Question about normalization of wavefunction

In summary, The integral for normalizing sin(\theta)e^(-i\phi) in polar coordinates with bounds 0 to 2pi for phi and 0 to pi for theta is incorrect. The correct set up of the integral should be \int sin^{3}\theta*dv with the same bounds, and the answer will include a pi term. The reason for including the pi term is unclear.
  • #1
josecuervo
18
0

Homework Statement


how to set up integrals for normalization of sin([itex]\theta[/itex])e^(-i[itex]\phi[/itex])


Homework Equations





The Attempt at a Solution

 
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  • #2
Can you give any more detail as to what you're doing? In general a normalized wave function would work along the following lines

[itex]\int[/itex][itex]\Psi[/itex](x)*[itex]\Psi[/itex](x)dx=1

or

[itex]\int[/itex]|[itex]\Psi[/itex](x)|2dx=1

with the limits of integration being from -[itex]\infty[/itex] to +[itex]\infty[/itex]
 
  • #3
I know it needs to be switched to polar coordinates and your equation you posted is correct. I know that it is normalizable (1=int(psi^2dv) over all space, i just need a confirmation of the correct set up of the integral and the bounds. I've tried it several different ways with different bounds and I haven't gotten it to be one.
 
  • #4
anyone?
 
  • #5
You need to show what you've tried so that we can help you identify where you're going wrong. When you don't do that, many people will simply ignore your thread. See the https://www.physicsforums.com/showthread.php?t=94380.
 
  • #6
Ok I've set up the integral like this:[itex]\oint[/itex]sin[itex]^{3}[/itex][itex]\theta[/itex]*dv with the bounds being 0 to 2pi for phi, and 0 to pi for theta. I left out r because it will make the integral go to infinity, but I'm still getting a pi in the answer. when I left out the phi part of the integral I got 4/3. what am I doing wrong?
 
  • #7
For starters, how do you know you're doing something wrong?
 

FAQ: Question about normalization of wavefunction

What is the wavefunction normalization condition?

The wavefunction normalization condition states that the integral of the square of the wavefunction over all space must equal 1. In other words, the total probability of finding the particle in any location must equal 1.

Why is it important to normalize the wavefunction?

Normalizing the wavefunction ensures that the total probability of finding the particle in any location is equal to 1, which is necessary for the wavefunction to accurately describe the behavior of the particle. Without normalization, the wavefunction could potentially have infinite values, making it impossible to calculate probabilities.

How do you normalize a wavefunction?

To normalize a wavefunction, you must first find the integral of the square of the wavefunction over all space. This integral is known as the normalization constant. Then, divide the wavefunction by the square root of the normalization constant to obtain the normalized wavefunction.

Can a wavefunction be normalized to a value other than 1?

No, the wavefunction must be normalized to a value of 1 in order for it to accurately describe the behavior of the particle. Any other normalization would result in incorrect probabilities.

What is the physical significance of the wavefunction normalization?

The normalization of the wavefunction represents the conservation of probability. It ensures that the total probability of finding the particle in any location is always equal to 1, regardless of the state of the particle. This is a fundamental concept in quantum mechanics.

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