How do I properly normalize a wave function with given real functions?

[ReidMerrill]

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!
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[vela]

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/.

 

[ReidMerrill]

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

 

[vela]

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., $\sqrt{3}/\sqrt{6} = 1/\sqrt{2}$.)

 

[ReidMerrill]

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?

 

[vela]

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.

 

[ReidMerrill]

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 into the integral?