Last step of a Schrodinger derivation

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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=_qzs1DD3TcsC&pg=PA70&lpg=PA70&dq=derive+"wave+mechanics"+schrodinger&source=web&ots=h4lTW0ElfR&sig=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.
 

Answers and Replies

  • #2
George Jones
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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
tiny-tim
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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
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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
George Jones
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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.
 
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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
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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.
 

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