What are the properties of a normalized wave function?

[vorcil]

At time = 0 a particle is represented by the wave function

$$\Psi(x,0) = \left\{ \begin{array}{ccc} A\frac{x}{a}, & if 0 \leq x \leq a, \\ A\frac{b-x}{b-a}, & if a \leq x \leq b, \\ 0, & otherwise, \end{array} \right$$where A, a, and b are constants.

(a) Normalize $$\Psi$$ (that is, find A, in terms of a and b).

(b) where is the particle most likely to be found, at t =0?

(c) What is the probability of finding the particle to the left of a? Check your result in the limiting cases b=a and b = 2a

(d) what is the expectation value of x?

 

[vorcil]

My attempt:

no bloody idea where to start,

I know that to normalize a function,

$$\int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx =1$$

But have no idea how I am supposed to incorporate that into my question...

I need EXAMPLES
I have no idea how to solve something I've never seen before!

 

[vorcil]

ok I'm just going to start guessing here,

if someone would be kind enough to correct me on mistakes

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since I want to find A,
and the wave function $$\Psi(x,0) = A \frac{x}{a}$$

if I square it, it should = 1?
$$\Psi(x,0)^2 = A^2 \frac{x^2}{a^2} ==1$$

not too sure where to go from here

 

[vorcil]

the same with the middle equation,

$$\Psi(x,0)^2 =(A\frac{(b-x}{b-a})^2) =1$$

 

[vorcil]

found this

http://en.wikipedia.org/wiki/Normalisable_wave_function

so

$$\frac{A^2}{a^2} \int_0^a x^2 dx =1$$

the question says A, a and b are constants so I'm taking them outside the integral

$$\frac{A}{b-a} \int_a^b (b-x)^2 dx =1$$

 

[VelvetRebel]

You have the right idea but the two integrals added together should be equal to one I do believe. Also be sure to square the A/(b-a) on the second integral.

 

[vorcil]

When I am normalizing the wave function, do I add the equations from all the parts of the wave function?

i.e

$$\int_{-\infty}^{0} +\int_{0}^{a}+\int_{a}^{b}+\int_{b}^\infty$$

 

[vorcil]

so adding that to concept to my question,

i'd get

$$|\Psi(x,0)|^2 = \int_{-\infty}^0 0^2dx + \frac{A^2}{a^2}\int_0^a x^2 dx + \frac{A}{(b-a)} \int_a^b (b-x)^2 dx + \int_b^\infty 0^2dx =$$

 

[vorcil]

$$= |\Psi(x,0)|^2 = \frac{A^2}{a^2}\int_0^a x^2 dx + \frac{A^2}{(b-a)^2} \int_a^b (b-x)^2 dx$$
then uh simplifying and such i get,
(since A^2 is in both equations)

$$\left( A^2 \left[ \left( \frac{1}{a^2} \frac{x^3}{3} \right) + \left( \frac{1}{(b-a)^2} \int_a^b (b-x)^2 dx \right) \right] \right) =1$$

 

[VelvetRebel]

That is almost correct except A/(b-a) should be squared as well and then all of that will be equal to one. Then you should do each integral then just solve for A.

 

[vorcil]

how do I integrate (b-x)^2 ?

$$\int (b-x)(b-x)dx = \int (b^2 - 2bx + x^2 )dx= b^2 -\frac{2bx^2}{2} + \frac{x^3}{3}$$

? then i evaluate that at $$& \left equation \right|_{a}^{b}$$

 

[VelvetRebel]

For b², you can think of that as b²*x⁰. So when you take the integral of that, you would end up with b²*x¹. Other than that, everything else looks correct.

 

[vorcil]

$$\int (b^2 - 2bx + x^2) dx = b^2x - \frac{2bx^2}{2} + \frac{x^3}{3}$$

$$b^2x - bx^2 +\frac{x^3}{3}$$

what now?
I was shown that the answer to that integral was

$$-\frac{(b-x)^3}{3}$$

but how do i get that?
from

$$\int_a^b (b-x)^2 dx$$
to
$$\left -\frac{(b-x)^3}{3} \right|_a^b$$

 

[diggy]

$$\int (b^2 - 2bx + x^2) dx = b^2x - \frac{2bx^2}{2} + \frac{x^3}{3}$$

$$b^2x - bx^2 +\frac{x^3}{3}$$

what now?
I was shown that the answer to that integral was

$$-\frac{(b-x)^3}{3}$$

but how do i get that?
from

$$\int_a^b (b-x)^2 dx$$
to
$$\left -\frac{(b-x)^3}{3} \right|_a^b$$

They used u=(b-x) => du=-dx

You are on a roll and almost there -- keep up the good work!

 

[vorcil]

embarassing

but I can't integrate that, I don't know how

can someone please explain to me how to integrate it?

$$\int_a^b (b-x)^2 dx$$

:(

 

[diggy]

embarassing

but I can't integrate that, I don't know how

can someone please explain to me how to integrate it?

$$\int_a^b (b-x)^2 dx$$

:(

You don't have to do it the way they did it. Just expand (b-x)², and make three simple integrals. You should be able to do those.

 

[vorcil]

$$\int_b^a (b-x)^2dx \\$$

$$\int_b^a (b-x)(b-x)dx \\$$

$$\int_b^a (b^2 +x^2 -2bx)dx \\$$

$$b^2x + \frac{x^3}{3} + \frac{-2bx^2}{2} \\$$

i Don't see how that equation,

$$b^2x + \frac{x^3}{3} + \frac{-2bx^2}{2}$$
equals

$$\frac{(b-x)^3}{3}$$

can someone just explain please

 

[diggy]

They do it by making the "u" substitution I wrote in the earlier post. The integral becomes -u²du

Integrate that, then put the b-x back into u.