Schrödinger equation and equivalence principle

1. Jul 6, 2005

hellfire

May be this is a silly question, but if one converts the nonrelativistic Schrödinger equation for a free particle to an uniformly accelerated frame, is the result the same as the Schrödinger equation for a particle within a gravitational potential? I was trying some simple calculations but did not have any success.

2. Jul 6, 2005

dextercioby

Hold on a second, what do you mean by "writing the SE in a uniformy accelerated frame" ?

I've never seen accelerated frames of reference in (nonrelativistic) quantum mechanics.

Daniel.

3. Jul 6, 2005

hellfire

Make a coordinate change:
x' = x - 1/2 a t2
t' = t

Me neither. Thats why I am not sure whether the question is meaninful at all.

Last edited: Jul 6, 2005
4. Jul 6, 2005

dextercioby

You can't really do that, since in QM everything (every observable, that is) except time is a densly defined selfadjoint linear operator on a separable Hilbert space. You can use the coordinate representation in order to make things less abstract, but i still don't see how you can fit this nonewtonian piece of dynamics into quantum mechanics.
I don't know how you can fit it into ordinary classical lagrangian/ hamiltonian dynamics, actually.

I don't know many.

Daniel.

5. Jul 6, 2005

MalleusScientiarum

My guess is that you would go through the standard routine of finding the classical Hamiltonian and turning it into a quantum operator. So you would start out with a new Lagrangian:
$$\mathcal{L} = \sum_{\imath} \frac{1}{2} m (\dot{q})^2 + V(q)$$
where $$q' = q - 1/2 a t^2$$ and figure out the Hamiltonian from the definition of generalized momentum and such.

However, this seems like an unpleasant choice, as it would make your whole Hamiltonian time-dependent, which makes the Schrodinger equation a whole new beast to solve.

6. Jul 7, 2005

seratend

See the nice paper arxiv quant-ph/0105074, Pravabati Chingangbam and Pankaj Sharan, 2001: Pseudo forces in QM.

Seratend.

7. Jul 7, 2005

hellfire

I will take:

$$\psi = e^{\lambda} \bar{\psi}$$
$$x = \bar{x} + \frac{1}{2} a t^2$$

Then I assume this is the way to proceed:

$$L = \frac{1}{2} m \dot{x}^2 = \frac{1}{2} m (\dot{\bar{x}} + at)^2$$

$$H = \bar{p} \dot{\bar{x}} - L$$

with

$$\bar{p} = \frac{\partial L}{\partial \dot{\bar{x}}} = m (\dot{\bar{x}} + at)$$

Thus:

$$H = \frac{1}{2} m (\dot{\bar{x}}^2 - a^2 t^2)$$

and

$$i \hbar \frac{\partial \bar{\psi}}{\partial t} = \left( \frac{- \hbar^2}{2m} \frac{\partial^2}{\partial \bar{x}^2} - a^2t^2 \right) e^{\lambda} \bar{\psi}$$

At the end, and according to page 4 of that paper, I should make a choice for $$\lambda$$ such that the last expression reduces to the SE in a gravitational potential for $$\bar{\psi}$$, right?

8. Jul 9, 2005

seratend

Note exactly. Beginning of section V says that you recover the equivalence principle if you multiply the wave function by an ad hoc phase (potential -mgX'). While, with group symmetry, you recover this solution in a more formal way (cf formula 46 vs 41 p4).

Seratend.

9. Jul 10, 2005

hellfire

There was an error in the last equation of my last post. It should be:

$$i \hbar \frac{\partial \bar{\psi}}{\partial t} = \left( \frac{- \hbar^2}{2m} \frac{\partial^2}{\partial \bar{x}^2} + at \frac{\partial}{\partial \bar{x}} \right) e^{\lambda} \bar{\psi}$$

It looks weird...

Last edited: Jul 11, 2005
10. Jul 11, 2005

seratend

Not so weird : ).
Hint: use the formula (a+b)^2=a^2+2ab+b^2

Seratend.

11. Jul 11, 2005

hellfire

I am sorry, but it seams that the formula was wrong again (I get confused with p and pbar). I have corrected my previous post. The Hamiltonian:

$$H = \frac{1}{2} m (\dot{\bar{x}}^2 - a^2 t^2)$$

is equivalent to:

$$H = \frac{\bar{p}^2}{2m} - at\bar{p}$$

with $$\bar{p} = m(\dot{\bar x} + at)$$

(May be someone could check that and the SE in my previous post). The question is how to proceed then...

Last edited: Jul 11, 2005
12. Jul 11, 2005

seratend

Apply the derivative operator in the hamiltonian H to exp(lamba) using the expression (42) page 4 and verify you recover 42 (correct the errors you have): (40) => (43) when psi(x)=exp(lambda(x',t)), expression (42)

Seratend

13. Jul 11, 2005

MalleusScientiarum

The point being that it such a change only complicates the Schrodinger equation more than it, in general, already is.

14. Jul 11, 2005

seratend

It is just an introduction to the expression of the SE in different frames (earth is a rotating frame => validity of the approximations we are doing in the lab): a constantly accelerated frame introduce the additional potential U(X')= mgX' in the unitary evolution of the state. We can see how it may change the eigenvalues of the hamiltonian.

Seratend.

15. Jul 11, 2005

MalleusScientiarum

I can see already that they will have a time dependence.

16. Jul 11, 2005

hellfire

I made the calculations with some more care and it seams it works. Taking:

$$\psi = e^{\lambda} \bar{\psi}$$
$$x = \bar{x} + \frac{1}{2} a t^2$$

I get the hamiltonian I mentioned above and the SE:

$$i \hbar \frac{\partial}{\partial t} (e^{\lambda} \bar{\psi}) = \left( \frac{- \hbar^2}{2m} \frac{\partial^2}{\partial \bar{x}^2} + i \hbar at \frac{\partial}{\partial \bar{x}} \right) e^{\lambda} \bar{\psi}$$

This can be written as:

$$i \hbar \frac{\partial \bar{\psi}}{\partial t} = \left( \frac{- \hbar^2}{2m} \frac{\partial^2}{\partial \bar{x}^2} + V \right) \bar{\psi}$$

With:

$$\lambda = \frac{i m a}{\hbar} \left( \bar{x} t + \frac{1}{6}a t^3 \right)$$

which gives $$V = m a \bar{x}$$

Note, however, that my $$\lambda$$ differs from the one in quant-ph/0105074 by one sign. Probably I have missed some sign somewhere, but I think I can conclude that the phase factor can be chosen properly to fit with the equivalence principle.

17. Jul 11, 2005

seratend

Very good.

In fact you have choosen $x = \bar{x} + \frac{1}{2} a t^2$ instead of $\bar{x} = x + \frac{1}{2} a t^2$ (see (39)) hence the difference of sign in $\lambda$.

Seratend.

18. Jul 11, 2005

hellfire

OK! Thank you for your help.