Newton's Laws and QM

[amol12882]

How can I prove Newtons Laws of Motion using Shrodinger's Equation ?

[Aeschylus]

First of all I'll say it's axiomatic to quantum theory that the operators of observables have the same relationship to each other as the observables do in classical physics

 

For exmaple the momentum operator and the part of the Hamiltionan operator correspondign to kinetic enrgy are related by:

 

[math]-\frac{\hbar^2}{2m}\nabla^2 = \frac{1}{2m}\hat{P}^2[/math]

 

which compares to:

 

[math]K.E. = \frac{p^2}{2m}[/math]

 

Actually showing the clasical limit of quantum mechanics is alot harder and a lot more advanced, but I imagine you would start to see it if you choose a large mass for the object you want to describe.

[fuhrerkeebs]

You can't prove Newton's laws of motion using Schrodingers equation. In fact, Schrodingers equation directly contradicts Newton's laws of motion. But, if you solve the time-dependent schrodingers equation for a certain potential, you should find that the evolution of the wave function with regards to time should have higher regions NEAR where Newton's laws predict them. It's much easier to show Newton's laws are approximations using GR, however...

[amol12882]

I have tried to solve the problem of "Stone fall under gravity",

I took V = mgy in Time dependent Shrodinger's Equation and try to find,

Psi(y,t), in this problem I have neglected the wave travelling in +ve 'Y' Direction because stone will fall downwards.

I got a normalization constant as "i*hbar/y" then |Psi(y,t)|^2 is infinite at y=0,

which says that stone will be at ground and it is quite obvious isn't it?

[fuhrerkeebs]

No no no...you must not have done it right...your not going to get a complex valued normalization constant, nor are there supposed to be infinities in the solution.

[fuhrerkeebs]

Your equation should have been:

 

[math]-\frac{\hbar^2}{2m} \frac{d\psi^2}{dx^2} + mgx\psi = i\hbar \frac{d\psi}{dt}[/math]

 

I'm not gonna try and solve that right now...

[fuhrerkeebs]

But you have to remember that this won't work with a stone anyways, Schrodingers equation describes electron movement...

[amol12882]

It's My mistake, Normalization constant is "-i*hbar/y", after we do

Psi(y,t)xPsi*(y,t) then that comes out be hbar^2/y^2.

Why Stone can not be treat as electron in a potential mgy?

[Aeschylus]

Schroedinger's equations describes any particle, all you need to do is simply feed the informtation such as the potential and mass into the Schroedinger equation and solve it in the usual manner.

 

rememebr tho' make sure the wavefunction meets it's boundary conditions.

[fuhrerkeebs]

Amol...what was the solution you found? Oh, and the stone cannot be treated as a particle because a particle is a microscopic object and a stone is a macroscopic one.

[Aeschylus]

That is probably what those who originally formulated the Copenhagen interpretatipn would of said, but it has been the orthodox view in QM fdor many years now that QM applies at all scales (we do not see quantum effects at the macroscopic level mainly due to statistical effects).

[amol12882]

Fuhrerkeebs,

You are right that normalization constant can not be imaginary. What I have done is while normalization I took limit as y to 0 therefore I got imaginary Constant but It can be corrected if I take limit as 0 to y.

I got

Psi(y,t) = A*Sin(ky)+B*Cos(ky) ; k is constant

Using Boundry condition as

Psi = 1 at y=0; because Stone should be at ground

and dPsi/dt = 0 at y=0;

I got B=1 and A=0;

After Solving I got Normalization constant as "2/y"

Is it right?

[amol12882]

Aeschylus Is right. Today QM can also applicable to Macroscopic Word.

Hence I think we can prove Newton's Laws of Motion using Shrodinger's Equation.

[Severian]

The Schrodinger Equation is:

 

[math]i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2 m} \frac{\partial^2 \Psi}{\partial x^2} + V(x) \Psi[/math]

 

Now, since the momentum operator is [math]\hat p = -i \hbar \frac{\partial}{\partial x}[/math] and the energy operator is similarly [math]\hat E = i \hbar \frac{\partial}{\partial t}[/math], the Schrodinger equation is just:

 

[math]\hat E \Psi = \frac{\hat p^2}{2m} \Psi +V(x) \Psi[/math]

 

This is just conservation of energy. [math]p=mv[/math] (remember Shrodinger Eq is non-relativistic), so we have

 

[math]E=\frac{1}{2}mv^2+V(x)[/math]

 

Carry on as usual....

 

(Can anyone see where I fudged slightly...)

 

Extra credit for anyone who can write down the relativistic version!

[fuhrerkeebs]

(Can anyone see where I fudged slightly...)

 

You didn't include the negative sign on the momentum operator.

 

Extra credit for anyone who can write down the relativistic version!

 

Wouldn't that just be dirac's equation?

[Aeschylus]

I can see a couple of fudges:

 

 

You've change the energy operator into the Hamiltonian operator in the second equation(For historical reasons the operator whose eigenvalues are the energy of the system is called the Hamiltonian operator represented as a 'H' plus the usual notation to indicate that it is a Hermitian operator) which describes the total energy of the system.

 

You've also tacitly assumed that the Hamiltonian is time-independent, i.e. that the total energy of the system is conserved (time-depenednt Hamiltonians are extermely difficult to deal with in general and solutions are usually obtained perturbatively).

 

Also anohter very minor fudge is that your delaing with the one-diemsnional case, but the extension to 3 dimensions is obvious.

 

You've also forgetten to square the momentum operator in your restaemnt of the Shroedinger equation, but I assume that's just a typing error.

 

edited to add: and I've missed the most obvious thing, as F.K points out you've lost a minus sign!

[Aeschylus]

The relativstic version of this I guess would be:

 

[math]\hat{H}\Psi = \sqrt{\hat{P}^2 c^2 + m^2c^4}\Psi[/math]

 

Which is not actually a very useful equation and is only half-way to the deriving the Dirac equation as the Dirac equation can't be derived directly from the Schroedinger equation and the Lorentz transformation, you've got to do a little extra work. Infact it looks more like the Klein gordon equation which is:

 

 

[math]\hbar^2\frac{\partial^2\psi}{\partial t^2} + [c^2\hat{P}^2 + m^2c^4]\psi = 0[/math]

[Severian]

You didn't include the negative sign on the momentum operator.

 

Well spotted' date=' but that and the p[sup']2[/sup] were just typos though. And Aeschylus's H is really just notation. (I'll correct them.)

 

I was really meaning that I switched from operators to normal quantities without a by-your-leave. It isn't a big problem - we could just consider ourselves continually collapsing the wavefunction into four-momentum eigenstates, ie making lots of observations, watching the motion, but to be rigorous we should really solve the differential equation for a superposition of plane wave states....

 

The time dependence is also a little bit of a fudge (as has been pointed out) but since I made V(x) explicitly time independent, it is fair enough I think).

 

Wouldn't that just be dirac's equation?

 

It would be for a fermion, but I was thinking of the Klein-Gordon Equation (for bosons), which is more relevent in this case.

 

(Edit: Aeschylus and I cross-posted here)