Shifting of the wall problem II

[DrDu]

I forgot this effect.

So even if the operator can be calculated it doesn't help, simply b/c it's the time evolution operator of a DIFFERENT physical system!

Thanks to DrDu for the hint.

With the danger of repeating myself: I would call the ansatz you did chose for the time dependent wavefunction a development into a basis of adiabatic wavefunctions, namely the instantaneous eigenfunctions $u_n(x,t)$ of $\frac{1}{2m(t)} p^2$. Compare now your expression for $\dot{u}_n$ with the effect of applying $\dot{\lambda}xp$ (which is kind of a nonadiabatic coupling) onto $u_n$.

 

[tom.stoer]

Compare now ...

not now; tomorrow ;-)

 

[DrDu]

I was thinking about your initial approach. I may not be the most efficient method to calculate the time dependence but it should work in principle
After scaling $u_n(x)=\sin(\pi n x/L_0)$, so the adiabatic basis functions aren't time dependent. But the energies $E_n$ are time dependent due to the time varying effective mass.
One can change to an interaction picture so that $V\propto \{x,p\}/2$. This is not completely trivial as the zeroth order hamiltonian $p^2/2m(t)$ is time dependent. But apparently, this can be taken care off by modifying the effective mass. (To be more precise, in the transformation to the interaction picture, new terms $\propto p^2$ appear. So this process has to be repeated.)
Now doing first order perturbation theory, the wavefunction becomes
$\psi(t)=\exp(i \int_0^t dt\, E_n(t)) u_n+ \sum_m \langle u_m|V|u_n \rangle |u_m\rangle \int_0^t dt''\,\exp(i \int_0^{t''}dt'\,(E_m(t')-E_n(t')))\ \dot{\lambda}(t'')$.
The matrix elements of V fall off very slowly $\sim 1/m$ as $Vu_n$ does not vanish at the boundary. What saves us is the rapidly oscillating time integral which ascertains that the wavefunction falls off smoothly to zero at the boundaries. This involves some variant of the Riemann-Lebesque lemma. The decisive point is that the limits t -> 0 and m to infinity don't commute. If this were not the case, we could replace the exponentials in the first order correction by 1.
Basically, the boundary violating cos is replaced by a truncated Fourier series. This truncated cos will fall off to 0 near the boundaries on a length scale $\propto 1/t$.

 

[tom.stoer]

I think your approach to consider the unitary transformation is the best one:
- to see that the time-derivative is modified, and that it's not the same as simply introducing a time-dep. mass!
- to map the different intervals to one single interval, i.e. of uses a unique Hilbert space!
- to identify the new interaction potential

 

[stevendaryl]

I think your approach to consider the unitary transformation is the best one:
- to see that the time-derivative is modified, and that it's not the same as simply introducing a time-dep. mass!
- to map the different intervals to one single interval, i.e. of uses a unique Hilbert space!
- to identify the new interaction potential

This approach with time-dependent masses seems very artificial to me. If you think of the moving, impenetrable walls as the limiting case of a time-dependent potential:

V(x) = 0 for 0 \leq x \leq L(t)
V(x) = V_0 for x < 0 or x > L(t)

(then you take the limit as V_0 \Rightarrow \infty)

That seems a lot less artificial, and there's no worry about the Hilbert space, or the mass, changing underneath you.

 

[DrDu]

That seems a lot less artificial, and there's no worry about the Hilbert space, or the mass, changing underneath you.

A mathematical physicist would answer you that your limit is ill defined or that it will take a lot of work to give sense to it.

 

[stevendaryl]

A mathematical physicist would answer you that your limit is ill defined or that it will take a lot of work to give sense to it.

Really? I would think it would be less work than the approach that is being discussed.

 

[tom.stoer]

This approach with time-dependent masses seems very artificial to me.

The time-dependet mass plus the additional interaction term are consequences of the rescaling implemented as a unitary transformation between Hilbert spaces; these two formulations are strictly equivalent mathematically.

 

[tom.stoer]

I would like to present a sketch of the approach via the unitary transformation; please have a look.

We start with the scale transformation U generated by G:

U[\lambda] = e^{iG\lambda}
G = (px+xp)/2

with time-dep. lambda, such that

x^\prime = UxU^\dagger = x\,e^\lambda
p^\prime = UpU^\dagger = p\,e^{-\lambda}

results in a fixed interval

x^\prime \in [0,L_0]

The problem is equivalent to a Hamiltonian

H_1 = \frac{p^2}{2me^{2\lambda}}

with time-dep. mass plus an additional term

H_2 = - iU\partial_tU^\dagger = -G\dot{\lambda}

The ansatz for a solution of the time-dep. problem is

|n,t\rangle = e^{-i\int_0^t d\tau \, E_n(\tau)}\,|n\rangle
|\psi,t\rangle = \sum_n a_n(t) \, |n,t\rangle

with momentum eigenstates

p|n\rangle = k_n|n\rangle
E_n = \frac{k_n^2}{2me^{2\lambda}}

We have to solve the following coupled differential equations:

i\dot{a}_m + \dot{\lambda}\,\sum_n\langle m|G|n\rangle\,a_n = 0

which is formally

i\dot{A} = -\dot{\lambda}gA

with

A = (a_n)
g = \langle m|G|n\rangle = (k_m + k_n) \, \langle m|x|n \rangle \, / \, 2

The formal solution is

A(t) = e^{i\int_0^t d\tau \,\dot{\lambda} g}\,A_0 = e^{i(\lambda - \lambda_0) g}\,A_0

and therefore

|\psi,t\rangle = \sum_n e^{-i\int_0^t d\tau \, E_n(\tau)} \left[ e^{i(\lambda - \lambda_0) g}\,A_0 \right]_n |n\rangle

This basically corresponds to a kind of interaction picture with "free, time-dep." H_1 plus interaction term H_2 ~ G. The first term does not cause transitions between "instantaneous eigenstates".

The "instantaneous eigenstates" of the full, time-dep. Hamiltonian are constructed as follows

(H-E)|u,t\rangle = 0
|u,t\rangle = \sum_n u_n(t)|n,t\rangle

with eigenvalue equation for E

(E_m - E)u_m - \dot{\lambda}\sum_n e^{i\int_0^t d\tau\,(E_m-E_n)} \, \langle m|G|n\rangle \, u_n = 0

 

[DrDu]

Wonderful. The use of momentum eigenfunctions gives easy expressions for the matrix elements g.
But I have some remarks:
1. |n,t> and |-n,t> must for all t combine into a sin function which vanishes at the boundaries.
2. In the differential equation for the a's, don't you have to use the time dependent momentum functions, i.e. <m,t|G|n,t> instead of <m|G|n>? As I tried to show in my last post, the time dependent oscillations are important in preventing boundary condition violation. Namely, the elements <m|x|n> would behave something like 1/(m+n). Together with the prefactor (n+m) you can see that all elements are of order 1.

 

[tom.stoer]

Wonderful.

Thanks

|n,t> and |-n,t> must for all t combine into a sin function

Yes, you are right; this constrains the coefficients a_n and u_n.

In the differential equation for the a's, don't you have to use the time dependent momentum functions, i.e. <m,t|G|n,t> instead of <m|G|n>?

Yes, again you are right; I overlooked this. It affects the definition of g.

=> We have to solve the following coupled differential equations:

i\dot{a}_m + \dot{\lambda}\,\sum_n\langle m,t|G|n,t\rangle\,a_n = 0

which is formally

i\dot{A} = -\dot{\lambda}gA

with

A = (a_n)
g = \langle m,t|G|n,t\rangle = e^{i\int_0^t d\tau\,(E_m-E_n)} \, \langle m|G|n\rangle = (k_m + k_n) \, e^{i\int_0^t d\tau\,(E_m-E_n)} \, \langle m|x|n \rangle \, / \, 2

This is what I can see at a first glance, but I will double check tonight.

 

[tom.stoer]

... in preventing boundary condition violation ...

Think about a related problem L²[S¹] instead of L²[0,1], i.e. the same problem but now with periodic boundary conditions. In this case something like an operator x simply does not exist, b/c u(x) = 1 ist nice but x * u(x) isn't [in our case x is not a desaster b/c x * sin() ist still OK]. So on S¹ it's harder to define the problem b/c of G ~ px+xp, am I right? How do you define x, G and the scaling on a circle?

 

[DrDu]

This looks good. Assuming n to be small and given that $E_m=\pi^2 m^2/2L^2M$ the oscillating integral will act as a cut off effectively when $E_nt\approx 1$, i.e. $k_n=\sqrt{2ME}\approx\sqrt{2M/t}$ or $\lambda=\sqrt{t/M}$ . That's the length scale where the term $x/2\, \cos( \pi x/L)$ falls off to 0 at the boundary.

 

[DrDu]

Think about a related problem L²[S¹] instead of L²[0,1], i.e. the same problem but now with periodic boundary conditions. In this case something like an operator x simply does not exist, b/c u(x) = 1 ist nice but x * u(x) isn't [in our case x is not a desaster b/c x * sin() ist still OK]. So on S¹ it's harder to define the problem b/c of G ~ px+xp, am I right? How do you define x, G and the scaling on a circle?

That's true, on a circle, phi always runs from 0 to 2π. The scaling is effected scaling the radial coordinate. Instead of scaling the mass, you scale the moment of inertia $mr^2$

 

[tom.stoer]

That's true, on a circle, phi always runs from 0 to 2π. The scaling is effected scaling the radial coordinate. Instead of scaling the mass, you scale the moment of inertia $mr^2$

What I mean is that on the circle the naive position operator x is ill-defined, and that therefore the operator xp+px is problematic, too. Have a look at the ideas in (6) in

http://arxiv.org/pdf/quant-ph/0010064.pdf

 

[tom.stoer]

One hint:

\exp\left[i\int_0^t d\tau\,(E_m-E_n)\right]= \exp\left[\frac{i(k_m^2 - k_n^2)}{2m}\,\int_0^t d\tau\,e^{-2\lambda}\right]<br /> <br />

 

[DrDu]

What I mean is that on the circle the naive position operator x is ill-defined, and that therefore the operator xp+px is problematic, too. Have a look at the ideas in (6) in

http://arxiv.org/pdf/quant-ph/0010064.pdf

There are numerous articles which deal with how to define an approximate angle operator. The relevance is not quite clear to me, as usually you can live without this operator. As I tried to show, you also don't need an angular scaling operator.