Shifting of the wall problem II

[tom.stoer]

I though there's a simple ansatz for a particle in an infinite square well with a moving wall, i.e. in the interval [0,L(t)].

The eigenfunctions and eigenvalues for

\left[-\frac{1}{2m}\frac{\partial^2}{\partial x^2} - E_n\right]\,u_n(x) = 0

are

u_n(x) = \sin\left(\frac{2\pi n}{L}x\right)
E_n = \frac{1}{2m}\left(\frac{2\pi n}{L}\right)^2

When making L time-dependent these equations remain valid. Therefore it seems a good idea to make the following ansatz for the time-dependent Schroedinger equation:

i\frac{d}{dt}\psi(x,t) = \left[-\frac{1}{2m}\frac{\partial^2}{\partial x^2} - E_n\right]\,\psi(x,t)
\psi(x,t) = \sum_n\phi_n(t)\,u_n(x,t)

When acting with the Hamiltonian on psi on the r.h.s. all what happens is that the eigenvalue becomes time-dep. This is fine. When acting with i∂_t on psi we get two terms; there is the time-derivative of phi, which is OK. But the other term

\dot{u}_n(x,t) \sim x\,\cos\left(\frac{2\pi n}{L}x\right)\,\frac{\dot{L}}{L^2}

is problematic b/c the cosine explicitly violates the boundary conditions for x=0 and x=L.

What is wrong with my ansatz? And what would be an alternative?

 

[maajdl]

I suggest to add a cos term to the trial solution and see what happens.
Furthermore, the time-derivative you obtained is not a trig function anymore, nor is it for the other derivatives, because L depends on t.
This suggests keeping all the terms of the Fourier series.
I do not expect a simple solution.

Could it help a little bit to consider an interval [-L(t), L(t)] ?
You might change the variables in order to have a fixed boundary, but a modified Schrodinger equation.
Could also help to assume a linear expansion L(t)=Lo+vt, instead of a general case.

 

[tom.stoer]

I suggest to add a cos term to the trial solution and see what happens.

The cosine is forbidden b/c of the boundary conditions.

Furthermore, the time-derivative you obtained is not a trig function anymore, nor is it for the other derivatives, because L depends on t.

It is a sine in x for each t, that's the important thing; it solves the time-indep. Schroedinger equation for each t.

You might change the variables in order to have a fixed boundary, but a modified Schrodinger equation.

That does not help, unfortunately; define

\chi(t) = \frac{x}{L(t)}

\chi \in [0,1]

and use

u_n(\chi(t)) = \sin(2\pi n\chi(t))

On the r.h.s you get a term like

-\frac{1}{2mL^2(t)} \frac{\partial^2}{\partial \chi^2}

so this just creates a "time dep. mass".

On the l.h.s. you again have to evaluate

\frac{d}{dt}\sin(2\pi n\chi(t)) \sim \cos(2\pi n\chi)\,\dot{\chi}

So this results in exactly the same problematic term; again the time-evolution seems to violate the boundary condition.

 

[DrDu]

That does not help, unfortunately; define

A time dependent mass is not a problem. To avoid the problem with the cos term you have to implement the time dependent scaling of the x variable as a unitary transformation The generator of the scaling is something like $1/2 \{p,x\}$. If the scaling is time dependent there will be an extra term from the transformation of id/dt which will cancel the cosine term.

 

[tom.stoer]

I do not start with a time-dependent mass; it's an artifact of the scaling.

I have to start with a Hamiltonian H(x,p) where the interval [0,L] is time-dep., but where the mass is constant. This is the physical setup. Now I want to introduce the time-dependency of the interval into the time-evolution of the wave-function.

As far as I understand you correctly you propose something like

U[\lambda(t)] = e^{-i(px+xp)\lambda/2}
H \to H^\prime[\lambda] = UHU^\dagger
|\psi\rangle \to |\psi^\prime\rangle = U|\psi\rangle

Of course this unitary trf. maps eigenstates of H onto eigenstates of H', but is this really a "physical" time evolution? Is this not just a trivial rescaling? This would mean that the time-dependent Hamiltonian H' is exactly diagonalized by the rescaled wave functions.

The Hamiltonian with time-dependent mass is not really the problem; the problem is my Ansatz to use the old eigenfunctions but with time-dep. L(t).

 

[DrDu]

Of course it is only a trivial rescaling. But the point I want to make is that you also have to scale id/dt:
$U(\lambda(t))^+ i d/dt U(\lambda(t)) =i d/dt +U^+ i dU/d\lambda \dot{\lambda}$
The second term is proportional to the generator of the scaling, i.e. it contains a p which transforms a sin into a cos. This will cancel the boundary violating terms from d/dt.

 

[maajdl]

On the r.h.s you get a term like
-\frac{1}{2mL^2(t)} \frac{\partial^2}{\partial \chi^2}
so this just creates a "time dep. mass".

You are right, but you are underestimating the impact.
This leads to a differential equation with non-constant coefficients.
Such equations can be very hard to solve and may lead to resonances with oscillating boundaries.

On the l.h.s. you again have to evaluate
\frac{d}{dt}\sin(2\pi n\chi(t)) \sim \cos(2\pi n\chi)\,\dot{\chi}
So this results in exactly the same problematic term; again the time-evolution seems to violate the boundary condition.

There you are wrong.
The time derivative brings up an additional term proportional to the time derivative of L(t).

 

[maajdl]

This might interest you:

http://solar.physics.montana.edu/tarrl/pubs/Lucas.Tarr.Reed.College.Thesis.pdf

 

[tom.stoer]

You are right, but you are underestimating the impact.

I don't. That's why I am asking.

There you are wrong. The time derivative brings up an additional term proportional to the time derivative of L(t).

Please read carefully. χ is defined as x/L, so dχ/dt ~ dL/dt. So this term is contained in my formula.

 

[maajdl]

This also might interest you:

http://highenergy.phys.ttu.edu/~akc...pers/DoescherAndRice-MovingWallsAJP001246.pdf

 

[tom.stoer]

This might interest you:

http://solar.physics.montana.edu/tarrl/pubs/Lucas.Tarr.Reed.College.Thesis.pdf

Sounds very interesting. Thanks for the link!

 

[maajdl]

Please read carefully. χ is defined as x/L, so dχ/dt ~ dL/dt. So this term is contained in my formula.

I didn't read carefully.
I was going back to the Schrodinger equation and figuring out what happens when changing variables in this way:

t' = t
x' = x/L(t)

Doing so, the time derivative on the LHS splits into two terms.
The Schrodinger equation is deeply modified.

 

[maajdl]

Note that the physics must be interesting.
For example, when the well is narrowing, "the energy must be pumped up", by analogy with classical mechanics.
Therefore, it implies transition from lower states to higher states.
Could be approached also by perturbation theory.

 

[tom.stoer]

Doing so, the time derivative on the LHS splits into two terms.
The Schrodinger equation is deeply modified.

Yes, I mentioned this in my first post.

 

[tom.stoer]

Note that the physics must be interesting.
For example, when the well is narrowing, "the energy must be pumped up", by analogy with classical mechanics.
Therefore, it implies transition from lower states to higher states.
Could be approached also by perturbation theory.

I agree, except for perturbation theory; one should try to find better solutions ;-)

The two references you provided are very interesting.

Anyway - what bothers me most is that the time evolution results in terms violating the boundary condition (look at eq. 2.6 in the 1st paper); this sounds counter-intuitive.

 

[maajdl]

Anyway - what bothers me most is that the time evolution results in terms violating the boundary condition (look at eq. 2.6 in the 1st paper); this sounds counter-intuitive.

That's because you didn't find the correct solution.
This will involve transitions between "energies" and this is how -logically- the boundary conditions will be satisfied: mixing of the "pseudo eigenstates".

 

[tom.stoer]

That's because you didn't find the correct solution.

It's in the paper you are mentioning: eq. 2.6 says that the time derivative of the solution (to be constructed) contains these unwanted terms.

In 2.5 everything is fine

\sum_n b_n\,E_n\,u_n(x)\,e^{\ldots}

But in 2.6 we have

\sum_n [\ldots]\,u_n(x)\,e^{\ldots} + \sum_n [\ldots] \, \dot{u}_n(x)\,e^{\ldots}

The \dot{u}_n-term violates the boundary conditions.

So infinitesimal time-evolution generates terms which are no longer manifest elements of the Hilbert space. In order to show that 2.10 is equivalent to the original Schrödinger eq. one must show that this apparent violation of boundary conditions somehow cancels after solving for b_n and re-summation for \psi.

That was the point where I decided to ask whether one can see or prove that these terms cancel in the very end.

 

[Jilang]

Note that the physics must be interesting.
For example, when the well is narrowing, "the energy must be pumped up", by analogy with classical mechanics.
Therefore, it implies transition from lower states to higher states.
Could be approached also by perturbation theory.

maajdl. I agree with this you in this regard. Quantum or not, it's all wave equations (and not a dissimilar model to some wind instruments). Without the energy input how can anything other then total internal reflection result?

 

[maajdl]

That was the point where I decided to ask whether one can see or prove that these terms cancel in the very end.

Sorry for not reading with due attention.
I believe the problem you submitted is not an easy one.
However, it is obvious (just by changing the variables) that this problem is not equivalent to a Schrodinger equation in the usual sense.
For example, the "equivalent Hamiltonian" should not be hermitian, and energy need not be conserved.

In addition, although I find it very interesting, I also thing this is an unphysical problem.
How can we expect to find such a system in nature?
Some system might approach it however.
However, such a system should then be governed by a "standard Schrodinger equation",
and the moving boundary would be then an approximation to such a system.
After all, to push the boundary energy is needed, and this would appear in a full Hamiltonian for the composite system.

Non-hermitian hamiltonian are not necessarily un-physical.
They can be useful approximations.
An example I know (from long ago) is the approximate to interaction between an atom and the electromagnetic field (for a Stark effect for example). In this case, to approximate the spontaneous decay of excited states, it is possible to add a non-hermitian termthat will lead to a decay of the density operator.
This is then a simplified model of an atom, which is convenient to approximate for example the infleunce of an electric field on the energy levels and their decay.

In the case here, the moving boundary is equivalent to a Schrodinger equation with a non-hermitian Hamiltonian.

 

[tom.stoer]

Thanks for your comments.

I agree that it's somehow artificial. Anyway, I was surprised by the complexity.

 

[DrDu]

? To be an element of the Hilbert space, the functions have only to be square integrable. The boundary conditions on the other hand are dictated by the Hamiltonian.

 

[DrDu]

In the case here, the moving boundary is equivalent to a Schrodinger equation with a non-hermitian Hamiltonian.

I really can't follow your conclusions. The scaling can be implemented via a unitary transformation, so clearly, the scaled hamiltonian remains hermitian.

 

[tom.stoer]

? To be an element of the Hilbert space, the functions have only to be square integrable. The boundary conditions on the other hand are dictated by the Hamiltonian.

I am not sure what you mean.

We start with L²[0,1], but in addition we require u(x<0) = u(x>1) = 0; b/c we want u(x) to be continuous we fix the boundary conditions u(0) = u(1) = 0 (I agree that this is somehow dictated by the Hamiltonian).

So physically we do no longer deal with the full Hilbert space L² but with a subset, i.e. a subspace L² \ span{cos(nπx)}; we exclude all these terms from the basis. This is the starting point.

Now for the moving wall [0,L(t)] we can rescale to [0,1] which introduces a time-dep. mass term.

Of course we expect that the time evolution generated by U = exp(-iHt) respects the boundary conditions - regardless where they are coming from. Unfortunately we find that an infinitesimal time evolution creates a term ~ x cos(nπx) which violates the boundary condition for x=L. So it seems that our Hamiltonian generates a time evolution which violates the boundary conditions dictated by this Hamiltonian. This would be a desaster.

Of course the problem is more subtle b/c we need not care about the single terms but only about the complete sum over all terms. For this sum (for the solution for a given L(t)) one has to prove that the time evolution of a function ψ (respecting the boundary conditions for t=0) respects this boundary conditions for later times t.

Perhaps my formulation was missleading, but the problem remains.

 

[tom.stoer]

I really can't follow your conclusions. The scaling can be implemented via a unitary transformation, so clearly, the scaled hamiltonian remains hermitian.

I agree.

The rescaled Hamiltonian has a time-dep. mass term but fixed boundaries [0,1]. The time dependence does not affect the self-adjointness.

 

[tom.stoer]

Perhaps it makes more sense to study the rescaled Hamiltonian h(t) with time-dep. mass on a fixed interval and the time evolution operator

U(t_2,t_1) = T\exp\left[-i\int_{t_1}^{t_2} d\tau\, h(\tau)\right]

I guess the Dyson series can be computed b/c the time dependence is trivial and

[h(t_1),h(t_2)] = 0

 

[maajdl]

I agree.

The rescaled Hamiltonian has a time-dep. mass term but fixed boundaries [0,1]. The time dependence does not affect the self-adjointness.

How do the boundary conditions come into play regarding self-adjointness?

My remark was about the PDE obtained when changing variable in the Schrodinger equation.
Doing so, you get -indeed- a modified mass, but also a additional term.
The equivalent Hamiltonian, I am not sure it is self-adjoint, altough I did not check.
Intuitively I would say it is not.
To be check.
Any candidate to check that?

In summary, self-adjointness could be check into two point of view.

 

[DrDu]

I am a bit lazy to do the full calculations, but the scaled Hamiltonian contains, beside the mass dependent kinetic energy term, a term $\sim\dot{\lambda}(t) (xp+px)/2$ which comes from transforming the time derivative. Of course, this term will transform a sin into a cos when isolated, but you should not argue like this. Rather, you have a new hamiltonian $\frac{a(t)}{2} p^2 +b(t) (xp+px)/2$ which generates the time evolution and whose domain are twice differentiable functions which vanish at the boundary.
Maybe it would be a good idea to complete the square and introduce new momenta $\tilde{p}=p+b(t)/a(t) x$.
This can be accomplished via a unitary transformation $\exp{i c(t) x^2/2}$ where $c\propto b/a$. Summa summarum you get an equation for a time dependent truncated harmonic oscillator. You can then use all the usual methods for solving a time dependent Schroedinger equation.
It might be possible for special time dependencies of the moving wall that the spring constant of the oscillator k(t) becomes proportional to the mass m(t). Introducing the new time $\tau=t/m(t)$, the Hamiltonian becomes time independent and one can look for stationary solutions.

 

[stevendaryl]

I though there's a simple ansatz for a particle in an infinite square well with a moving wall, i.e. in the interval [0,L(t)].

The eigenfunctions and eigenvalues for

\left[-\frac{1}{2m}\frac{\partial^2}{\partial x^2} - E_n\right]\,u_n(x) = 0

are

u_n(x) = \sin\left(\frac{2\pi n}{L}x\right)
E_n = \frac{1}{2m}\left(\frac{2\pi n}{L}\right)^2

When making L time-dependent these equations remain valid. Therefore it seems a good idea to make the following ansatz for the time-dependent Schroedinger equation:

i\frac{d}{dt}\psi(x,t) = \left[-\frac{1}{2m}\frac{\partial^2}{\partial x^2} - E_n\right]\,\psi(x,t)
\psi(x,t) = \sum_n\phi_n(t)\,u_n(x,t)

When acting with the Hamiltonian on psi on the r.h.s. all what happens is that the eigenvalue becomes time-dep. This is fine. When acting with i∂_t on psi we get two terms; there is the time-derivative of phi, which is OK. But the other term

\dot{u}_n(x,t) \sim x\,\cos\left(\frac{2\pi n}{L}x\right)\,\frac{\dot{L}}{L^2}

is problematic b/c the cosine explicitly violates the boundary conditions for x=0 and x=L.

What is wrong with my ansatz? And what would be an alternative?

I'm not sure that there is anything wrong with it, except for the fact that your ansatz only describes the wave function in the region [0,L]. Outside of that region, \psi(x,t) = 0. At the boundary between the regions, \psi must be continuous, but I don't see that its derivatives need to be.

 

[stevendaryl]

I'm not sure that there is anything wrong with it, except for the fact that your ansatz only describes the wave function in the region [0,L]. Outside of that region, \psi(x,t) = 0. At the boundary between the regions, \psi must be continuous, but I don't see that its derivatives need to be.

An alternative approach might be to consider the step potential

V(x) = 0 for 0 \leq x \leq L
V(x) = +V_0 outside that region.

Then you can take the limit as V_0 \Rightarrow \infty

 

[tom.stoer]

Perhaps it makes more sense to study the rescaled Hamiltonian h(t) with time-dep. mass on a fixed interval and the time evolution operator

U(t_2,t_1) = T\exp\left[-i\int_{t_1}^{t_2} d\tau\, h(\tau)\right]

I guess the Dyson series can be computed b/c the time dependence is trivial and

[h(t_1),h(t_2)] = 0

This idea is useless :-(

The time-dep. mass in the Hamiltonian is not the problem. There is another effect due to the unitary trf. of the time derivative which changes the time-dep. Schroedinger eq.:

I am a bit lazy to do the full calculations, but the scaled Hamiltonian contains, beside the mass dependent kinetic energy term, a term $\sim\dot{\lambda}(t) (xp+px)/2$ which comes from transforming the time derivative.

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.

 

[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 &lt; 0 or x &gt; 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 defintion 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.