Why Does QM Wave Functions Use Pi?

[CuriosOne]

Why Does QM Wave Functions Use Pi?

Does this mean waves travel in perfect partial circles of pi? Or fractions of pi?

 

 

[timo]

Wave functions are solutions of linear differential equations. Mathematically, solutions to linear differential equations contain sine and cosine (or, equivalently, exponential functions with imaginary exponents). The Pi comes in from them.

 

Example: Imagine the Schroedinger equation for a free particle would be f(x) = -f''(x), where f''(x) is the 2nd derivative of f(x) with respect to x. A possible solution for this is f(x) = sin(x) (normalization ignored for this example). The wavelength of this wave is 2*Pi.

[CuriosOne]

9 hours ago, timo said:

Wave functions are solutions of linear differential equations. Mathematically, solutions to linear differential equations contain sine and cosine (or, equivalently, exponential functions with imaginary exponents). The Pi comes in from them.

 

Example: Imagine the Schroedinger equation for a free particle would be f(x) = -f''(x), where f''(x) is the 2nd derivative of f(x) with respect to x. A possible solution for this is f(x) = sin(x) (normalization ignored for this example). The wavelength of this wave is 2*Pi.

Does -f"(x) imply -1 ?

Does linear differential reffer to a "radius"?

I get this idea becuase sin and cos and pi have relationships to frequencies and "cycles" to the radian or radius which ever it uses.

In a sense using pi makes QM waves look "perfect" predictable, neat and organized, truly its not like that..Or is it???

When you say normalization is ignored, is that for ""All QM waves?""

...I hope so! After all the electron isn't that easy to figure out, plus the nature of its movement is not yet confirmed...

Is this information correct?

[timo]

I don't think I understand anything you just wrote. But I'll give it a try:

- "Does -f"(x) imply -1 ?": -f''(x) means -1*f''(x), if that was your question. It does not mean that f(x) = -1 or f''(x)=-1, if that was the question.

- "Does linear differential reffer to a "radius"?": Differential equations are a special type of equations that relate functions and their derivatives. They one of the most important mathematical concepts in physics. 'Linear" is just a mathematical property of this equation.

- "When you say normalization is ignored, is that for ""All QM waves?"" ": Usually, wave functions must meet a requirement that they are "normalized". In my example, I ignored this requirement because it is irrelevant for the point I wanted to highlight.

- "After all the electron isn't that easy to figure out, plus the nature of its movement is not yet confirmed... ": I kind of disagree with this statement. Moreover, I think it has really little to do with the question why the constant Pi shows up in the context of QM.

 

I think my first attempt to answer your question did not go very well. So let me try an alternative story:

In many important cases, wave functions can be expressed as sine and cosine functions. The Pi comes from the 2*Pi periodicity of these functions (if you'd express the arguments in terms of degrees you'd get 360s popping up in the equations, I guess). For example, if a wave function looks like f(x) = sin(x), it has the wave length (the distance after which it repeats itself) of 2*Pi. If the function should have the wave length 1, the function would look like f(x) = sin(2*Pi*x).

[CuriosOne]

1 hour ago, timo said:

I don't think I understand anything you just wrote. But I'll give it a try:

- "Does -f"(x) imply -1 ?": -f''(x) means -1*f''(x), if that was your question. It does not mean that f(x) = -1 or f''(x)=-1, if that was the question.

- "Does linear differential reffer to a "radius"?": Differential equations are a special type of equations that relate functions and their derivatives. They one of the most important mathematical concepts in physics. 'Linear" is just a mathematical property of this equation.

- "When you say normalization is ignored, is that for ""All QM waves?"" ": Usually, wave functions must meet a requirement that they are "normalized". In my example, I ignored this requirement because it is irrelevant for the point I wanted to highlight.

- "After all the electron isn't that easy to figure out, plus the nature of its movement is not yet confirmed... ": I kind of disagree with this statement. Moreover, I think it has really little to do with the question why the constant Pi shows up in the context of QM.

 

I think my first attempt to answer your question did not go very well. So let me try an alternative story:

In many important cases, wave functions can be expressed as sine and cosine functions. The Pi comes from the 2*Pi periodicity of these functions (if you'd express the arguments in terms of degrees you'd get 360s popping up in the equations, I guess). For example, if a wave function looks like f(x) = sin(x), it has the wave length (the distance after which it repeats itself) of 2*Pi. If the function should have the wave length 1, the function would look like f(x) = sin(2*Pi*x).

-1 "does answer my question" I "really" needed to know that, thnXxxx..

I can now understand why linear equations are so important now, afterall signal processing uses the same ideas.

"The 360 degrees popping up all the time is a "very important concept" and I truly think its the framework of much confusion..""Unless I'm confusing this wave function with dirivitives used in calculus"" becuase "again" if the function has wave length of 1 the function would look like:"

f(x)= sin(2*Pi*x) which looks very similar to the circumference of a circle to its diameter, or radian or radius or 1 cycle completed...

In this framework, do we know the initial start of the electron like a parked  car on the road ready to drive off??