# Question On Schrodinger

1. Dec 30, 2004

### Raparicio

Dear friends,

How can I transform this

$$\frac{\partial (\Psi \Psi^*)}{\partial t}=- \frac{\nabla^2 \hbar^2}{2m} (\Psi \Psi^*)+\vec{v}(\Psi \Psi^*)$$

To this?

$$\frac{\partial (\Psi)}{\partial t}=- \frac{\nabla^2 \hbar^2}{2m} (\Psi )+\vec{v}(\Psi)$$

... at one day of the new year... best wishes for all

R.Aparicio

2. Dec 30, 2004

### dextercioby

1.Let $\Psi^{*}(\vec{r},t)=1$
2.If you're struggling to find the Schroedinger's equation from the continuity law for the probability density and the probability current density,i'll say it again:
FORGET IT!!!!!

Besides,Born gave this function:
$$\rho(\vec{r},t)=:\Psi^{*}(\vec{r},t)\Psi(\vec{r},t)$$
,the physical significance of localization density probability at "t",and he found the probability density current knowing Schroedinger' equation.
Without knowing the Schroedinger's equation,u cannot find the probability density current,i.e.u cannot solve this equation
$$\frac{\partial[\Psi^{*}(\vec{r},t)\Psi(\vec{r},t)]}{\partial t}=-\nabla\cdot \vec{j}(\vec{r},t)$$

for 'j' as a function of psi and psi star.

Daniel.

3. Dec 30, 2004

### Raparicio

1.Let $\Psi^{*}(\vec{r},t)=1$

And this form, first ecuation transforms in second?

2.If you're struggling to find the Schroedinger's equation from the continuity law for the probability density and the probability current density,i'll say it again:
FORGET IT!!!!!

Why? In quantums mechanics, schrödinger is a postulate. If anyone could derivate it from continuity law, scrödinger law will be a part of the all fórmula.

Besides,Born gave this function:
$$\rho(\vec{r},t)=:\Psi^{*}(\vec{r},t)\Psi(\vec{r},t)$$
,the physical significance of localization density probability at "t",and he found the probability density current knowing Schroedinger' equation.

It's good time to rethink it!!!

Without knowing the Schroedinger's equation,u cannot find the probability density current,i.e.u cannot solve this equation
$$\frac{\partial[\Psi^{*}(\vec{r},t)\Psi(\vec{r},t)]}{\partial t}=-\nabla\cdot \vec{j}(\vec{r},t)$$

for 'j' as a function of psi and psi star.

Daniel.

Thanks for your time. There's a title advance of my thinking in:

http://www.usuarios.lycos.es/Rufianin/articulo.pdf

page 8, but there's a little mathematical error, that was to elimitante the nabla without including a constant. From Continuity and transport ecuation, coult be de derivation of the schrödinger ecuation, and this form, the Heisemberg formula is little completed. (the only problem is that is in Spanish languaje)

R. Aparicio.

4. Dec 30, 2004

### dextercioby

And this form, first ecuation transforms in second?

Why didn't u try to see???

Why? In quantums mechanics, schrödinger is a postulate. If anyone could derivate it from continuity law, scrödinger law will be a part of the all fórmula.

Schroedinger's equation is a postulate in the Schroedinger's picture.And it cannot be derived.The IV-th postulate of QM in the Schroedinger picture in the Dirac formulation states that state vectors' dynamics obey this law:
$$\frac{\partial |\Psi(t)\rangle}{\partial t}=\frac{1}{i\hbar}\hat{H}|\Psi(t)\rangle$$

Tell me please how would you deduce this equation starting from the continuity law???

It's good time to rethink it!!!

I have no reason to rethink it.I've given you the historically correct version.I'm fully convinced that my arguments are correct.

Thanks for your time. There's a title advance of my thinking in:
http://www.usuarios.lycos.es/Rufianin/articulo.pdf
page 8, but there's a little mathematical error, that was to elimitante the nabla without including a constant. From Continuity and transport ecuation, coult be de derivation of the schrödinger ecuation, and this form, the Heisemberg formula is little completed. (the only problem is that is in Spanish languaje)
R. Aparicio.

I'll discuss you article in my next post.I have to translate it first.

Daniel.

5. Dec 30, 2004

### dextercioby

Let's take it methodically.From the first pages i didn't understand anything,as they were written in Spanish and i thought the idea of picking up a dictionary to translate it was horrible.
On page 7 you have two labeled equations.The one labeled with (A) seems very known to me.It is the continuity law/equation for mass volumic density of the fluid or sometimes called "first equation for fluid dynamics".The formula (B),where did u get it??Did u invent it...??Did u check its history??It comes out of the blue,at least in your article.Can u give a justification for it??
Carry on.What do you mean by this formula??
$$\vec{v}=\frac{\nabla i\hbar}{2m}$$
You claim 'v' is the velocity vector for the particle...I assume quantum particle.Do you know that in QM context the 'velocity' is a rather blurred concept??First of all,it accepts definiton within the Heisenberg picture.What does Heisenberg picture have to do with Schroedinger picture,as long as in your article i haven't seen the unitary operators making the change from one picture to another.
Do you have any idea that nabla is a differential operator?If you did,u might have figured out that
$$\nabla (i\hbar)=\vec{0}$$,as the gradient of a constant...

All considered,my guess is that,beginning with page 8,all Hell breakes loose... :grumpy:

Daniel.

Last edited: Dec 30, 2004
6. Dec 30, 2004

### Raparicio

Let's take it methodically.From the first pages i didn't understand anything,as they were written in Spanish and i thought the idea of picking up a dictionary to translate it was horrible.

I'm sorry about it. I can't translate all it, becouse it could be similar to arapahoes speaking english.

On page 7 you have two labeled equations.The one labeled with (A) seems very known to me.It is the continuity law/equation for mass volumic density of the fluid or sometimes called "first equation for fluid dynamics".The formula (B),where did u get it??Did u invent it...??Did u check its history??It comes out of the blue,at least in your article.Can u give a justification for it??

It's the result of have the schródinger's formula, apply to it the conjugate complex, and rest it. It's the continuity formula in QM, if I'm not wrong.

Carry on.What do you mean by this formula??
$$\vec{v}=\frac{\nabla i\hbar}{2m}$$
You claim 'v' is the velocity vector for the particle...I assume quantum particle.Do you know that in QM context the 'velocity' is a rather blurred concept??

If you compare the continuity formula with his definition, by analogy, this will be a velocity. And more, if you see his unities, they are unities of velocity. I think it's velocity of particle and wave.

First of all,it accepts definiton within the Heisenberg picture.What does Heisenberg picture have to do with Schroedinger picture,as long as in your article i haven't seen the unitary operators making the change from one picture to another.

Becouse there's no necessary. It's what I explain in spanish previous to this: the existence (verificated) of "quantum aether". It's the explanation of the 4 forces, and the unification of the 2 great theories.

Do you have any idea that nabla is a differential operator?If you did,u might have figured out that
$$\nabla (i\hbar)=\vec{0}$$,as the gradient of a constant...

Why do you think that planck constant is really constant? There are evidences that it's constant only in a few materials, in a very restringed conditions, and is the projection of a moment vector in the z axe, not?

All considered,my guess is that,beginning with page 8,all Hell breakes loose... :grumpy:
Daniel.

Its becouse of this that I need your aportation and knowledge. And I'm very agreed to your time and dedication.

R. Aparicio.

7. Dec 30, 2004

### dextercioby

I'm sorry about it. I can't translate all it, becouse it could be similar to arapahoes speaking english.

In that case,"Houston,we have a problem".

It's the result of have the schródinger's formula, apply to it the conjugate complex, and rest it. It's the continuity formula in QM, if I'm not wrong.

Aha...Here's the deal.For a free particle,Schroedinger's equation:
$$\frac{\partial \Psi(\vec{r},t)}{\partial t} =\frac{1}{i\hbar} (-\frac{\hbar^{2}}{2m})\nabla^{2} \Psi(\vec{r},t)$$
leads to the continuity law for the localization probability density.I state that the inverse won't hold.Think logically:U cannot prove A starting from B,if B uses a result given by A.Do you agree???It's like chasing your tail...'A' is Schroedinger's eq and 'B' is the continuity law:u can only go from A->B,because u cannot go from B->A without breaking the elementary laws of logics.Mathematics and physics are based on logics.

If you compare the continuity formula with his definition, by analogy, this will be a velocity. And more, if you see his unities, they are unities of velocity. I think it's velocity of particle and wave.

Of course it corresponds with the units.But iff
$$\hat{\vec{v}}=:\frac{-i\hbar\nabla}{m}\hat{1}$$

Do u see any difference between my formula and yours??Mine is a densly defined unbounded selfadjoint vector operator,while yours is something totally different.

Becouse there's no necessary. It's what I explain in spanish previous to this: the existence (verificated) of "quantum aether". It's the explanation of the 4 forces, and the unification of the 2 great theories.

I don't like this word.U know which.I don't know what u're taking about.Please supply credible references (preferably in English/German/French) to both the theory and the experiment behind this concept.Else,i'm entitled to think it's all ballooney.

Why do you think that planck constant is really constant? There are evidences that it's constant only in a few materials, in a very restringed conditions, and is the projection of a moment vector in the z axe, not?

I'm sorry.I won't agree.I cannot agree.From the triplet of fundamental constants,apparently,or should i say to my knowledge,noone has produced any piece of evidence (theoretical and/or experimental) that [itex] h/\hbar [/tex] would change with space (nonhomogenous scalar field) or/and with time (time variable scalar field).There are both stories and experimental evidence that 'G' and 'c' would.

Its becouse of this that I need your aportation and knowledge. And I'm very agreed to your time and dedication.

R. Aparicio.

You're welcome.

Daniel.

Last edited: Dec 30, 2004
8. Dec 31, 2004

### Philcorp

I'll have to agree that it is maybe foolish to derive the Schrodinger eqn. form the continuity eqn. It is certainly not the way to go about things. You can however derive the continuity equation from the schrodnger equation. You can write the wavefunction as an amplitude times a complex phase, ie $$\psi=fe^{i\phi}$$. If you use this in your continuity equation and compare it to the classical one for a fluid you get a velocity whcih looks like $$v=\frac{\hbar}{m}\nabla\phi$$. This is often doen in areas such as BEC, which is where i have seen it the most. I suppose you could also interpert this as the velocity of the probability current of the quantum system as it evolves.

9. Jan 2, 2005

### Raparicio

Dear friends

Sorry for my mistakes. I'm learning so much with your aportations. I'm only a young student with a very great imagination. I understand the mistake about the diference of the quantic operator and my "no-operator" velocity.

By another hand, i am curious about the operator. If you apply a scalar, is this form, not?

$$v=\frac{\hbar}{m}\nabla\phi$$

With a vector, could it be this form?

$$\vec v =\frac{\hbar}{m}\nabla \vec A}$$

My best reggards, and a happy year of physics.

R. Aparicio.

10. Jan 2, 2005

### Dr Transport

No, the operator $$\nabla$$ is a vector operator. Applying it to a scalar results in a vector, applying it to a vector such as $$\vec{A}$$ , makes the resulting quantity a tensor of order 2. the correct manner to apply this is

$$\vec{v}\phi=\frac{\hbar}{m}\vec{\nabla}\phi$$

and

$$\vec v\vec{A} =\frac{\hbar}{m}\vec{\nabla} \vec A}$$

even in QWuantum mechanics you muxt keep the vectors and scalars straight. To make a scalar from the second operation, just apply the vector operator $$\vec{v}$$ using a dot-product to obtain

$$\vec v\cdot\vec{A} =\frac{\hbar}{m}\vec{\nabla} \cdot\vec A}$$.

11. Jan 2, 2005

### dextercioby

If u had read carefully the post written by philcorp,u would have noticed that
$$\langle x,y,z|\hat{v}|\psi\rangle=-\frac{i\hbar}{m}\nabla [e^{i\phi(\vec{r})}]=\frac{\hbar}{m}[\nabla \phi(\vec{r})] e^{i\phi(\vec{r})}$$

For a vector operator it won't hold.In the LHS u have a vector,but in the RHS u have a gradient from a vector,which is a second rank tensor.

Daniel.

12. Jan 3, 2005

### Raparicio

Thanks For All!!!