Schrödinger equation and equivalence principle

[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.

 

[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.

 

[hellfire]

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

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

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

 

[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.

 

[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.

 

[seratend]

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

Seratend.

 

[hellfire]

Thank you for your answers. So, putting things together (please correct me if I am wrong).

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?

 

[seratend]

Thank you for your answers. So, putting things together (please correct me if I am wrong).

I will take:

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

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.

 

[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...

 

[seratend]

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

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

It looks weird...

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

Seratend.

 

[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...

 

[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

 

[MalleusScientiarum]

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

 

[seratend]

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

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.

 

[MalleusScientiarum]

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

 

[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.

 

[seratend]

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

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.

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.

 

[hellfire]

OK! Thank you for your help.