Wave equation, psi with dots and things like that....

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Discussion Overview

The discussion revolves around the Schrödinger equation and the mathematical representation of wave functions, particularly focusing on the derivatives of sine wave functions and the implications of using dots to denote time derivatives. Participants explore the confusion surrounding the correct forms of these derivatives and the application of calculus concepts, such as the chain rule.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant, Joao, expresses confusion about the derivatives of sine wave functions and their representation as psi and psi with dots.
  • Another participant suggests that dots above a function typically denote time derivatives and mentions the importance of calculus in understanding these concepts.
  • It is proposed that the derivative of Ψ = Sin(2πft) is Ψ(dot) = 2πf cos(2πft), emphasizing that this is not simply cos(2πft).
  • Participants discuss the chain rule as a crucial concept for understanding the steepness of the sine function's graph when modified.

Areas of Agreement / Disagreement

There is no consensus on the understanding of the derivatives and their implications, as the discussion reflects varying levels of familiarity with calculus concepts. Some participants agree on the importance of the chain rule, while others express ongoing confusion.

Contextual Notes

Participants acknowledge a lack of foundational knowledge in calculus, which may limit their understanding of the mathematical representations discussed.

Who May Find This Useful

This discussion may be useful for individuals seeking to understand the mathematical foundations of wave functions, particularly those new to calculus or struggling with the concepts of derivatives in the context of physics.

Joao
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Hi everyone! I'm a psychologist form Brazil, so sorry for the bad English and for the lack of knowledge in math!

I ve been trying to understand the Schrödinger equation and, as predicted, it's very hard!

Please, help me with this:

A sine wave function can be written as:

F (x) = sin (x)

And that can be represented as psi.

The the derivative of that function can be written as:

F (x) = cos (x)

And that can be represented as psi dot.

Also, the derivative of psi dot can be written as:

F (x) = - sin (x)

And that's psi with two dots.

So far I understood!

But things get confusing here:

A wave equation can be written as:

F (x) = A sin (2π f x )

Where A stands for the amplitude, 2π f x stands for the period.

But, in this video:



It is stated, at 8 mins, that the correct wave function is, as expected:

F (x)= sin (2π f t)

And that's represented by psi

But the derivative of that is

F (x) = 2π f cos (2π f t)

And not

F (x) = cos (2π f t)

(And that's represented by psi dot)

To make things worst, the derivative of that last equation, psi dot, is represented by

## f (x) = - (2π f)^2 sin (2π f t) ##

And not just

F (x) = - sin (2π f t)

(And that's represented by psi with 2 dots)

Why is that?

Thanks a lot!
 
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@Joao you need a calculus course.

Dots above a function normally denote a time derivative. The normal convention for a derivative is either ##f'## or ##\frac{df}{dx}##.

Then, of course, you have "partial" derivatives ##\frac{\partial \psi}{\partial x}## and ##\frac{\partial \psi}{\partial t}## etc.

In answer to your question, there is a little thing called the chain rule, which is quite important!
 
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Thanks a lot! I will do that! So, that makes sense, the derivative of

Ψ = Sin (2πft)
Is
Ψ(dot) = 2πf cos (2πft)
And not
Ψ(dot) = cos (2πft)

And after I make the calculus course, I will understand why?

Really sorry to bother!
 
Joao said:
Thanks a lot! I will do that! So, that makes sense, the derivative of

Ψ = Sin (2πft)
Is
Ψ(dot) = 2πf cos (2πft)
And not
Ψ(dot) = cos (2πft)

And after I make the calculus course, I will understand why?

Really sorry to bother!

Yes. You could, say, draw a graph of ##\sin(2x)## and look at the slope. You'll see that it is steeper than the graph of ##\sin(x)##.

In fact, it's twice as steep at each point ##x##.

Which is what you get from the chain rule.
 
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PeroK said:
Yes. You could, say, draw a graph of ##\sin(2x)## and look at the slope. You'll see that it is steeper than the graph of ##\sin(x)##.

In fact, it's twice as steep at each point ##x##.

Which is what you get from the chain rule.
Thanks a lot for your time! I'll do the calculus course! =)))))

Happy 2018!
 

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