Schrödinger equation and equivalence principle

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hellfire

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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

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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

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

dextercioby said:
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.
 
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dextercioby

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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. :redface:

Daniel.
 
M

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:
[tex]\mathcal{L} = \sum_{\imath} \frac{1}{2} m (\dot{q})^2 + V(q) [/tex]
where [tex]q' = q - 1/2 a t^2 [/tex] 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.
 
See the nice paper arxiv quant-ph/0105074, Pravabati Chingangbam and Pankaj Sharan, 2001: Pseudo forces in QM.

Seratend.
 

hellfire

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

I will take:

[tex] \psi = e^{\lambda} \bar{\psi}[/tex]
[tex]x = \bar{x} + \frac{1}{2} a t^2[/tex]

Then I assume this is the way to proceed:

[tex]L = \frac{1}{2} m \dot{x}^2 = \frac{1}{2} m (\dot{\bar{x}} + at)^2[/tex]

[tex]H = \bar{p} \dot{\bar{x}} - L[/tex]

with

[tex]\bar{p} = \frac{\partial L}{\partial \dot{\bar{x}}} = m (\dot{\bar{x}} + at)[/tex]

Thus:

[tex]H = \frac{1}{2} m (\dot{\bar{x}}^2 - a^2 t^2)[/tex]

and

[tex]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}[/tex]

At the end, and according to page 4 of that paper, I should make a choice for [tex]\lambda[/tex] such that the last expression reduces to the SE in a gravitational potential for [tex]\bar{\psi}[/tex], right?
 
hellfire said:
Thank you for your answers. So, putting things together (please correct me if I am wrong).

I will take:

[tex] \psi = e^{\lambda} \bar{\psi}[/tex]
[tex]x = \bar{x} + \frac{1}{2} a t^2[/tex]
...
At the end, and according to page 4 of that paper, I should make a choice for [tex]\lambda[/tex] such that the last expression reduces to the SE in a gravitational potential for [tex]\bar{\psi}[/tex], 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

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

[tex]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}[/tex]

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

[tex]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}[/tex]

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

Seratend.
 

hellfire

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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:

[tex]H = \frac{1}{2} m (\dot{\bar{x}}^2 - a^2 t^2)[/tex]

is equivalent to:

[tex]H = \frac{\bar{p}^2}{2m} - at\bar{p}[/tex]

with [tex]\bar{p} = m(\dot{\bar x} + at)[/tex]

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

MalleusScientiarum

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

MalleusScientiarum

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

hellfire

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

[tex] \psi = e^{\lambda} \bar{\psi}[/tex]
[tex]x = \bar{x} + \frac{1}{2} a t^2[/tex]

I get the hamiltonian I mentioned above and the SE:

[tex]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}[/tex]

This can be written as:

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

With:

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

which gives [tex]V = m a \bar{x}[/tex]

Note, however, that my [tex]\lambda[/tex] 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.
 
hellfire said:
[tex]x = \bar{x} + \frac{1}{2} a t^2[/tex]

Note, however, that my [tex]\lambda[/tex] 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 [itex]x = \bar{x} + \frac{1}{2} a t^2[/itex] instead of [itex] \bar{x} = x + \frac{1}{2} a t^2[/itex] (see (39)) hence the difference of sign in [itex]\lambda [/itex].

Seratend.
 

hellfire

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OK! Thank you for your help.
 

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