Normalizing a wave function

  • #1

Homework Statement


"assume that the three real functions ψ1,ψ2, and ψ3 are normalized and orthogonal. Normalize the following function"

ψ1 - ψ21/(sqrt2) + ψ3sqrt(3)/sqrt(6)



Homework Equations


This is for a physical chemistry class. I haven't seen an example like this. All that is in our textbook is that integral[ /ψ/2dT] needs to equal 1 which is accomplished by adjusting N so N2 Integral[ /ψ/2dx] =1

The Attempt at a Solution


Is this correct? What range do I integrate it over? no x values are given.


Any help will be greatly appreciated!
[/B]
 

Answers and Replies

  • #2
vela
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Homework Statement


"assume that the three real functions ψ1,ψ2, and ψ3 are normalized and orthogonal. Normalize the following function"

ψ1 - ψ21/(sqrt2) + ψ3sqrt(3)/sqrt(6)



Homework Equations


This is for a physical chemistry class. I haven't seen an example like this. All that is in our textbook is that integral[ /ψ/2dT] needs to equal 1 which is accomplished by adjusting N so N2 Integral[ /ψ/2dx] =1

The Attempt at a Solution

[/b]
Is this correct? What range do I integrate it over? no x values are given.
Is what correct? You haven't shown any work.

The readability of your post would improve if you use LaTeX. Here's a primer: https://www.physicsforums.com/help/latexhelp/.
 
  • #3
Is what correct? You haven't shown any work.

The readability of your post would improve if you use LaTeX. Here's a primer: https://www.physicsforums.com/help/latexhelp/.
I don't know why all my subscripts disappeared.

ψ1 - ψ2 1/(sqrt6) + ψ3 sqrt(3)/sqrt(6)

In the book it says that to normalize a function you need to adjust N so that
N2 Integral [/ψ/2 dx] =1

I don't know how I'd apply that to this question
 
  • #4
vela
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In this problem, you have ##\psi(x) = N\left(\psi_1 - \frac{1}{\sqrt{6}} \psi_2 + \frac{\sqrt{3}}{\sqrt{6}}\psi_3\right)##. (Are you sure about that state? It's written a bit strangely, i.e., ##\sqrt{3}/\sqrt{6} = 1/\sqrt{2}##.)
 
  • #5
In this problem, you have ##\psi(x) = N\left(\psi_1 - \frac{1}{\sqrt{6}} \psi_2 + \frac{\sqrt{3}}{\sqrt{6}}\psi_3\right)##. (Are you sure about that state? It's written a bit strangely, i.e., ##\sqrt{3}/\sqrt{6} = 1/\sqrt{2}##.)

That's how it's written in the book.

Since x isn't in the function
In this problem, you have ##\psi(x) = N\left(\psi_1 - \frac{1}{\sqrt{6}} \psi_2 + \frac{\sqrt{3}}{\sqrt{6}}\psi_3\right)##. (Are you sure about that state? It's written a bit strangely, i.e., ##\sqrt{3}/\sqrt{6} = 1/\sqrt{2}##.)

That's just how it's written. So when I integrate that with respect to x i get N(ψ1 - ψ2 1/(sqrt6) + ψ3 sqrt(3)/sqrt(6))x +C
Did I integrate that correctly? And if so what do I do from here?
 
  • #6
vela
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Remember that ##\psi_1##, ##\psi_2##, and ##\psi_3## are functions of ##x##. You said the normalization condition is
$$\int_{-\infty}^\infty \psi^2\,dx = 1.$$ You want to substitute the expression you're given for ##\psi## and evaluate the integral.

Think about what it means when you're told that the ##\psi_i##'s are normalized and orthogonal to each other.
 
  • #7
Remember that ##\psi_1##, ##\psi_2##, and ##\psi_3## are functions of ##x##. You said the normalization condition is
$$\int_{-\infty}^\infty \psi^2\,dx = 1.$$ You want to substitute the expression you're given for ##\psi## and evaluate the integral.
Ohhh! I did it as an indefinite integral. So, just to clarify, I need to plug in the whole original equation in to the integral?
 

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