# Shifting of the wall problem II

1. Aug 12, 2014

### 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?

Last edited: Aug 12, 2014
2. Aug 12, 2014

### 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.

3. Aug 12, 2014

### tom.stoer

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

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

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.

4. Aug 12, 2014

### DrDu

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.

5. Aug 12, 2014

### 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).

6. Aug 12, 2014

### 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.

7. Aug 12, 2014

### maajdl

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.

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

8. Aug 12, 2014

### maajdl

9. Aug 12, 2014

### tom.stoer

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

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

10. Aug 12, 2014

### maajdl

11. Aug 12, 2014

### tom.stoer

12. Aug 12, 2014

### maajdl

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.

13. Aug 12, 2014

### 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.

14. Aug 12, 2014

### tom.stoer

Yes, I mentioned this in my first post.

15. Aug 12, 2014

### tom.stoer

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.

16. Aug 12, 2014

### maajdl

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".

17. Aug 12, 2014

### tom.stoer

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.

18. Aug 12, 2014

### Jilang

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?

19. Aug 12, 2014

### maajdl

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.

Last edited: Aug 12, 2014
20. Aug 12, 2014

### tom.stoer

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

21. Aug 13, 2014

### DrDu

22. Aug 13, 2014

### DrDu

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

23. Aug 13, 2014

### tom.stoer

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.

24. Aug 13, 2014

### tom.stoer

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.

25. Aug 13, 2014

### 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

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