# Normalise the Wave function

• October 16th 2010, 07:50 AM
bugatti79
Normalise the Wave function
Hi All,

I am going through notes our lecturer gave us, am having some problems with it
Given the wave function $\Psi(x,t)=Ae^{-\lambda \mid x \mid} e^{-i\omega t}$ where A, lambda and omega are psoitive real constants and

$\mid x \mid$ either (x greater and equal to 0) and (-x less than or equal to 0)

1) Normalise Psi, 2) Determine the expectation values of x and x^2

I can derive to far as $1=A^2(\int^{0}_{-\infty} e^{2\lambda x} dx + \int^{\infty}_{0} e^{-2\lambda x} dx)$ Then he changes the integral such that

= $-A^2\int^{0}_{-\infty} e^{-2\lambda x} dx + \int^{\infty}_{0} e^{-2\lambda x} dx = 2A^2\int^{\infty}_{0} e^{-2\lambda x} dx$
Where did the first termof the last expression come out of?

Thanks
• October 16th 2010, 08:46 AM
CaptainBlack
Quote:

Originally Posted by bugatti79

I can derive to far as $1=A^2(\int^{0}_{-\infty} e^{2\lambda x} dx + \int^{\infty}_{0} e^{-2\lambda x} dx)$ Then he changes the integral such that...

Change the variable in the first integral to $x'=-x$ to get to the final result.

CB
• October 17th 2010, 07:55 AM
bugatti79
ok, i get that, thanks. The next line becomes

$1=2A^2\frac{1}{-2\lambda}\int^{\infty}_{0} e^{-2\lambda x} d(-2\lambda x)= \frac{-A^2}{\lambda}e^{-2\lambda x} = \frac{A^2}{\lambda}$ upon substitution of the limits.
He seems to have integrated the integral but then he changes the dx to $d(-2\lambda x})$ Why? I dont understand this?

Thanks
• October 17th 2010, 08:18 AM
CaptainBlack
Quote:

Originally Posted by bugatti79
ok, i get that, thanks. The next line becomes

$1=2A^2\frac{1}{-2\lambda}\int^{\infty}_{0} e^{-2\lambda x} d(-2\lambda x)= \frac{-A^2}{\lambda}e^{-2\lambda x} = \frac{A^2}{\lambda}$ upon substitution of the limits.
He seems to have integrated the integral but then he changes the dx to $d(-2\lambda x})$ Why? I dont understand this?

Thanks

Just to be bloody minded and obscurantist, you can integrate it straight off without doing that (OK may be that is how he was taught to do such things, I would do it as shown below).

$\displaystyle 1= 2A^2\int_0^{\infty}e^{-2\lambda x}\;dx=2A^2 \left[\dfrac{1}{(-2\lambda)}e^{-2\lambda x} \right]_{x=0}^{\infty}=\dfrac{A^2}{\lambda}$

CB
• October 17th 2010, 12:14 PM
bugatti79
yes, Im more comfortable with that too. Next I have a query wrt integration by parts using reduction formulae

$=2\lambda\int^{\infty}_{0} x^2 e^{-2\lambda x} dx$
Setting up the formulas I get

$I_{n}=\int x^n e^{-ax}=-\frac{1}{a}x^n e^{-ax}+\frac{n}{a}\int x^{n-1}e^{-ax}dx =- \frac{1}{a}x^n e^{-ax}+\frac{n}{a}I_{n-1}$

Assuming the above is correct, do i proceed to calculate I2, I1 and I0? Im not sure where I put in the limits?

Thanks
• October 17th 2010, 03:05 PM
CaptainBlack
Quote:

Originally Posted by bugatti79
yes, Im more comfortable with that too. Next I have a query wrt integration by parts using reduction formulae

$=2\lambda\int^{\infty}_{0} x^2 e^{-2\lambda x} dx$
Setting up the formulas I get

$I_{n}=\int x^n e^{-ax}=-\frac{1}{a}x^n e^{-ax}+\frac{n}{a}\int x^{n-1}e^{-ax}dx =- \frac{1}{a}x^n e^{-ax}+\frac{n}{a}I_{n-1}$

Assuming the above is correct, do i proceed to calculate I2, I1 and I0? Im not sure where I put in the limits?

Thanks

You should be able to calculate I_0 directly as an integral. The use the reduction formula to findI_1 then I_2

CB
• October 17th 2010, 03:09 PM
bugatti79
Quote:

Originally Posted by CaptainBlack
You should be able to calculate I_0 directly as an integral. The use the reduction formula to findI_1 then I_2

CB

DO I sub in the limits in I_0 only and then find I_1 and I_2?
• October 17th 2010, 03:36 PM
CaptainBlack
Quote:

Originally Posted by bugatti79
DO I sub in the limits in I_0 only and then find I_1 and I_2?

No the recurence is for indefinite integrals so you have (check the algebra here please):

$I_0=-\frac{1}{a}e^{-ax}+c_0$

$I_1=-\frac{1}{a}xe^{-ax}-\frac{n}{a^2}e^{-ax}+c_1$

(you can express $c_1$ in terms of $c_0$ if you like but there is no need)

and so on for $I_2$, now apply the limits of integration.

There is another way to evaluate the integral, which is to observe that for some quadratic $ax^2+bx+c$ that:

$\dfrac{d}{dx}[(ax^2+bx+c)e^{-\lambda x}]=x^2e^{-\lambda x}$

doing the differentiation will then allow you to find the quadratic and so the required integral.

CB
• October 19th 2010, 12:20 AM
bugatti79
Thanks,

I will respond to this when I get the chance
• October 20th 2010, 01:11 PM
bugatti79
Hi CB. IM stil not sure how you work out the integral. Here is what I got so far.

$I_0=-\frac{1}{a}e^{-ax} +C_0$
$I_1=-\frac{1}{a}xe^{-ax} + \frac{1}{a}(-\frac{1}{a}e^{-ax} +C_0)$
$I_2=-\frac{1}{a}x^2e^{-ax}+\frac{2}{a}(-\frac{1}{a}xe^{-ax}-\frac{1}{a^2}e^{-ax}+\frac{c_0}{a})$

I have calcd I0 and I1 and substituted into I2. Assuming the above algebra is corect, what is the next step? I'm thinking that since its a definite integral we can eliminate the C0 therefore the above becomes

$I_2=\left[-\frac{1}{a}x^2e^{-ax}+\frac{2}{a}(-\frac{1}{a}xe^{-ax}-\frac{1}{a^2}e^{-ax}) \right]^{\infty}_{0}$

Thanks
• October 22nd 2010, 04:48 AM
bugatti79
Following on from last thread, I calculate I2 to be equal to $\frac{1}{(2\lambda)^3}$.
Assuming this is correct, what do I do with this value in relation to $2A^2\int^{\infty}_{0} e^{-2\lambda x} dx$

My lecture notes goes on to let $t=2\lambda x$??? I dont understand the overall methodology of integration by recursion with definite integrals.

Any help appreciated.

Thanks
• October 22nd 2010, 10:53 AM
CaptainBlack
Quote:

Originally Posted by bugatti79
Following on from last thread, I calculate I2 to be equal to $\frac{1}{(2\lambda)^3}$.
Assuming this is correct, what do I do with this value in relation to $2A^2\int^{\infty}_{0} e^{-2\lambda x} dx$

My lecture notes goes on to let $t=2\lambda x$??? I dont understand the overall methodology of integration by recursion with definite integrals.

Any help appreciated.

Thanks

Goodness knows, $I_2$ comes from post #5 which is unrelated to the question in post #1.

CB
• October 23rd 2010, 10:47 AM
bugatti79
CB,

It turns out I was mislead by my notes and with some typos and apologies if I mislead you! :-) I just solved the following

$=2\lambda\int^{\infty}_{0} x^2 e^{-2\lambda x} dx$ using integration by parts without any recursion. I dont know what the recursion calcs where for. If i find out I will inform you.

Thanks