Normalization of Wavefunction Integration

[brinraeven]

Homework Statement

[/B]

Determine the value that A (assumed real) must have if the wavefunction is to be correctly normalised, i.e. the volume integral of |Ψ|2 over all space is equal to unity.

Homework Equations

Integration by parts

(I think?)

The Attempt at a Solution

So, I've managed to do the integration correctly (I think), but I don't know where to go from here. I have the solution manual, but it skips most steps, so I don't know how to get there. It seems like it has something to do with writing out e^-x as a sum, since factorials play into it, but I don't know how.

My work:

Solution manual:

 

[Dick]

While I didn't check every step - it looks like you got it generally correct. But you haven't evaluated the definite integral yet. In the end, you should just get a number not a function of $r$. And the mysterious factorials in the book solution are coming from the formula $\int_0^\infty u^n e^{-u} du = n!$. It might be handy to remember that, so you don't have to do integration by parts over and over again.

 

[Ray Vickson]

Homework Statement

[/B]
View attachment 212088
Determine the value that A (assumed real) must have if the wavefunction is to be correctly normalised, i.e. the volume integral of |Ψ|2 over all space is equal to unity.

Homework Equations

Integration by parts

View attachment 212089 View attachment 212090 (I think?)

The Attempt at a Solution

So, I've managed to do the integration correctly (I think), but I don't know where to go from here. I have the solution manual, but it skips most steps, so I don't know how to get there. It seems like it has something to do with writing out e^-x as a sum, since factorials play into it, but I don't know how.

My work:
View attachment 212091
View attachment 212092
View attachment 212093
View attachment 212094
View attachment 212095
Solution manual:
View attachment 212096

Have not checked your details, but note that you are doing a lot more work than you need to. You want to compute an integral of the form
$$\int_0^{\infty} (a_2 r^2 + a_3 r^3 + a_4 r^4) e^{-cr} \, dr$$
with known constants $a_2, a_3, a_4$ and $c > 0$.

It is easy to compute $F(c) = \int_0^{\infty} e^{-cr} \, dr .$ Then, since $(\partial / \partial c)^n e^{-cr} = (-1)^n r^n e^{-rc}$ we have
$$ \int_0^{\infty} r^n e^{-rc} \, dr = (-1)^n \int_0^{\infty} \left(\frac{\partial}{\partial c} \right)^n e^{-cr} \, dr = (-1)^n \left(\frac{d}{dc} \right)^n F(c).$$
This avoids integration by parts and most other complications.

 

[brinraeven]

Thanks so much! I had no idea that was a property. So, I tried to work my first term using that property, but I'm not sure I did it right.

 

[brinraeven]

Have not checked your details, but note that you are doing a lot more work than you need to. You want to compute an integral of the form
$$\int_0^{\infty} (a_2 r^2 + a_3 r^3 + a_4 r^4) e^{-cr} \, dr$$
with known constants $a_2, a_3, a_4$ and $c > 0$.

It is easy to compute $F(c) = \int_0^{\infty} e^{-cr} \, dr .$ Then, since $(\partial / \partial c)^n e^{-cr} = (-1)^n r^n e^{-rc}$ we have
$$ \int_0^{\infty} r^n e^{-rc} \, dr = (-1)^n \int_0^{\infty} \left(\frac{\partial}{\partial c} \right)^n e^{-cr} \, dr = (-1)^n \left(\frac{d}{dc} \right)^n F(c).$$
This avoids integration by parts and most other complications.

This is also great. I attempted to work it out using that, and I am closer to a solution, but still not sure I computed it correctly.

 

[Dick]

Thanks so much! I had no idea that was a property. So, I tried to work my first term using that property, but I'm not sure I did it right.
View attachment 212105

Not right. You aren't substituting consistently. Change the variable to $v=uz$ and write it completely in terms of $v$. Then think about it.

 

[Dick]

BTW you can get the correct answer out of your first effort. But you didn't actually evaluate the definite integral. You need to substitute $r=0$ and then think about the limit as $r \rightarrow \infty$.

 

[brinraeven]

BTW you can get the correct answer out of your first effort. But you didn't actually evaluate the definite integral. You need to substitute $r=0$ and then think about the limit as $r \rightarrow \infty$.

I figured that, but was having a ton of trouble envisioning it. Well, then I went back after you said this, and gave them all the common denominator of z^3. The only term that does not have an r in the numerator becomes -8a^3/z^3, which gives me the correct answer, but negative, but that doesn't matter since the final answer is +- a square root anyway. How does having r in the numerator make all the other terms disappear? Doesn't that mean all the terms go to infinity?

But wait, it's infinity minus -8a^3/z^3, but wouldn't that equal infinity?

 

[Dick]

I figured that, but was having a ton of trouble envisioning it. Well, then I went back after you said this, and gave them all the common denominator of z^3. The only term that does not have an r in the numerator becomes -8a^3/z^3, which gives me the correct answer, but negative, but that doesn't matter since the final answer is +- a square root anyway. How does having r in the numerator make all the other terms disappear? Doesn't that mean all the terms go to infinity?

But wait, it's infinity minus -8a^3/z^3, but wouldn't that equal infinity?

Take the example of $\int_0^\infty u^2e^{-u}du$. The indefinite integral is $(-u^2-2u-2)e^{-u}$. You have to evaluate it between the limits of $0$ and $\infty$. The $0$ part is easy, it just gives you $-2$. For the upper limit you can't just put $u=\infty$, you need to figure out the limit of $(-u^2-2u-2)e^{-u}$ as $u$ approaches $\infty$. I claim that $\lim_{u \to \infty}u^n e^{-u}=0$ for any $n$. You can prove it with l'Hopital's rule, or if this is a physics course you may not have to prove it. So the difference is $0-(-2)=2$ and that's the value of the definite integral.

 

[brinraeven]

Thank you everyone! I did end up doing it the long way, only because I couldn't quite get the z^3 on the bottom for the other method, but I will keep practicing it, because it is infinitely easier than the long way.

 

[Ray Vickson]

Thank you everyone! I did end up doing it the long way, only because I couldn't quite get the z^3 on the bottom for the other method, but I will keep practicing it, because it is infinitely easier than the long way.

$$F(c) = \int_0^{\infty} e^{-rc} \, dr = \frac{1}{c}, $$
so
$$ \begin{array}{rcl}
\displaystyle \int_0^{\infty} r e^{-rc} \, dr &=&\displaystyle -\frac{d}{dc} \frac{1}{c} = \frac{1}{c^2} \\
\displaystyle \int_0^{\infty} r^2 e^{-rc} \, dt &=&\displaystyle -\frac{d}{dc} \frac{1}{c^2} = \frac{2}{c^3} \\
\displaystyle \int_0^{\infty} r^3 e^{-rc} \, dt &=&\displaystyle -\frac{d}{dc} \frac{2}{c^3} = \frac{3!}{c^4}
\end{array}
$$
and so forth. By induction we have $\int_0^{\infty} r^n e^{-rc} \, dr = n!/c^{n+1}.$ As I said already: almost no work is needed, and the chances of making an error are almost zero.