# Schrodinger equation

1. Mar 23, 2006

### UrbanXrisis

I think I copied the wrong notes or something because my notes do not follow.

I am trying to find the probability of finding a particle in a box length L in the area $$\frac{L}{3}-\frac{\partial}{2}$$ to $$\frac{L}{3}+\frac{\partial}{2}$$

basically we have the following wave funtion:

$$\Phi(x,t)=\sqrt{\frac{2}{3}} \Phi_1 - \sqrt{\frac{1}{3}} \Phi_2$$

so we got the absolute square of the function which gave us probability, then we integrated from $$\frac{L}{3}-\frac{\partial}{2}$$ to $$\frac{L}{3}+\frac{\partial}{2}$$

this gave us the probability, however, we used a short cut, our teacher split P(robability) into 4 parts..

P=I1+I2+I3+I4

then said that I=integrand * del

and came up with a short cut:

$$P=\frac{4 \partial}{3L} *\frac{3}{4}+\frac{2}{3} \frac{\partiao}{L} \frac{3}{4}+2 \sqrt{\frac{2}{9}} \frac{2}{L} \sqrt{\frac{3}{4} } \sqrt{\frac{3}{4} }cos\left( \frac{E_2-E_1}{\hbar}t \right)$$

$$=\frac{3}{2} \frac{\partial}{L} \left(1+cos\left( \frac{E_2-E_1}{\hbar}t \right) \right)$$

i dont understand this idea of I=integrand * del, could someone direct me to a site that explains this or help me out with this concept?

Last edited: Mar 23, 2006
2. Mar 23, 2006

### nrqed

I am pretty sure that everybody who reads your post will stop at this line and wonder what you mean by using the partial derivative symbol there!! Can you explain?

3. Mar 23, 2006

### UrbanXrisis

it is trying to set a finit size, because if it was just L/3, you cant integrate a point because it would just be zero. so i think that this is supposed to be to show the +/- some amount

i cant explain it myself, i think it is just a variable, but these are the notes I copied but I'm not sure what my prof means. but if you read the whole thing, you might be able to see something that I dont. My prof said that I=integrand * del was very important. basically, it is just a quick short cut to integrating the probability. not sure, that's why i need some feedback, thanks

4. Mar 23, 2006

### nrqed

Ok, I see now. But it is a very unconventional notation. It's an infinitesimal distance. Normally, people would write dx, say, if it is along the x axis.

What he is saying is that (this is true for any integral):

$$\int_{a-dx/2}^{a+dx/2} f(x) dx = f(a) dx [/itex] This is also a bad notation because there are two dx here playing different roles, but that's the way it is usually presented in books. Using a better notation: [tex] \int_{a-\delta/2}^{a+\delta/2} f(x) dx = f(a) \delta [/itex] where $\delta$ is an infinitesimal distance. The key point is that if you integrate over a very small (infinitesimal) interval, you just get th efunction evaluated at a point in your interval times the infinitesimal interval. Pat (edit: arghh! I kept making typos...) Last edited: Mar 23, 2006 5. Mar 23, 2006 ### UrbanXrisis I dont understand the rule of I=integrand * del When I took the absolute square of the wave funtion, it gave me 4 different terms, each one distinguised by I1, I2, I3, and I4. There was some shortcut way of integrating them in the distance dx. That is what the second part was about, which I dont understand 6. Mar 23, 2006 ### Doc Al ### Staff: Mentor I suspect that the professor meant to use deltas, not partial derivative signs: [tex]\frac{L}{3}-\frac{\delta}{2}$$ to $$\frac{L}{3}+\frac{\delta}{2}$$

Realize that taking the integral is equivalent to finding the area under a curve. If you are integrating from $L/3 - \delta/2$ to $L/3 + \delta/2$, where the delta is small enough, then the function doesn't vary much in that interval. So its "height" is just f(L/3) times its "width" of delta gives you the area equal to f(L/3) * delta. Make sense?

[Edit: Looks like nrqed beat me to it!]