Normalize function - quantum chemistry

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The discussion centers on the normalization of the function f(r) = Nexp{-alpha*r}, where alpha is a positive constant and r is considered a vector. Participants clarify that the argument inside the exponential must be a scalar, implying that the equation should involve a scalar product, such as α·r. This distinction is crucial for proper integration and normalization of the function. The conversation emphasizes the need to correctly interpret vector and scalar quantities in quantum chemistry equations. Understanding these nuances is essential for accurate mathematical treatment in quantum mechanics.
kanciara
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Homework Statement
Normalize function f(r) = Nexp{-alpha*r}
Where alpha is positive const and r is a vector
Relevant Equations
f(r)=N*exp{-alpha*r}
Normalize function f(r) = Nexp{-alpha*r}
Where alpha is positive const and r is a vector

I was just wondering if the fact that we have a vector value in our equation changes anything about the solution
 
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kanciara said:
Homework Statement:: Normalize function f(r) = Nexp{-alpha*r}
Where alpha is positive const and r is a vector
Relevant Equations:: f(r)=N*exp{-alpha*r}

Normalize function f(r) = Nexp{-alpha*r}
Where alpha is positive const and r is a vector

I was just wondering if the fact that we have a vector value in our equation changes anything about the solution
Why do you think ##r## is a vector? Make sure you're not confusing vector ##\vec r## with its magnitude ##r##.
 
kanciara said:
Homework Statement:: Normalize function f(r) = Nexp{-alpha*r}
Where alpha is positive const and r is a vector
Relevant Equations:: f(r)=N*exp{-alpha*r}

Normalize function f(r) = Nexp{-alpha*r}
Where alpha is positive const and r is a vector

I was just wondering if the fact that we have a vector value in our equation changes anything about the solution
The argument inside the exponential needs to be a scalar, so it would have to be something like ## \alpha \cdot \textbf{r}##. It should be clear by context. I've seen ##\textbf{k} \cdot \textbf{x}## in a wavefunction but never written with a radial variable.

If it is a scalar product then you will have something like
##\int N e^{ \alpha _r r + \alpha _{ \theta } \theta + \alpha _{ \phi } \phi }## (or some such) which you should be able to separate out and integrate individually.

-Dan
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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