−iΩ⋅Lt∕ℏψ(r,t).

The (time-dependent unitary transformation) operator UG=e−iΩ⋅Lt∕ℏapplied to the wave function gives the new wave function in the frame undergoing rotational acceleration. As we shall learn in Sec. (2.7)[see Problem 2.19, Eq. (2.112)], the Hamiltonian of the system in the rotating frame is given by

(2.189)H′(r,t)=UGHUG†+iℏ∂UG∂tUG†=UGHUG†+Ω⋅L.

Coriolis and centrifugal terms will in general be present in this Hamiltonian.

In the problem below, you will introduce transformations of the form r→r′=r+ξ(t)into the Schrödinger equation for general ξ(t)and determine the solution in the transformed frame. Then you will compare your result with the application of a unitary transformation of the wave function.

Problem 2.31(a)

Consider the Schrödinger equation iℏ∂ψ∂t=−ℏ22m∇2ψ+V(r,t)ψ. Make the following transformation:

r→r′=r+ξ(t),t→t′=t.

Consider explicitly the following three cases:

r→r′=r+R,r→r′=r+vt,r→r′=r+at2∕2,

Write the wave function as ψ(r,t)=u(r′,t)ei[f(r′,t)], and choose

f(r′,t)=(m∕ℏ)−ξ˙⋅r′+12∫t(ξ˙(t′′))2dt′′.

Find the equation satisfied by u(r′,t).

Show that UGψ, with v(t)=ξ˙(t), is consistent with your results in (a) and (b).

2.2 Rotations and Angular Momenta

Angular momentum operators have been defined in Section 1.3 on the basis of the commutation rules (1.3-1). Another important aspect of angular momentum operators is their close relationship to rotations.

Consider once more the coordinate rotation (2.1-5), but instead of a finite rotation angle α about the z axis, let the angle be an infinitesimal one, δα. In that case, to first order,

(2.2-1)x′=x+yδα,y′=−xδα+y,z′=z,

or

(2.2-2)r′=r(δα,z)r,

where

(2.2-3)r(δα,z)=(1δα0−δα10001).

The operator PR(δδ, z) associated with the coordinate transformation R(δα, z) gives

(2.2-4)PR(δα,z)f(r)=f[R−1(δα,z)r].

Since R(δα, z) is an orthogonal matrix, R−1(δα, z) is simply the transpose of R(δα, z), so that

(2.2-5)PR(δα,z)f(r)=f(x−yδα,y+xδα,z).

To first order in δα,

(2.2-6)PR(δα,z)f(r)=f(r)−yδα∂f(r)∂x+xδα∂f(r)∂y=f(r)+δα(x∂∂y−y∂∂x)f(r).

But according to (1.1-3c),

(2.2-7)x∂∂y−y∂∂x=iLz,

therefore,

(2.2-8)PR(δα,z)f(r)=(1+i δα Lz)f(r).

For a finite rotation through an angle α, it is merely necessary to subdivide α into n δα in such a way that α remains constant as n → ∞ and δα → 0. Then

(2.2-9)PR(α,z)=limn→∞,δα→0PR n(δα,z)=limn→∞,δα→0(1+iδαLz)n=eiαLz

in which the exponential operator is to be understood as a power series

(2.2-10)eiαLz=1+iαLz+12!(iαLz)2+⋯.

The development leading to (2.2-9) may be repeated for any rotation axis whose direction is specified by a unit vector

(2.2-11)PR(δω,n^)=1+iδω  n^⋅L,

(2.2-12)PR(ω,n^)=eiωn^⋅L,

where δω and ω are the infinitesimal and finite angles of rotation, respectively. The connection between rotation and orbital angular momentum is contained in (2.2-12). As an extension it is possible to define a general angular momentum operator J (which includes L as a special case) as a vector operator that satisfies

(2.2-13)PR(ω,n^)=eiωn^⋅J,

and whose components Jx, Jy, and Jz are Hermitian. It follows at once from the latter property that PR is a unitary operator. Also, by considering two rotations in succession and then reversing the order it may be shown that the commutation rules J × J = iJ follow directly from (2.2-13).

If

(2.2-14)[ℋ,PR(α,z)]=0

or

(2.2-15)PR(α,z)ℋPR−1(α,z)=ℋ,

which shows that if an operator commutes with Jz, it is invariant under a rotation about the z axis. Generalizing this result to a rotation about an arbitrary axis, it may be said that an operator which commutes with all the components of J is invariant under rotations. Such an operator is a scalar operator. Conversely, a scalar operator commutes with all components of J.

It is possible to express R(ω,

4.5.2 Diffusion on the Brownian Time Scale

The Smoluchowski equation for non-interacting particles is a relatively simple equation, which can be solved without having to resort to the preceding mathematical subsection.

Let us first calculate the structure factor from eq.(4.8). The pdf P(X = r) in eq.(4.8) is the equilibrium pdf, which is equal to Peq(r) ≡ 1/V, with V the volume of the system. The fortunate fact that allows the explicit evaluation of the integral in eq.(4.8) is that the function exp{ik · r} × Peq is an eigenfunction of the Smoluchowski operator. The Smoluchowski equation (4.42,43 without the external potential reduces to,

(4.62)∂∂tP(r,t)=D0∇r2P(r,t).

The initial condition here is,

(4.63)P(r,t)=δ(r−r0),

where r0 = r(t = 0) is the initial value of the position coordinate of the Brownian particle. It is easily verified that,

LˆS0(exp{ik·r}×Peq)≡D0∇r2(exp{ik·r}×Peq)         =−D0k2(exp{ik·r}×Peq)

Since the operator exponential is formally defined by its Taylor expansion, this implies that,

exp{LˆS0t}(Exp{ik·r}×Peq)=exp{−D0k2t}(exp{ik·r}×Peq)

According to eq.(4.8), the dynamic structure factor is thus simply equal to,

(4.64)Ss(k,t)=exp[−D0k2t}.

Alternatively, the dynamic structure factor can be calculated from eq.(4.7), which reads for the present case,

(4.65)Ss(k,t)=∫ dr01Vexp{ik·r0}∫ drexp{−ik·r}P(r,t)|r0,t=0).

The integral with respect to r is nothing but the Fourier transform of the conditional pdf with respect to r, which is easily calculated from the Fourier transformed Smoluchowski equation (4.62),

(4.66)∂∂tP(k,t)=−D0k2P(k,t).

The initial condition for the Fourier transform follows from eq.(4.63),

P(k,t=0)=∫ drexp{−ik·r}δ(r−r0)=exp{−ik·r0}.

The solution of eq.(4.66) subject to this initial condition is,

(4.67)P(k,t)=exp{−ik·r0}exp{−D0K2t}.

Substitution of this result into eq.(4.65) immediately leads to the result (4.64) for the dynamic structure factor.

The solution of the Smoluchowski equation can be obtained from the expression (4.67) for its Fourier transform, by Fourier inversion,

(4.68)P(r,t|r0,t=0)=1(2π)3∫ dkexp{ik·r}P(k,t)      =1(4πD0t)3/2exp[−|r−r0|24D0t],

which is precisely the result that we found on the basis of the Langevin equation in chapter 2 (see eq.(2.39)).

An expression for the mean squared displacement <| r0 − r(t) |2> can be found in several alternative ways. The first method is simply the integration of | r0 − r |2 × the pdf in eq.(4.68). Secondly, the mean squared displacement can be found from the dynamic structure factor by expanding the defining equation (4.47) with respect to the wavevector (see also exercise 3.9),

(4.69)Ss(k,t)=1−16k2<|r(t=0)−r(t)|2>+⋯.

Comparison with eq.(4.64) for Ss yields,

(4.70)<|r(t=0)−r(t)|2=6D0t,

in accordance with the result obtained in chapter 2 (see eq.(2.21)). A third way to calculate the mean squared displacement is directly from the equation of motion for the pdf. Multiplying both sides of eq.(4.62) with r and r2, respectively, and integration with respect to r yields (see exercise 4.1),

(4.71)ddt<r(t)>=0,

(4.72)ddt<r2(t)>=6D0.

Integration leads to the result in eq.(4.70).