Hello,
For number 1, when you square a wave function, you don't just multiply it with itself, rather you multiply it by its complex conjugate.
That'll allow you to solve numbers two and three aswell.
Given that a particle is illustrated by
1.) Normalize the wave func.
2.) Find and
3.) Sketch and explain what the plot illustrates.
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For 1.), we need this condition to be met:
So we have:
Not sure if I did this correctly. Used Maple to help me through it. I know we have a value for h that I can plug in, but what about the E and t?
For 2.),
Again, not sure if this is right. I had difficulty with finding .
I know:
although what do I use?
Then, for 3.), that is graphing it, can't plot it with so many variables, as
.
Thanks topsquark (or whoever else knows this stuff).
Okay, how is this:
We have:
Yay? Nay? I was fortunate that the t dropped out, or I wouldn't know how to solve it. I'm going to hope the same happens when I solve for <x> and <p>.
EDIT: Okay, solved for <x>, although not sure if it's right.
Can anyone confirm if this is right?
I need help with the set-up for <p> please!
Are we meant to use the C I found in the first part for <x> and <p>? How should I attack the sketching part? I have the problem with there being too many variables.
This is correct.
Recall that the operator p is, in the position representation
So
Please note the position of the operator in the expression. Technically you got it wrong in the <x> calculation. However, since the operator is simply "x" there was no problem with the order of the terms. The <p> calculation demonstrates that you do need to be careful of this in general.
For an arbitrary operator ,
Also, once you have found a normalization constant you keep it. That is, after all, the meaning of "constant."
-Dan
Note two things:
1) <p> must be a real number. As this integral winds up with a complex coefficient, it had better be 0.
2) The integral you derived contains an odd integrand over even limits. Thus the integral has a value of 0. You didn't evaluate your result over the endpoints.
-Dan