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Author: Subject: Help with PChem homework plz!
BromicAcid
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shocked.gif posted on 10-10-2005 at 17:44
Help with PChem homework plz!


I need homework help cuz' suck at pchem, help plz?!?!

Evaluate the communicator [(d/dx) –x, (d/dx) + x] by applying the operators to an arbitrary function f(x).

Evaluate the communicator [d/dr, 1/r] by applying the operators to an arbitrary function f(r).

4. What is the difference between probability and probability density?

5. Calculate the probability that a particle in a one-dimensional box of length a is found between 0.31a and 0.35a when it is described by the following wave functions:
a) [(2/a)]½ sin ([pi]x/a)
b) [(2/a)] ½ sin (3[pi]x/a)
What would you expect for a classical particle? Compare your results in the two cases with the classical.

Anyone can help me ?
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[*] posted on 11-10-2005 at 11:24


Here we go,

Use commutator properties to simplify first problem:

[d/dx-x,d/dx+x]=[d/dx,d/dx+x]-[x,d/dx+x] now since [d/dx,d/dx]=[x,x]=0 we get
=>[d/dx,x]-[x,d/dx] since [x,d/dx]=-[d/dx,x] we get
=>2[d/dx,x] which we solve using f(x)
=>2[d/dx,x]f(x)=2d/dx(xf(x))-2x(d/dx(f(x)) since d/dx(xf(x))=f(x)+x(d/dx(f(x)) we get
=>2(f(x)+x(d/dx(f(x)))-2x(d/dx(f(x))=2f(x)
so, [d/dx-x,d/dx+x]=2

Second problem,
[d/dr, 1/r]f(r)=d/dr(f(r)/r)-(d/dr(f(r)))/r
d/dr(f(r)/r)=(d/dr(f(r)))/r-f(r)/r^2 so plugging this in we get
=>[d/dr, 1/r]=-1/r^2

The difference between probability and probability density. Probabilty is given for a single event, say there is a 1 in 6 chance in rolling a 6 on a die. A probability density is given over a continous range of values. So to clarify this imagine you released a particle at some point a into a diffusing medium where it would under go Brownian motion. At some time later there would be a probability of still finding the particle at point a and another probablity of finding the particle at point a+dx. It is possible to find a function which describes the probability of finding our Brownian particle at every point in space. Clear? Probably not, I can't explain anything without pictures.

To further clarify this let us consider your next problem. We can get the probability density by taking the complex conjugate of the wavefunction times the wavefunction, (psi*)(psi). Since this is a fully real wavefunction this means just square the wavefunction. So our probability function is given as:
(a) P(x)=(2/a)sin^2(pi*x/a) If you plot this from 0 to a you should find the max probabilty to be at a/2 and zero at 0,a. Now the probability of finding the particle in some small range say a/2+dx is given as P(a/2)dx, so the probability of finding the particle in the asked for range is merely the integral of P(x) over this range. So let's try it:

(2/a)int{sin^2(pi*x/a)dx}=-sin(pi*x/a)cos(pi*x/a)/pi+x/a

Now plug in the limits 0.31a and 0.35a and we get P=0.22125-0.16202=0.059229. Notice if we integrate from 0 to a we get 1. This is what is known as a normalized wavefunction, meaning the particle has to be somewhere. Part b can be solved in a similar manner.

Now for a classical particle it will be equally probable to find it anywhere in the box. This gives a probabilty distribution of P(x)=1/a so the probabilty of finding in the given range will be 0.35-0.31=0.04.

Hope this helps.
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BromicAcid
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biggrin.gif posted on 11-10-2005 at 11:55
OMG THANK YOU


Thank you so much! That's a big help!



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