![image0.png](https://www.dummies.com/wp-content/uploads/397231.image0.png)
if you remember that, for a free particle,
![image1.png](https://www.dummies.com/wp-content/uploads/397232.image1.png)
That equation gives you this form for
![image2.png](https://www.dummies.com/wp-content/uploads/397233.image2.png)
Because
![image3.png](https://www.dummies.com/wp-content/uploads/397234.image3.png)
the equation turns into
![image4.png](https://www.dummies.com/wp-content/uploads/397235.image4.png)
In fact, now that the right side of the equation is in terms of the radius vector r, you can make the left side match:
![image5.png](https://www.dummies.com/wp-content/uploads/397236.image5.png)
That’s the solution to the Schrödinger equation, but it’s unphysical. Why? Trying to normalize this equation in three dimensions, for example, gives you the following, where A is a constant:
![image6.png](https://www.dummies.com/wp-content/uploads/397237.image6.png)
(Remember that the asterisk symbol [*] means the complex conjugate. A complex conjugate flips the sign connecting the real and imaginary parts of a complex number. The limits on the integral just mean to integrate over all of space, like this:
![image7.png](https://www.dummies.com/wp-content/uploads/397238.image7.png)
Thus, the integral diverges and you can’t normalize
![image8.png](https://www.dummies.com/wp-content/uploads/397239.image8.png)
as written here. So what do you do here to get a physical particle?
The key to solving this problem is realizing that if you have a number of solutions to the Schrödinger equation, then any linear combination of those solutions is also a solution. In other words, you add various wave functions together so that you get a wave packet, which is a collection of wave functions of the form
![image9.png](https://www.dummies.com/wp-content/uploads/397240.image9.png)
such that
-
The wave functions interfere constructively at one location.
-
They interfere destructively (go to zero) at all other locations.
![image10.png](https://www.dummies.com/wp-content/uploads/397241.image10.png)
However, for a free particle, the energy states are not separated into distinct bands; the possible energies are continuous, so people write this summation as an integral:
![image11.png](https://www.dummies.com/wp-content/uploads/397242.image11.png)
So what is
![image12.png](https://www.dummies.com/wp-content/uploads/397243.image12.png)
It’s the three-dimensional analog of
![image13.png](https://www.dummies.com/wp-content/uploads/397244.image13.png)
That is, it’s the amplitude of each component wave function. You can find
![image14.png](https://www.dummies.com/wp-content/uploads/397245.image14.png)
from the Fourier transform of
![image15.png](https://www.dummies.com/wp-content/uploads/397246.image15.png)
like this:
![image16.png](https://www.dummies.com/wp-content/uploads/397247.image16.png)
In practice, you choose
![image17.png](https://www.dummies.com/wp-content/uploads/397248.image17.png)
yourself. Look at an example, using the following form for
![image18.png](https://www.dummies.com/wp-content/uploads/397249.image18.png)
which is for a Gaussian wave packet (Note: The exponential part is what makes this a Gaussian wave form):
![image19.png](https://www.dummies.com/wp-content/uploads/397250.image19.png)
where a and A are constants. You can begin by normalizing
![image20.png](https://www.dummies.com/wp-content/uploads/397251.image20.png)
to determine what A is. Here’s how that works:
![image21.png](https://www.dummies.com/wp-content/uploads/397252.image21.png)
Okay. Performing the integral gives you
![image22.png](https://www.dummies.com/wp-content/uploads/397253.image22.png)
which means that the wave function is
![image23.png](https://www.dummies.com/wp-content/uploads/397254.image23.png)
You can evaluate this equation to give you the following, which is what the time-independent wave function for a Gaussian wave packet looks like in 3D:
![image24.png](https://www.dummies.com/wp-content/uploads/397255.image24.png)