Variational Derivation of Schrodinger Equation

In summary, Weinstock explains that Schrodinger derived the Schrodinger equation from variational principles. He also says that it is only after Schrodinger links his work with DeBroglie's work that he says the Schrodinger equation is admissible.
  • #1
bolbteppa
309
41
In reading Weinstock's Calculus of Variations, on pages 261 - 262 he explains how Schrodinger apparently first derived the Schrodinger equation from variational principles.

Unfortunately I don't think page 262 is showing so I'll explain the gist of it:

"In his initial paper" he considers the reduced Hamilton-Jacobi equation

[itex]\frac{1}{2m}[(\frac{\partial S}{\partial x})^2 \ + \ (\frac{\partial S}{\partial y})^2 \ + \ (\frac{\partial S}{\partial z})^2] \ + \ V(x,y,z) \ - \ E \ = \ 0[/itex]

for a single particle of mass [itex]m[/itex] in an arbitrary force field described by a potential [itex]V = V(x,y,z)[/itex]

& with a change of variables [itex]S \ = \ K\log(\Psi)[/itex], (where [itex]K[/itex] will turn out to be [itex] \frac{h}{2\pi}[/itex]) it reduces to

[itex] \frac{K^2}{2m}[(\frac{\partial \Psi}{\partial x})^2 \ + \ (\frac{\partial \Psi}{\partial y})^2 \ + \ (\frac{\partial \Psi}{\partial z})^2] \ + \ (V \ - \ E)\Psi^2 \ = 0[/itex].

Now instead of solving this he, randomly from my point of view, choosed to integrate over space

[itex]I \ = \ \int\int\int_\mathcal{V}(\frac{K^2}{2m}[(\frac{\partial \Psi}{\partial x})^2 \ + \ (\frac{\partial \Psi}{\partial y})^2 \ + \ (\frac{\partial \Psi}{\partial z})^2] \ + \ (V \ - \ E)\Psi^2)dxdydz[/itex]

& then extremizes this integral which gives us the Schrodinger equation.

Apparently as the book then claims on page 264 it is only after this derivation that he sought to connect his ideas to deBroglie's wave-particle duality.

Thus I have three questions,

1) What is the justification for Feynman's famous quote:

"Where did we get that [Schrödinger's equation] from? It's not possible to derive it from anything you know. It came out of the mind of Schrödinger"
The Feynman Lectures on Physics

in light of the above derivation. I note that all the derivations I've seen of the Schrodinger equation doing something like using operators such as [itex]i\frac{h}{2\pi}\frac{\partial}{\partial t} \ = \ E[/itex] to derive it always mention it's merely heuristic, yet what Schrodinger apparently originally did seems like a roundabout way of solving the Hamilton-Jacobi equation with no heuristic-ness in sight. What subtleties am I missing here? Why would I be a fool to arrogantly 'correct' someone who says Schrodinger is not derivable from anything you know?

2) Is the mathematical trick Schrodinger has used something you can use to solve problems?

3) Why can't you use this exact derivation in the relativistic case?
 
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  • #2
There's no subtlety, one has to take what Schroedinger wrote for granted. Likewise for all other articles in 1925-1926.
 
  • #3
This is not the Schrodinger equation. The signs are wrong.
 
  • #4
bolbteppa said:
"In his initial paper" he considers the reduced Hamilton-Jacobi equation

[itex]\frac{1}{2m}[(\frac{\partial S}{\partial x})^2 \ + \ (\frac{\partial S}{\partial y})^2 \ + \ (\frac{\partial S}{\partial z})^2] \ + \ V(x,y,z) \ - \ E \ = \ 0[/itex]

for a single particle of mass [itex]m[/itex] in an arbitrary force field described by a potential [itex]V = V(x,y,z)[/itex]

& with a change of variables [itex]S \ = \ K\log(\Psi)[/itex], (where [itex]K[/itex] will turn out to be [itex] \frac{h}{2\pi}[/itex]) it reduces to

[itex] \frac{K^2}{2m}[(\frac{\partial \Psi}{\partial x})^2 \ + \ (\frac{\partial \Psi}{\partial y})^2 \ + \ (\frac{\partial \Psi}{\partial z})^2] \ + \ (V \ - \ E)\Psi^2 \ = 0[/itex].

Now instead of solving this he, randomly from my point of view, choosed to integrate over space

[itex]I \ = \ \int\int\int_\mathcal{V}(\frac{K^2}{2m}[(\frac{\partial \Psi}{\partial x})^2 \ + \ (\frac{\partial \Psi}{\partial y})^2 \ + \ (\frac{\partial \Psi}{\partial z})^2] \ + \ (V \ - \ E)\Psi^2)dxdydz[/itex]

& then extremizes this integral which gives us the Schrodinger equation.

No, this does not give you the Schrodinger equation. You have to CHOOSE [itex]K[/itex] to be the pure IMAGINARY number [itex]- i \hbar[/itex].

1) What is the justification for Feynman's famous quote:

"Where did we get that [Schrödinger's equation] from? It's not possible to derive it from anything you know. It came out of the mind of Schrödinger"

Now Feynman would tell you that POSTULATING [itex]K = - i \hbar[/itex] is EQUIVALENT to postulating the Schrodinger equation [itex]i \hbar \partial_{ t } \Psi = H \Psi[/itex].

2) Is the mathematical trick Schrodinger has used something you can use to solve problems?

Do you mean the “variation methods”? We use them all over the places.

3) Why can't you use this exact derivation in the relativistic case?

You can. Start with the relativistic HJ equation
[tex]\left( \frac{ \partial S }{ \partial t } \right)^{ 2 }- \left( \frac{ \partial S }{ \partial x } \right)^{ 2 } = m^{ 2 }[/tex]

Sam
 
  • #5
On page 276, 14 pages after he explains what I posted, & after he goes through how Schrodinger linked his work with DeBroglie's work, it is only then that Weinstock says:

In a more complete study of quantum mechanics than the present one the admissibility of complex eigenfunctions [itex]\Psi[/itex] is generally shown to be necessary. If [itex]\Psi[/itex] is complex, the quantity [itex]|\Psi|^2[/itex] is employed as the position probability-density function inasmuch as [itex]\Psi^2[/itex] is not restricted to real nonnegative values.

Apparently Schrodinger was able to do what I have posted using real-valued functions & have [itex]K[/itex] as I have defined it, without [itex]i[/itex]. If you're following what Weinstock is saying he shows how the hydrogen atom energy levels are explainable without complex numbers, i.e. he is able to derive a physical interpretation of the eigenvalues (discrete energy levels) of the Schrodinger equation that were in accord with experiment (see Section 11.3 Page 279 on). As far as I understand it it is in trying to find a physical interpretation of the eigenfunctions that one is forced into complex numbers, though apparently, according to the book, it can be shown to be necessary.

If you guys are telling me that all this is just 'postulated as definition' & the book says it isn't that it's all derivable then what can I do?

samalkhaiat said:
Do you mean the “variation methods”? We use them all over the places.

I meant the trick of taking one differential equation, doing a change of variables, randomly integrating the entire thing & then extremizing this result as a means to find a solution to your original differential equation? Seems like there are subtle assumptions involved, for example that the function you're seeking is originally postulated to be the minimum since you're working with the Hamilton-Jacobi equation thus justifying the extremization of the integral of the resulting differential equation. Would like to hear people's thoughts on it, what other assumptions they see built in & in general just a discussion on it's validity.
 
  • #7
bolbteppa said:
Apparently Schrodinger was able to do what I have posted using real-valued functions & have [itex]K[/itex] as I have defined it, without [itex]i[/itex].

For bound particles, the radial time-independent wave function can always be taken to be real.
 

FAQ: Variational Derivation of Schrodinger Equation

What is the Schrodinger equation?

The Schrodinger equation is a mathematical equation that describes the evolution of a quantum system over time. It is a fundamental equation in quantum mechanics and is used to calculate the probability of finding a particle in a particular state at a specific time.

2. What is variational derivation?

Variational derivation is a mathematical technique used to find the most accurate approximation to a given function. It involves comparing the function to a set of known functions and finding the one that minimizes the difference between the two.

3. Why is the Schrodinger equation important?

The Schrodinger equation is important because it allows us to accurately describe and predict the behavior of quantum systems. It has been used to explain a wide range of phenomena, from the behavior of atoms and molecules to the properties of materials and the structure of the universe.

4. How is the Schrodinger equation derived using variational methods?

The Schrodinger equation can be derived using variational methods by minimizing the difference between the actual energy of a quantum system and a trial wave function. This leads to a set of equations known as the variational principles, which can be solved to obtain the Schrodinger equation.

5. What are the applications of the variational derivation of the Schrodinger equation?

The variational derivation of the Schrodinger equation has many applications in physics and chemistry. It is used to calculate the properties of molecules, atoms, and materials, as well as to study the behavior of quantum systems in general. It is also used in fields such as quantum computing and quantum information theory.

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