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

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SUMMARY

The discussion centers on the understanding of the Schrödinger equation and its derivatives, specifically focusing on the wave function represented as psi. The sine wave function is expressed as F(x) = sin(2πft), with its first derivative as F(x) = 2πf cos(2πft), highlighting the importance of the chain rule in calculus. The confusion arises from the misinterpretation of derivatives, particularly the second derivative, which is represented as F(x) = - (2πf)² sin(2πft). The conversation emphasizes the necessity of calculus knowledge to grasp these concepts fully.

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  • Understanding of wave functions in physics
  • Basic knowledge of calculus, including derivatives
  • Familiarity with the chain rule in differentiation
  • Concept of sine and cosine functions
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  • Study the Schrödinger equation in quantum mechanics
  • Learn about the chain rule in calculus
  • Explore wave functions and their applications in physics
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Students of physics, particularly those studying quantum mechanics, as well as individuals seeking to strengthen their calculus skills and understanding of wave functions.

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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