Last step of a Schrodinger derivation

In summary, the conversation discusses the Schrodinger equation and its derivation. The participants mention a preferred derivation from a specific textbook and discuss the coefficients and usage of the term 'motivation'. They also mention that the equation is not a true derivation, but rather induced through observation and experimentation. The equation has been corrected in the past and its origins are not fully known.
  • #1
bluestar
80
0
I have looked at several derivations of the Schrodinger equation but the one I like the best is from Piravonu Mathews and K. Venkatesan in their book ‘A Textbook of Quantum Mechanics’. I follow their logic and algebra up until the last step were they arrive at the Schrodinger equation for one dimension with no external potential field. This is equation 2.8 on page 37 of this book. If you don’t have a copy of the book Google has scanned the book and below is a link to that scan so you can copy paste it into your browser.

http://books.google.com/books?id=_q...ig=0e3r6SrD_LzYo3G0KPeuflefxyw&hl=en#PPA37,M1

Specifically, I don’t understand how they arrived at the variables that are multiplied against the partial derivatives on both sides of the equation.

Some guidance would greatly be appreciated.
 
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  • #2
The two equations before 2.8 are

[tex]
\frac{\partial \psi}{\partial t} = - i \omega \psi
[/tex]

and

[tex]
\frac{\partial^2 \psi}{\partial x^2} = - \frac{2m}{\hbar} \omega \psi.
[/tex]

Solve the first equation for [itex]\omega \psi[/itex] and substitute into the second.

Note that this is really motivation for the Schrodinger equation, not a derivation.
 
  • #3
bluestar said:
Specifically, I don’t understand how they arrived at the variables that are multiplied against the partial derivatives on both sides of the equation.

Hi bluestar! :smile:

From the line above it:

ih∂ψ/∂t = hωψ = (h²/2m)(2mωψ/h) = (h²/2m)∂²ψ/∂x² :smile:
 
  • #4
Hi George Jones, you have helped my before and I appreciate you and Tiny-Tim helping me with this problem. Tiny-tim, your equation cleared up the confusion and has shown me that the coefficients are correct with respect to the equation. Although this example presumes to know what the coefficients were beforehand and you filled in the middle so I could understand. So if the Schrodinger equation coefficients were not known beforehand then I guess a true derivation would be required to come up with the proper values.

George, I’m not sure I understand your usage of the term ‘motivation’. I guess I should also admit I’m not sure why the book’s treatment of the Schrodinger equation is not a true derivation.
 
  • #5
bluestar said:
George, I’m not sure I understand your usage of the term ‘motivation’. I guess I should also admit I’m not sure why the book’s treatment of the Schrodinger equation is not a true derivation.

Schrodinger's equation isn't derived, just as Newton's second law isn't derived. We look for equations that describe phenomena in nature, but we don't derive all of them, we induce some of them.
 
  • #6
George Jones said:
Schrodinger's equation isn't derived, just as Newton's second law isn't derived. We look for equations that describe phenomena in nature, but we don't derive all of them, we induce some of them.

A derivation might exist, and give more insight into the assumptions underlying the equation. Equations have also been corrected in the past by means of new derivations.
 
  • #7
I have never seen a true derivation of Schrodinger equation. Usually the book-writer do a derivation for a free particle, and then he just ASSUME that it is good for a non-free particle in other words for a particle in a force field. My opinion is that it is the same story in here.

Also, I don`t think that Schrodinger has just written his famous equation, but he worked very hard, and than he thought that this equation might be right (also I don`t think that this equation is the only one or the first one that he had created for that purpose) and then he published it. After publishing, the equation had to go through series of experimental testing, and just after that it became a fundamental equation of quantum mechanics.
 

Related to Last step of a Schrodinger derivation

1. What is the last step in a Schrodinger derivation?

The last step in a Schrodinger derivation is to solve the time-independent Schrodinger equation, which gives the wave function and energy eigenvalues for a quantum mechanical system.

2. What is the significance of the last step in a Schrodinger derivation?

The last step in a Schrodinger derivation is crucial because it allows us to determine the possible energy states and corresponding wave functions for a quantum system, which helps us understand its behavior and make predictions.

3. How is the last step in a Schrodinger derivation performed?

The last step in a Schrodinger derivation involves applying mathematical techniques, such as separation of variables or perturbation theory, to solve the Schrodinger equation and obtain the wave function and energy eigenvalues for the system.

4. What information can be obtained from the last step of a Schrodinger derivation?

The last step of a Schrodinger derivation provides us with the wave function, which describes the spatial distribution of a particle, and the corresponding energy eigenvalues, which represent the possible energy levels for the system.

5. Are there any limitations to the last step of a Schrodinger derivation?

Yes, the last step of a Schrodinger derivation is based on the assumptions and approximations made in the Schrodinger equation, so it may not accurately describe all quantum systems. Additionally, for more complex systems, it may be difficult or impossible to solve the Schrodinger equation analytically, requiring the use of numerical methods.

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